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Question 1 of 101
2017 soldiers were lined up in a single file. The commander ordered them to number off 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, $\dots$ starting from the first soldier. He then ordered them to number off 1, 2, 3, 4, 1, 2, 3, 4, 1, $\dots$ starting from the last soldier. How many soldiers had the same number both times?
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Consider the following table where the first row stands for the first numbering of the soldiers \begin{equation} \begin{array}{*{15}{|c}|c|} 1 & 2 & 3 & 4 & 5 & 1 & 2 & 3 & 4 & 5 & 1 & 2 & 3 & 4 & 5\\\hline 1 & 4 & 3 & 2 & 1 & 4 & 3 & 2 & 1 & 4 & 3 & 2 & 1 & 4 & 3 \end{array} \end{equation} \begin{equation} \begin{array}{*{13}{|c}|c|} 1 & 2 & 3 & 4 & 5 & 1 & 2 & 3 & 4 & 5 & 1 & 2 & 3 & \cdots \\\hline 2 & 1 & 4 & 3 & 2 & 1 & 4 & 3 & 2 & 1 & 4 & 3 & 2 & \cdots \end{array} \end{equation} \begin{align*} 2017 \div 5 &= 403 \text{R} 2 \\ 2017 \div 4 &= 504 \text{R} 1 \\ \end{align*} Every 20 soldiers there are 4 with the same numbers. \begin{equation} 2017 \div 20 = 100 \text{R} 17 \end{equation} So, $100\times 4 = 400$. But there are 4 more soldiers with the same numbers amongst the last 17 soldiers. So, $400+4=404$
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Consider the following table where the first row stands for the first numbering of the soldiers \begin{equation} \begin{array}{*{15}{|c}|c|} 1 & 2 & 3 & 4 & 5 & 1 & 2 & 3 & 4 & 5 & 1 & 2 & 3 & 4 & 5\\\hline 1 & 4 & 3 & 2 & 1 & 4 & 3 & 2 & 1 & 4 & 3 & 2 & 1 & 4 & 3 \end{array} \end{equation} \begin{equation} \begin{array}{*{13}{|c}|c|} 1 & 2 & 3 & 4 & 5 & 1 & 2 & 3 & 4 & 5 & 1 & 2 & 3 & \cdots \\\hline 2 & 1 & 4 & 3 & 2 & 1 & 4 & 3 & 2 & 1 & 4 & 3 & 2 & \cdots \end{array} \end{equation} \begin{align*} 2017 \div 5 &= 403 \text{R} 2 \\ 2017 \div 4 &= 504 \text{R} 1 \\ \end{align*} Every 20 soldiers there are 4 with the same numbers. \begin{equation} 2017 \div 20 = 100 \text{R} 17 \end{equation} So, $100\times 4 = 400$. But there are 4 more soldiers with the same numbers amongst the last 17 soldiers. So, $400+4=404$
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=5^\circ$ and diameter $=$ 8 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 8+\frac{1}{2}\times 2 \times \pi \times r + \frac{5}{360}\times 2 \times \pi \times r_1 \\ &= 8+\frac{1}{2}\times 2\times \frac{22}{7}\times 4 + \frac{1}{72}\times 2 \times \frac{22}{7} \times 8 \\ &= 8+\frac{88}{7}+\frac{44}{63} \\ &= 21\frac{17}{63} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=5^\circ$ and diameter $=$ 10 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 10+\frac{1}{2}\times 2 \times \pi \times r + \frac{5}{360}\times 2 \times \pi \times r_1 \\ &= 10+\frac{1}{2}\times 2\times \frac{22}{7}\times 5 + \frac{1}{72}\times 2 \times \frac{22}{7} \times 10 \\ &= 10+\frac{110}{7}+\frac{55}{63} \\ &= 26\frac{37}{63} \text{ cm} \end{align*}
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\begin{align*} &\phantom{=} 10+\frac{1}{2}\times 2 \times \pi \times r + \frac{5}{360}\times 2 \times \pi \times r_1 \\ &= 10+\frac{1}{2}\times 2\times \frac{22}{7}\times 5 + \frac{1}{72}\times 2 \times \frac{22}{7} \times 10 \\ &= 10+\frac{110}{7}+\frac{55}{63} \\ &= 26\frac{37}{63} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=5^\circ$ and diameter $=$ 12 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 12+\frac{1}{2}\times 2 \times \pi \times r + \frac{5}{360}\times 2 \times \pi \times r_1 \\ &= 12+\frac{1}{2}\times 2\times \frac{22}{7}\times 6 + \frac{1}{72}\times 2 \times \frac{22}{7} \times 12 \\ &= 12+\frac{132}{7}+\frac{22}{21} \\ &= 31\frac{19}{21} \text{ cm} \end{align*}
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\begin{align*} &\phantom{=} 12+\frac{1}{2}\times 2 \times \pi \times r + \frac{5}{360}\times 2 \times \pi \times r_1 \\ &= 12+\frac{1}{2}\times 2\times \frac{22}{7}\times 6 + \frac{1}{72}\times 2 \times \frac{22}{7} \times 12 \\ &= 12+\frac{132}{7}+\frac{22}{21} \\ &= 31\frac{19}{21} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=5^\circ$ and diameter $=$ 14 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 14+\frac{1}{2}\times 2 \times \pi \times r + \frac{5}{360}\times 2 \times \pi \times r_1 \\ &= 14+\frac{1}{2}\times 2\times \frac{22}{7}\times 7 + \frac{1}{72}\times 2 \times \frac{22}{7} \times 14 \\ &= 14+22+\frac{11}{9} \\ &= 37\frac{2}{9} \text{ cm} \end{align*}
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\begin{align*} &\phantom{=} 14+\frac{1}{2}\times 2 \times \pi \times r + \frac{5}{360}\times 2 \times \pi \times r_1 \\ &= 14+\frac{1}{2}\times 2\times \frac{22}{7}\times 7 + \frac{1}{72}\times 2 \times \frac{22}{7} \times 14 \\ &= 14+22+\frac{11}{9} \\ &= 37\frac{2}{9} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=5^\circ$ and diameter $=$ 16 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 16+\frac{1}{2}\times 2 \times \pi \times r + \frac{5}{360}\times 2 \times \pi \times r_1 \\ &= 16+\frac{1}{2}\times 2\times \frac{22}{7}\times 8 + \frac{1}{72}\times 2 \times \frac{22}{7} \times 16 \\ &= 16+\frac{176}{7}+\frac{88}{63} \\ &= 42\frac{34}{63} \text{ cm} \end{align*}
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\begin{align*} &\phantom{=} 16+\frac{1}{2}\times 2 \times \pi \times r + \frac{5}{360}\times 2 \times \pi \times r_1 \\ &= 16+\frac{1}{2}\times 2\times \frac{22}{7}\times 8 + \frac{1}{72}\times 2 \times \frac{22}{7} \times 16 \\ &= 16+\frac{176}{7}+\frac{88}{63} \\ &= 42\frac{34}{63} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=5^\circ$ and diameter $=$ 18 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 18+\frac{1}{2}\times 2 \times \pi \times r + \frac{5}{360}\times 2 \times \pi \times r_1 \\ &= 18+\frac{1}{2}\times 2\times \frac{22}{7}\times 9 + \frac{1}{72}\times 2 \times \frac{22}{7} \times 18 \\ &= 18+\frac{198}{7}+\frac{11}{7} \\ &= 47\frac{6}{7} \text{ cm} \end{align*}
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\begin{align*} &\phantom{=} 18+\frac{1}{2}\times 2 \times \pi \times r + \frac{5}{360}\times 2 \times \pi \times r_1 \\ &= 18+\frac{1}{2}\times 2\times \frac{22}{7}\times 9 + \frac{1}{72}\times 2 \times \frac{22}{7} \times 18 \\ &= 18+\frac{198}{7}+\frac{11}{7} \\ &= 47\frac{6}{7} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=5^\circ$ and diameter $=$ 20 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 20+\frac{1}{2}\times 2 \times \pi \times r + \frac{5}{360}\times 2 \times \pi \times r_1 \\ &= 20+\frac{1}{2}\times 2\times \frac{22}{7}\times 10 + \frac{1}{72}\times 2 \times \frac{22}{7} \times 20 \\ &= 20+\frac{220}{7}+\frac{110}{63} \\ &= 53\frac{11}{63} \text{ cm} \end{align*}
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\begin{align*} &\phantom{=} 20+\frac{1}{2}\times 2 \times \pi \times r + \frac{5}{360}\times 2 \times \pi \times r_1 \\ &= 20+\frac{1}{2}\times 2\times \frac{22}{7}\times 10 + \frac{1}{72}\times 2 \times \frac{22}{7} \times 20 \\ &= 20+\frac{220}{7}+\frac{110}{63} \\ &= 53\frac{11}{63} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=5^\circ$ and diameter $=$ 22 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 22+\frac{1}{2}\times 2 \times \pi \times r + \frac{5}{360}\times 2 \times \pi \times r_1 \\ &= 22+\frac{1}{2}\times 2\times \frac{22}{7}\times 11 + \frac{1}{72}\times 2 \times \frac{22}{7} \times 22 \\ &= 22+\frac{242}{7}+\frac{121}{63} \\ &= 58\frac{31}{63} \text{ cm} \end{align*}
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\begin{align*} &\phantom{=} 22+\frac{1}{2}\times 2 \times \pi \times r + \frac{5}{360}\times 2 \times \pi \times r_1 \\ &= 22+\frac{1}{2}\times 2\times \frac{22}{7}\times 11 + \frac{1}{72}\times 2 \times \frac{22}{7} \times 22 \\ &= 22+\frac{242}{7}+\frac{121}{63} \\ &= 58\frac{31}{63} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=6^\circ$ and diameter $=$ 8 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 8+\frac{1}{2}\times 2 \times \pi \times r + \frac{6}{360}\times 2 \times \pi \times r_1 \\ &= 8+\frac{1}{2}\times 2\times \frac{22}{7}\times 4 + \frac{1}{60}\times 2 \times \frac{22}{7} \times 8 \\ &= 8+\frac{88}{7}+\frac{88}{105} \\ &= 21\frac{43}{105} \text{ cm} \end{align*}
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\begin{align*} &\phantom{=} 8+\frac{1}{2}\times 2 \times \pi \times r + \frac{6}{360}\times 2 \times \pi \times r_1 \\ &= 8+\frac{1}{2}\times 2\times \frac{22}{7}\times 4 + \frac{1}{60}\times 2 \times \frac{22}{7} \times 8 \\ &= 8+\frac{88}{7}+\frac{88}{105} \\ &= 21\frac{43}{105} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=6^\circ$ and diameter $=$ 10 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 10+\frac{1}{2}\times 2 \times \pi \times r + \frac{6}{360}\times 2 \times \pi \times r_1 \\ &= 10+\frac{1}{2}\times 2\times \frac{22}{7}\times 5 + \frac{1}{60}\times 2 \times \frac{22}{7} \times 10 \\ &= 10+\frac{110}{7}+\frac{22}{21} \\ &= 26\frac{16}{21} \text{ cm} \end{align*}
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\begin{align*} &\phantom{=} 10+\frac{1}{2}\times 2 \times \pi \times r + \frac{6}{360}\times 2 \times \pi \times r_1 \\ &= 10+\frac{1}{2}\times 2\times \frac{22}{7}\times 5 + \frac{1}{60}\times 2 \times \frac{22}{7} \times 10 \\ &= 10+\frac{110}{7}+\frac{22}{21} \\ &= 26\frac{16}{21} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=6^\circ$ and diameter $=$ 12 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 12+\frac{1}{2}\times 2 \times \pi \times r + \frac{6}{360}\times 2 \times \pi \times r_1 \\ &= 12+\frac{1}{2}\times 2\times \frac{22}{7}\times 6 + \frac{1}{60}\times 2 \times \frac{22}{7} \times 12 \\ &= 12+\frac{132}{7}+\frac{44}{35} \\ &= 32\frac{4}{35} \text{ cm} \end{align*}
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\begin{align*} &\phantom{=} 12+\frac{1}{2}\times 2 \times \pi \times r + \frac{6}{360}\times 2 \times \pi \times r_1 \\ &= 12+\frac{1}{2}\times 2\times \frac{22}{7}\times 6 + \frac{1}{60}\times 2 \times \frac{22}{7} \times 12 \\ &= 12+\frac{132}{7}+\frac{44}{35} \\ &= 32\frac{4}{35} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=6^\circ$ and diameter $=$ 14 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 14+\frac{1}{2}\times 2 \times \pi \times r + \frac{6}{360}\times 2 \times \pi \times r_1 \\ &= 14+\frac{1}{2}\times 2\times \frac{22}{7}\times 7 + \frac{1}{60}\times 2 \times \frac{22}{7} \times 14 \\ &= 14+22+\frac{22}{15} \\ &= 37\frac{7}{15} \text{ cm} \end{align*}
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\begin{align*} &\phantom{=} 14+\frac{1}{2}\times 2 \times \pi \times r + \frac{6}{360}\times 2 \times \pi \times r_1 \\ &= 14+\frac{1}{2}\times 2\times \frac{22}{7}\times 7 + \frac{1}{60}\times 2 \times \frac{22}{7} \times 14 \\ &= 14+22+\frac{22}{15} \\ &= 37\frac{7}{15} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=6^\circ$ and diameter $=$ 16 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 16+\frac{1}{2}\times 2 \times \pi \times r + \frac{6}{360}\times 2 \times \pi \times r_1 \\ &= 16+\frac{1}{2}\times 2\times \frac{22}{7}\times 8 + \frac{1}{60}\times 2 \times \frac{22}{7} \times 16 \\ &= 16+\frac{176}{7}+\frac{176}{105} \\ &= 42\frac{86}{105} \text{ cm} \end{align*}
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\begin{align*} &\phantom{=} 16+\frac{1}{2}\times 2 \times \pi \times r + \frac{6}{360}\times 2 \times \pi \times r_1 \\ &= 16+\frac{1}{2}\times 2\times \frac{22}{7}\times 8 + \frac{1}{60}\times 2 \times \frac{22}{7} \times 16 \\ &= 16+\frac{176}{7}+\frac{176}{105} \\ &= 42\frac{86}{105} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=6^\circ$ and diameter $=$ 18 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 18+\frac{1}{2}\times 2 \times \pi \times r + \frac{6}{360}\times 2 \times \pi \times r_1 \\ &= 18+\frac{1}{2}\times 2\times \frac{22}{7}\times 9 + \frac{1}{60}\times 2 \times \frac{22}{7} \times 18 \\ &= 18+\frac{198}{7}+\frac{66}{35} \\ &= 48\frac{6}{35} \text{ cm} \end{align*}
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\begin{align*} &\phantom{=} 18+\frac{1}{2}\times 2 \times \pi \times r + \frac{6}{360}\times 2 \times \pi \times r_1 \\ &= 18+\frac{1}{2}\times 2\times \frac{22}{7}\times 9 + \frac{1}{60}\times 2 \times \frac{22}{7} \times 18 \\ &= 18+\frac{198}{7}+\frac{66}{35} \\ &= 48\frac{6}{35} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=6^\circ$ and diameter $=$ 20 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 20+\frac{1}{2}\times 2 \times \pi \times r + \frac{6}{360}\times 2 \times \pi \times r_1 \\ &= 20+\frac{1}{2}\times 2\times \frac{22}{7}\times 10 + \frac{1}{60}\times 2 \times \frac{22}{7} \times 20 \\ &= 20+\frac{220}{7}+\frac{44}{21} \\ &= 53\frac{11}{21} \text{ cm} \end{align*}
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\begin{align*} &\phantom{=} 20+\frac{1}{2}\times 2 \times \pi \times r + \frac{6}{360}\times 2 \times \pi \times r_1 \\ &= 20+\frac{1}{2}\times 2\times \frac{22}{7}\times 10 + \frac{1}{60}\times 2 \times \frac{22}{7} \times 20 \\ &= 20+\frac{220}{7}+\frac{44}{21} \\ &= 53\frac{11}{21} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=6^\circ$ and diameter $=$ 22 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 22+\frac{1}{2}\times 2 \times \pi \times r + \frac{6}{360}\times 2 \times \pi \times r_1 \\ &= 22+\frac{1}{2}\times 2\times \frac{22}{7}\times 11 + \frac{1}{60}\times 2 \times \frac{22}{7} \times 22 \\ &= 22+\frac{242}{7}+\frac{242}{105} \\ &= 58\frac{92}{105} \text{ cm} \end{align*}
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\begin{align*} &\phantom{=} 22+\frac{1}{2}\times 2 \times \pi \times r + \frac{6}{360}\times 2 \times \pi \times r_1 \\ &= 22+\frac{1}{2}\times 2\times \frac{22}{7}\times 11 + \frac{1}{60}\times 2 \times \frac{22}{7} \times 22 \\ &= 22+\frac{242}{7}+\frac{242}{105} \\ &= 58\frac{92}{105} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=8^\circ$ and diameter $=$ 8 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 8+\frac{1}{2}\times 2 \times \pi \times r + \frac{8}{360}\times 2 \times \pi \times r_1 \\ &= 8+\frac{1}{2}\times 2\times \frac{22}{7}\times 4 + \frac{1}{45}\times 2 \times \frac{22}{7} \times 8 \\ &= 8+\frac{88}{7}+\frac{352}{315} \\ &= 21\frac{31}{45} \text{ cm} \end{align*}
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\begin{align*} &\phantom{=} 8+\frac{1}{2}\times 2 \times \pi \times r + \frac{8}{360}\times 2 \times \pi \times r_1 \\ &= 8+\frac{1}{2}\times 2\times \frac{22}{7}\times 4 + \frac{1}{45}\times 2 \times \frac{22}{7} \times 8 \\ &= 8+\frac{88}{7}+\frac{352}{315} \\ &= 21\frac{31}{45} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=8^\circ$ and diameter $=$ 10 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 10+\frac{1}{2}\times 2 \times \pi \times r + \frac{8}{360}\times 2 \times \pi \times r_1 \\ &= 10+\frac{1}{2}\times 2\times \frac{22}{7}\times 5 + \frac{1}{45}\times 2 \times \frac{22}{7} \times 10 \\ &= 10+\frac{110}{7}+\frac{88}{63} \\ &= 27\frac{1}{9} \text{ cm} \end{align*}
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\begin{align*} &\phantom{=} 10+\frac{1}{2}\times 2 \times \pi \times r + \frac{8}{360}\times 2 \times \pi \times r_1 \\ &= 10+\frac{1}{2}\times 2\times \frac{22}{7}\times 5 + \frac{1}{45}\times 2 \times \frac{22}{7} \times 10 \\ &= 10+\frac{110}{7}+\frac{88}{63} \\ &= 27\frac{1}{9} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=8^\circ$ and diameter $=$ 12 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 12+\frac{1}{2}\times 2 \times \pi \times r + \frac{8}{360}\times 2 \times \pi \times r_1 \\ &= 12+\frac{1}{2}\times 2\times \frac{22}{7}\times 6 + \frac{1}{45}\times 2 \times \frac{22}{7} \times 12 \\ &= 12+\frac{132}{7}+\frac{176}{105} \\ &= 32\frac{8}{15} \text{ cm} \end{align*}
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\begin{align*} &\phantom{=} 12+\frac{1}{2}\times 2 \times \pi \times r + \frac{8}{360}\times 2 \times \pi \times r_1 \\ &= 12+\frac{1}{2}\times 2\times \frac{22}{7}\times 6 + \frac{1}{45}\times 2 \times \frac{22}{7} \times 12 \\ &= 12+\frac{132}{7}+\frac{176}{105} \\ &= 32\frac{8}{15} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=8^\circ$ and diameter $=$ 14 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 14+\frac{1}{2}\times 2 \times \pi \times r + \frac{8}{360}\times 2 \times \pi \times r_1 \\ &= 14+\frac{1}{2}\times 2\times \frac{22}{7}\times 7 + \frac{1}{45}\times 2 \times \frac{22}{7} \times 14 \\ &= 14+22+\frac{88}{45} \\ &= 37\frac{43}{45} \text{ cm} \end{align*}
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\begin{align*} &\phantom{=} 14+\frac{1}{2}\times 2 \times \pi \times r + \frac{8}{360}\times 2 \times \pi \times r_1 \\ &= 14+\frac{1}{2}\times 2\times \frac{22}{7}\times 7 + \frac{1}{45}\times 2 \times \frac{22}{7} \times 14 \\ &= 14+22+\frac{88}{45} \\ &= 37\frac{43}{45} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=8^\circ$ and diameter $=$ 16 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 16+\frac{1}{2}\times 2 \times \pi \times r + \frac{8}{360}\times 2 \times \pi \times r_1 \\ &= 16+\frac{1}{2}\times 2\times \frac{22}{7}\times 8 + \frac{1}{45}\times 2 \times \frac{22}{7} \times 16 \\ &= 16+\frac{176}{7}+\frac{704}{315} \\ &= 43\frac{17}{45} \text{ cm} \end{align*}
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\begin{align*} &\phantom{=} 16+\frac{1}{2}\times 2 \times \pi \times r + \frac{8}{360}\times 2 \times \pi \times r_1 \\ &= 16+\frac{1}{2}\times 2\times \frac{22}{7}\times 8 + \frac{1}{45}\times 2 \times \frac{22}{7} \times 16 \\ &= 16+\frac{176}{7}+\frac{704}{315} \\ &= 43\frac{17}{45} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=8^\circ$ and diameter $=$ 18 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 18+\frac{1}{2}\times 2 \times \pi \times r + \frac{8}{360}\times 2 \times \pi \times r_1 \\ &= 18+\frac{1}{2}\times 2\times \frac{22}{7}\times 9 + \frac{1}{45}\times 2 \times \frac{22}{7} \times 18 \\ &= 18+\frac{198}{7}+\frac{88}{35} \\ &= 48\frac{4}{5} \text{ cm} \end{align*}
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\begin{align*} &\phantom{=} 18+\frac{1}{2}\times 2 \times \pi \times r + \frac{8}{360}\times 2 \times \pi \times r_1 \\ &= 18+\frac{1}{2}\times 2\times \frac{22}{7}\times 9 + \frac{1}{45}\times 2 \times \frac{22}{7} \times 18 \\ &= 18+\frac{198}{7}+\frac{88}{35} \\ &= 48\frac{4}{5} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=8^\circ$ and diameter $=$ 20 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 20+\frac{1}{2}\times 2 \times \pi \times r + \frac{8}{360}\times 2 \times \pi \times r_1 \\ &= 20+\frac{1}{2}\times 2\times \frac{22}{7}\times 10 + \frac{1}{45}\times 2 \times \frac{22}{7} \times 20 \\ &= 20+\frac{220}{7}+\frac{176}{63} \\ &= 54\frac{2}{9} \text{ cm} \end{align*}
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\begin{align*} &\phantom{=} 20+\frac{1}{2}\times 2 \times \pi \times r + \frac{8}{360}\times 2 \times \pi \times r_1 \\ &= 20+\frac{1}{2}\times 2\times \frac{22}{7}\times 10 + \frac{1}{45}\times 2 \times \frac{22}{7} \times 20 \\ &= 20+\frac{220}{7}+\frac{176}{63} \\ &= 54\frac{2}{9} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=8^\circ$ and diameter $=$ 22 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 22+\frac{1}{2}\times 2 \times \pi \times r + \frac{8}{360}\times 2 \times \pi \times r_1 \\ &= 22+\frac{1}{2}\times 2\times \frac{22}{7}\times 11 + \frac{1}{45}\times 2 \times \frac{22}{7} \times 22 \\ &= 22+\frac{242}{7}+\frac{968}{315} \\ &= 59\frac{29}{45} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 22+\frac{1}{2}\times 2 \times \pi \times r + \frac{8}{360}\times 2 \times \pi \times r_1 \\ &= 22+\frac{1}{2}\times 2\times \frac{22}{7}\times 11 + \frac{1}{45}\times 2 \times \frac{22}{7} \times 22 \\ &= 22+\frac{242}{7}+\frac{968}{315} \\ &= 59\frac{29}{45} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=9^\circ$ and diameter $=$ 8 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 8+\frac{1}{2}\times 2 \times \pi \times r + \frac{9}{360}\times 2 \times \pi \times r_1 \\ &= 8+\frac{1}{2}\times 2\times \frac{22}{7}\times 4 + \frac{1}{40}\times 2 \times \frac{22}{7} \times 8 \\ &= 8+\frac{88}{7}+\frac{44}{35} \\ &= 21\frac{29}{35} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 8+\frac{1}{2}\times 2 \times \pi \times r + \frac{9}{360}\times 2 \times \pi \times r_1 \\ &= 8+\frac{1}{2}\times 2\times \frac{22}{7}\times 4 + \frac{1}{40}\times 2 \times \frac{22}{7} \times 8 \\ &= 8+\frac{88}{7}+\frac{44}{35} \\ &= 21\frac{29}{35} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=9^\circ$ and diameter $=$ 10 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 10+\frac{1}{2}\times 2 \times \pi \times r + \frac{9}{360}\times 2 \times \pi \times r_1 \\ &= 10+\frac{1}{2}\times 2\times \frac{22}{7}\times 5 + \frac{1}{40}\times 2 \times \frac{22}{7} \times 10 \\ &= 10+\frac{110}{7}+\frac{11}{7} \\ &= 27\frac{2}{7} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 10+\frac{1}{2}\times 2 \times \pi \times r + \frac{9}{360}\times 2 \times \pi \times r_1 \\ &= 10+\frac{1}{2}\times 2\times \frac{22}{7}\times 5 + \frac{1}{40}\times 2 \times \frac{22}{7} \times 10 \\ &= 10+\frac{110}{7}+\frac{11}{7} \\ &= 27\frac{2}{7} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=9^\circ$ and diameter $=$ 12 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 12+\frac{1}{2}\times 2 \times \pi \times r + \frac{9}{360}\times 2 \times \pi \times r_1 \\ &= 12+\frac{1}{2}\times 2\times \frac{22}{7}\times 6 + \frac{1}{40}\times 2 \times \frac{22}{7} \times 12 \\ &= 12+\frac{132}{7}+\frac{66}{35} \\ &= 32\frac{26}{35} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 12+\frac{1}{2}\times 2 \times \pi \times r + \frac{9}{360}\times 2 \times \pi \times r_1 \\ &= 12+\frac{1}{2}\times 2\times \frac{22}{7}\times 6 + \frac{1}{40}\times 2 \times \frac{22}{7} \times 12 \\ &= 12+\frac{132}{7}+\frac{66}{35} \\ &= 32\frac{26}{35} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=9^\circ$ and diameter $=$ 14 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 14+\frac{1}{2}\times 2 \times \pi \times r + \frac{9}{360}\times 2 \times \pi \times r_1 \\ &= 14+\frac{1}{2}\times 2\times \frac{22}{7}\times 7 + \frac{1}{40}\times 2 \times \frac{22}{7} \times 14 \\ &= 14+22+\frac{11}{5} \\ &= 38\frac{1}{5} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 14+\frac{1}{2}\times 2 \times \pi \times r + \frac{9}{360}\times 2 \times \pi \times r_1 \\ &= 14+\frac{1}{2}\times 2\times \frac{22}{7}\times 7 + \frac{1}{40}\times 2 \times \frac{22}{7} \times 14 \\ &= 14+22+\frac{11}{5} \\ &= 38\frac{1}{5} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=9^\circ$ and diameter $=$ 16 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 16+\frac{1}{2}\times 2 \times \pi \times r + \frac{9}{360}\times 2 \times \pi \times r_1 \\ &= 16+\frac{1}{2}\times 2\times \frac{22}{7}\times 8 + \frac{1}{40}\times 2 \times \frac{22}{7} \times 16 \\ &= 16+\frac{176}{7}+\frac{88}{35} \\ &= 43\frac{23}{35} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 16+\frac{1}{2}\times 2 \times \pi \times r + \frac{9}{360}\times 2 \times \pi \times r_1 \\ &= 16+\frac{1}{2}\times 2\times \frac{22}{7}\times 8 + \frac{1}{40}\times 2 \times \frac{22}{7} \times 16 \\ &= 16+\frac{176}{7}+\frac{88}{35} \\ &= 43\frac{23}{35} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=9^\circ$ and diameter $=$ 18 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 18+\frac{1}{2}\times 2 \times \pi \times r + \frac{9}{360}\times 2 \times \pi \times r_1 \\ &= 18+\frac{1}{2}\times 2\times \frac{22}{7}\times 9 + \frac{1}{40}\times 2 \times \frac{22}{7} \times 18 \\ &= 18+\frac{198}{7}+\frac{99}{35} \\ &= 49\frac{4}{35} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 18+\frac{1}{2}\times 2 \times \pi \times r + \frac{9}{360}\times 2 \times \pi \times r_1 \\ &= 18+\frac{1}{2}\times 2\times \frac{22}{7}\times 9 + \frac{1}{40}\times 2 \times \frac{22}{7} \times 18 \\ &= 18+\frac{198}{7}+\frac{99}{35} \\ &= 49\frac{4}{35} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=9^\circ$ and diameter $=$ 20 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 20+\frac{1}{2}\times 2 \times \pi \times r + \frac{9}{360}\times 2 \times \pi \times r_1 \\ &= 20+\frac{1}{2}\times 2\times \frac{22}{7}\times 10 + \frac{1}{40}\times 2 \times \frac{22}{7} \times 20 \\ &= 20+\frac{220}{7}+\frac{22}{7} \\ &= 54\frac{4}{7} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 20+\frac{1}{2}\times 2 \times \pi \times r + \frac{9}{360}\times 2 \times \pi \times r_1 \\ &= 20+\frac{1}{2}\times 2\times \frac{22}{7}\times 10 + \frac{1}{40}\times 2 \times \frac{22}{7} \times 20 \\ &= 20+\frac{220}{7}+\frac{22}{7} \\ &= 54\frac{4}{7} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=9^\circ$ and diameter $=$ 22 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 22+\frac{1}{2}\times 2 \times \pi \times r + \frac{9}{360}\times 2 \times \pi \times r_1 \\ &= 22+\frac{1}{2}\times 2\times \frac{22}{7}\times 11 + \frac{1}{40}\times 2 \times \frac{22}{7} \times 22 \\ &= 22+\frac{242}{7}+\frac{121}{35} \\ &= 60\frac{1}{35} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 22+\frac{1}{2}\times 2 \times \pi \times r + \frac{9}{360}\times 2 \times \pi \times r_1 \\ &= 22+\frac{1}{2}\times 2\times \frac{22}{7}\times 11 + \frac{1}{40}\times 2 \times \frac{22}{7} \times 22 \\ &= 22+\frac{242}{7}+\frac{121}{35} \\ &= 60\frac{1}{35} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=10^\circ$ and diameter $=$ 8 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 8+\frac{1}{2}\times 2 \times \pi \times r + \frac{10}{360}\times 2 \times \pi \times r_1 \\ &= 8+\frac{1}{2}\times 2\times \frac{22}{7}\times 4 + \frac{1}{36}\times 2 \times \frac{22}{7} \times 8 \\ &= 8+\frac{88}{7}+\frac{88}{63} \\ &= 21\frac{61}{63} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 8+\frac{1}{2}\times 2 \times \pi \times r + \frac{10}{360}\times 2 \times \pi \times r_1 \\ &= 8+\frac{1}{2}\times 2\times \frac{22}{7}\times 4 + \frac{1}{36}\times 2 \times \frac{22}{7} \times 8 \\ &= 8+\frac{88}{7}+\frac{88}{63} \\ &= 21\frac{61}{63} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=10^\circ$ and diameter $=$ 10 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 10+\frac{1}{2}\times 2 \times \pi \times r + \frac{10}{360}\times 2 \times \pi \times r_1 \\ &= 10+\frac{1}{2}\times 2\times \frac{22}{7}\times 5 + \frac{1}{36}\times 2 \times \frac{22}{7} \times 10 \\ &= 10+\frac{110}{7}+\frac{110}{63} \\ &= 27\frac{29}{63} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 10+\frac{1}{2}\times 2 \times \pi \times r + \frac{10}{360}\times 2 \times \pi \times r_1 \\ &= 10+\frac{1}{2}\times 2\times \frac{22}{7}\times 5 + \frac{1}{36}\times 2 \times \frac{22}{7} \times 10 \\ &= 10+\frac{110}{7}+\frac{110}{63} \\ &= 27\frac{29}{63} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=10^\circ$ and diameter $=$ 12 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 12+\frac{1}{2}\times 2 \times \pi \times r + \frac{10}{360}\times 2 \times \pi \times r_1 \\ &= 12+\frac{1}{2}\times 2\times \frac{22}{7}\times 6 + \frac{1}{36}\times 2 \times \frac{22}{7} \times 12 \\ &= 12+\frac{132}{7}+\frac{44}{21} \\ &= 32\frac{20}{21} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 12+\frac{1}{2}\times 2 \times \pi \times r + \frac{10}{360}\times 2 \times \pi \times r_1 \\ &= 12+\frac{1}{2}\times 2\times \frac{22}{7}\times 6 + \frac{1}{36}\times 2 \times \frac{22}{7} \times 12 \\ &= 12+\frac{132}{7}+\frac{44}{21} \\ &= 32\frac{20}{21} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=10^\circ$ and diameter $=$ 14 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 14+\frac{1}{2}\times 2 \times \pi \times r + \frac{10}{360}\times 2 \times \pi \times r_1 \\ &= 14+\frac{1}{2}\times 2\times \frac{22}{7}\times 7 + \frac{1}{36}\times 2 \times \frac{22}{7} \times 14 \\ &= 14+22+\frac{22}{9} \\ &= 38\frac{4}{9} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 14+\frac{1}{2}\times 2 \times \pi \times r + \frac{10}{360}\times 2 \times \pi \times r_1 \\ &= 14+\frac{1}{2}\times 2\times \frac{22}{7}\times 7 + \frac{1}{36}\times 2 \times \frac{22}{7} \times 14 \\ &= 14+22+\frac{22}{9} \\ &= 38\frac{4}{9} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=10^\circ$ and diameter $=$ 16 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 16+\frac{1}{2}\times 2 \times \pi \times r + \frac{10}{360}\times 2 \times \pi \times r_1 \\ &= 16+\frac{1}{2}\times 2\times \frac{22}{7}\times 8 + \frac{1}{36}\times 2 \times \frac{22}{7} \times 16 \\ &= 16+\frac{176}{7}+\frac{176}{63} \\ &= 43\frac{59}{63} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 16+\frac{1}{2}\times 2 \times \pi \times r + \frac{10}{360}\times 2 \times \pi \times r_1 \\ &= 16+\frac{1}{2}\times 2\times \frac{22}{7}\times 8 + \frac{1}{36}\times 2 \times \frac{22}{7} \times 16 \\ &= 16+\frac{176}{7}+\frac{176}{63} \\ &= 43\frac{59}{63} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=10^\circ$ and diameter $=$ 18 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 18+\frac{1}{2}\times 2 \times \pi \times r + \frac{10}{360}\times 2 \times \pi \times r_1 \\ &= 18+\frac{1}{2}\times 2\times \frac{22}{7}\times 9 + \frac{1}{36}\times 2 \times \frac{22}{7} \times 18 \\ &= 18+\frac{198}{7}+\frac{22}{7} \\ &= 49\frac{3}{7} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 18+\frac{1}{2}\times 2 \times \pi \times r + \frac{10}{360}\times 2 \times \pi \times r_1 \\ &= 18+\frac{1}{2}\times 2\times \frac{22}{7}\times 9 + \frac{1}{36}\times 2 \times \frac{22}{7} \times 18 \\ &= 18+\frac{198}{7}+\frac{22}{7} \\ &= 49\frac{3}{7} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=10^\circ$ and diameter $=$ 20 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 20+\frac{1}{2}\times 2 \times \pi \times r + \frac{10}{360}\times 2 \times \pi \times r_1 \\ &= 20+\frac{1}{2}\times 2\times \frac{22}{7}\times 10 + \frac{1}{36}\times 2 \times \frac{22}{7} \times 20 \\ &= 20+\frac{220}{7}+\frac{220}{63} \\ &= 54\frac{58}{63} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 20+\frac{1}{2}\times 2 \times \pi \times r + \frac{10}{360}\times 2 \times \pi \times r_1 \\ &= 20+\frac{1}{2}\times 2\times \frac{22}{7}\times 10 + \frac{1}{36}\times 2 \times \frac{22}{7} \times 20 \\ &= 20+\frac{220}{7}+\frac{220}{63} \\ &= 54\frac{58}{63} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=10^\circ$ and diameter $=$ 22 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 22+\frac{1}{2}\times 2 \times \pi \times r + \frac{10}{360}\times 2 \times \pi \times r_1 \\ &= 22+\frac{1}{2}\times 2\times \frac{22}{7}\times 11 + \frac{1}{36}\times 2 \times \frac{22}{7} \times 22 \\ &= 22+\frac{242}{7}+\frac{242}{63} \\ &= 60\frac{26}{63} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 22+\frac{1}{2}\times 2 \times \pi \times r + \frac{10}{360}\times 2 \times \pi \times r_1 \\ &= 22+\frac{1}{2}\times 2\times \frac{22}{7}\times 11 + \frac{1}{36}\times 2 \times \frac{22}{7} \times 22 \\ &= 22+\frac{242}{7}+\frac{242}{63} \\ &= 60\frac{26}{63} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=12^\circ$ and diameter $=$ 8 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 8+\frac{1}{2}\times 2 \times \pi \times r + \frac{12}{360}\times 2 \times \pi \times r_1 \\ &= 8+\frac{1}{2}\times 2\times \frac{22}{7}\times 4 + \frac{1}{30}\times 2 \times \frac{22}{7} \times 8 \\ &= 8+\frac{88}{7}+\frac{176}{105} \\ &= 22\frac{26}{105} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 8+\frac{1}{2}\times 2 \times \pi \times r + \frac{12}{360}\times 2 \times \pi \times r_1 \\ &= 8+\frac{1}{2}\times 2\times \frac{22}{7}\times 4 + \frac{1}{30}\times 2 \times \frac{22}{7} \times 8 \\ &= 8+\frac{88}{7}+\frac{176}{105} \\ &= 22\frac{26}{105} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=12^\circ$ and diameter $=$ 10 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 10+\frac{1}{2}\times 2 \times \pi \times r + \frac{12}{360}\times 2 \times \pi \times r_1 \\ &= 10+\frac{1}{2}\times 2\times \frac{22}{7}\times 5 + \frac{1}{30}\times 2 \times \frac{22}{7} \times 10 \\ &= 10+\frac{110}{7}+\frac{44}{21} \\ &= 27\frac{17}{21} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 10+\frac{1}{2}\times 2 \times \pi \times r + \frac{12}{360}\times 2 \times \pi \times r_1 \\ &= 10+\frac{1}{2}\times 2\times \frac{22}{7}\times 5 + \frac{1}{30}\times 2 \times \frac{22}{7} \times 10 \\ &= 10+\frac{110}{7}+\frac{44}{21} \\ &= 27\frac{17}{21} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=12^\circ$ and diameter $=$ 12 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 12+\frac{1}{2}\times 2 \times \pi \times r + \frac{12}{360}\times 2 \times \pi \times r_1 \\ &= 12+\frac{1}{2}\times 2\times \frac{22}{7}\times 6 + \frac{1}{30}\times 2 \times \frac{22}{7} \times 12 \\ &= 12+\frac{132}{7}+\frac{88}{35} \\ &= 33\frac{13}{35} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 12+\frac{1}{2}\times 2 \times \pi \times r + \frac{12}{360}\times 2 \times \pi \times r_1 \\ &= 12+\frac{1}{2}\times 2\times \frac{22}{7}\times 6 + \frac{1}{30}\times 2 \times \frac{22}{7} \times 12 \\ &= 12+\frac{132}{7}+\frac{88}{35} \\ &= 33\frac{13}{35} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=12^\circ$ and diameter $=$ 14 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 14+\frac{1}{2}\times 2 \times \pi \times r + \frac{12}{360}\times 2 \times \pi \times r_1 \\ &= 14+\frac{1}{2}\times 2\times \frac{22}{7}\times 7 + \frac{1}{30}\times 2 \times \frac{22}{7} \times 14 \\ &= 14+22+\frac{44}{15} \\ &= 38\frac{14}{15} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 14+\frac{1}{2}\times 2 \times \pi \times r + \frac{12}{360}\times 2 \times \pi \times r_1 \\ &= 14+\frac{1}{2}\times 2\times \frac{22}{7}\times 7 + \frac{1}{30}\times 2 \times \frac{22}{7} \times 14 \\ &= 14+22+\frac{44}{15} \\ &= 38\frac{14}{15} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=12^\circ$ and diameter $=$ 16 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 16+\frac{1}{2}\times 2 \times \pi \times r + \frac{12}{360}\times 2 \times \pi \times r_1 \\ &= 16+\frac{1}{2}\times 2\times \frac{22}{7}\times 8 + \frac{1}{30}\times 2 \times \frac{22}{7} \times 16 \\ &= 16+\frac{176}{7}+\frac{352}{105} \\ &= 44\frac{52}{105} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 16+\frac{1}{2}\times 2 \times \pi \times r + \frac{12}{360}\times 2 \times \pi \times r_1 \\ &= 16+\frac{1}{2}\times 2\times \frac{22}{7}\times 8 + \frac{1}{30}\times 2 \times \frac{22}{7} \times 16 \\ &= 16+\frac{176}{7}+\frac{352}{105} \\ &= 44\frac{52}{105} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=12^\circ$ and diameter $=$ 18 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 18+\frac{1}{2}\times 2 \times \pi \times r + \frac{12}{360}\times 2 \times \pi \times r_1 \\ &= 18+\frac{1}{2}\times 2\times \frac{22}{7}\times 9 + \frac{1}{30}\times 2 \times \frac{22}{7} \times 18 \\ &= 18+\frac{198}{7}+\frac{132}{35} \\ &= 50\frac{2}{35} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 18+\frac{1}{2}\times 2 \times \pi \times r + \frac{12}{360}\times 2 \times \pi \times r_1 \\ &= 18+\frac{1}{2}\times 2\times \frac{22}{7}\times 9 + \frac{1}{30}\times 2 \times \frac{22}{7} \times 18 \\ &= 18+\frac{198}{7}+\frac{132}{35} \\ &= 50\frac{2}{35} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=12^\circ$ and diameter $=$ 20 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 20+\frac{1}{2}\times 2 \times \pi \times r + \frac{12}{360}\times 2 \times \pi \times r_1 \\ &= 20+\frac{1}{2}\times 2\times \frac{22}{7}\times 10 + \frac{1}{30}\times 2 \times \frac{22}{7} \times 20 \\ &= 20+\frac{220}{7}+\frac{88}{21} \\ &= 55\frac{13}{21} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 20+\frac{1}{2}\times 2 \times \pi \times r + \frac{12}{360}\times 2 \times \pi \times r_1 \\ &= 20+\frac{1}{2}\times 2\times \frac{22}{7}\times 10 + \frac{1}{30}\times 2 \times \frac{22}{7} \times 20 \\ &= 20+\frac{220}{7}+\frac{88}{21} \\ &= 55\frac{13}{21} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=12^\circ$ and diameter $=$ 22 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 22+\frac{1}{2}\times 2 \times \pi \times r + \frac{12}{360}\times 2 \times \pi \times r_1 \\ &= 22+\frac{1}{2}\times 2\times \frac{22}{7}\times 11 + \frac{1}{30}\times 2 \times \frac{22}{7} \times 22 \\ &= 22+\frac{242}{7}+\frac{484}{105} \\ &= 61\frac{19}{105} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 22+\frac{1}{2}\times 2 \times \pi \times r + \frac{12}{360}\times 2 \times \pi \times r_1 \\ &= 22+\frac{1}{2}\times 2\times \frac{22}{7}\times 11 + \frac{1}{30}\times 2 \times \frac{22}{7} \times 22 \\ &= 22+\frac{242}{7}+\frac{484}{105} \\ &= 61\frac{19}{105} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=15^\circ$ and diameter $=$ 8 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 8+\frac{1}{2}\times 2 \times \pi \times r + \frac{15}{360}\times 2 \times \pi \times r_1 \\ &= 8+\frac{1}{2}\times 2\times \frac{22}{7}\times 4 + \frac{1}{24}\times 2 \times \frac{22}{7} \times 8 \\ &= 8+\frac{88}{7}+\frac{44}{21} \\ &= 22\frac{2}{3} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 8+\frac{1}{2}\times 2 \times \pi \times r + \frac{15}{360}\times 2 \times \pi \times r_1 \\ &= 8+\frac{1}{2}\times 2\times \frac{22}{7}\times 4 + \frac{1}{24}\times 2 \times \frac{22}{7} \times 8 \\ &= 8+\frac{88}{7}+\frac{44}{21} \\ &= 22\frac{2}{3} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=15^\circ$ and diameter $=$ 10 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 10+\frac{1}{2}\times 2 \times \pi \times r + \frac{15}{360}\times 2 \times \pi \times r_1 \\ &= 10+\frac{1}{2}\times 2\times \frac{22}{7}\times 5 + \frac{1}{24}\times 2 \times \frac{22}{7} \times 10 \\ &= 10+\frac{110}{7}+\frac{55}{21} \\ &= 28\frac{1}{3} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 10+\frac{1}{2}\times 2 \times \pi \times r + \frac{15}{360}\times 2 \times \pi \times r_1 \\ &= 10+\frac{1}{2}\times 2\times \frac{22}{7}\times 5 + \frac{1}{24}\times 2 \times \frac{22}{7} \times 10 \\ &= 10+\frac{110}{7}+\frac{55}{21} \\ &= 28\frac{1}{3} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=15^\circ$ and diameter $=$ 12 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 12+\frac{1}{2}\times 2 \times \pi \times r + \frac{15}{360}\times 2 \times \pi \times r_1 \\ &= 12+\frac{1}{2}\times 2\times \frac{22}{7}\times 6 + \frac{1}{24}\times 2 \times \frac{22}{7} \times 12 \\ &= 12+\frac{132}{7}+\frac{22}{7} \\ &= 34 \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 12+\frac{1}{2}\times 2 \times \pi \times r + \frac{15}{360}\times 2 \times \pi \times r_1 \\ &= 12+\frac{1}{2}\times 2\times \frac{22}{7}\times 6 + \frac{1}{24}\times 2 \times \frac{22}{7} \times 12 \\ &= 12+\frac{132}{7}+\frac{22}{7} \\ &= 34 \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=15^\circ$ and diameter $=$ 14 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 14+\frac{1}{2}\times 2 \times \pi \times r + \frac{15}{360}\times 2 \times \pi \times r_1 \\ &= 14+\frac{1}{2}\times 2\times \frac{22}{7}\times 7 + \frac{1}{24}\times 2 \times \frac{22}{7} \times 14 \\ &= 14+22+\frac{11}{3} \\ &= 39\frac{2}{3} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 14+\frac{1}{2}\times 2 \times \pi \times r + \frac{15}{360}\times 2 \times \pi \times r_1 \\ &= 14+\frac{1}{2}\times 2\times \frac{22}{7}\times 7 + \frac{1}{24}\times 2 \times \frac{22}{7} \times 14 \\ &= 14+22+\frac{11}{3} \\ &= 39\frac{2}{3} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=15^\circ$ and diameter $=$ 16 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 16+\frac{1}{2}\times 2 \times \pi \times r + \frac{15}{360}\times 2 \times \pi \times r_1 \\ &= 16+\frac{1}{2}\times 2\times \frac{22}{7}\times 8 + \frac{1}{24}\times 2 \times \frac{22}{7} \times 16 \\ &= 16+\frac{176}{7}+\frac{88}{21} \\ &= 45\frac{1}{3} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 16+\frac{1}{2}\times 2 \times \pi \times r + \frac{15}{360}\times 2 \times \pi \times r_1 \\ &= 16+\frac{1}{2}\times 2\times \frac{22}{7}\times 8 + \frac{1}{24}\times 2 \times \frac{22}{7} \times 16 \\ &= 16+\frac{176}{7}+\frac{88}{21} \\ &= 45\frac{1}{3} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=15^\circ$ and diameter $=$ 18 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 18+\frac{1}{2}\times 2 \times \pi \times r + \frac{15}{360}\times 2 \times \pi \times r_1 \\ &= 18+\frac{1}{2}\times 2\times \frac{22}{7}\times 9 + \frac{1}{24}\times 2 \times \frac{22}{7} \times 18 \\ &= 18+\frac{198}{7}+\frac{33}{7} \\ &= 51 \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 18+\frac{1}{2}\times 2 \times \pi \times r + \frac{15}{360}\times 2 \times \pi \times r_1 \\ &= 18+\frac{1}{2}\times 2\times \frac{22}{7}\times 9 + \frac{1}{24}\times 2 \times \frac{22}{7} \times 18 \\ &= 18+\frac{198}{7}+\frac{33}{7} \\ &= 51 \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=15^\circ$ and diameter $=$ 20 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 20+\frac{1}{2}\times 2 \times \pi \times r + \frac{15}{360}\times 2 \times \pi \times r_1 \\ &= 20+\frac{1}{2}\times 2\times \frac{22}{7}\times 10 + \frac{1}{24}\times 2 \times \frac{22}{7} \times 20 \\ &= 20+\frac{220}{7}+\frac{110}{21} \\ &= 56\frac{2}{3} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 20+\frac{1}{2}\times 2 \times \pi \times r + \frac{15}{360}\times 2 \times \pi \times r_1 \\ &= 20+\frac{1}{2}\times 2\times \frac{22}{7}\times 10 + \frac{1}{24}\times 2 \times \frac{22}{7} \times 20 \\ &= 20+\frac{220}{7}+\frac{110}{21} \\ &= 56\frac{2}{3} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=15^\circ$ and diameter $=$ 22 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 22+\frac{1}{2}\times 2 \times \pi \times r + \frac{15}{360}\times 2 \times \pi \times r_1 \\ &= 22+\frac{1}{2}\times 2\times \frac{22}{7}\times 11 + \frac{1}{24}\times 2 \times \frac{22}{7} \times 22 \\ &= 22+\frac{242}{7}+\frac{121}{21} \\ &= 62\frac{1}{3} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 22+\frac{1}{2}\times 2 \times \pi \times r + \frac{15}{360}\times 2 \times \pi \times r_1 \\ &= 22+\frac{1}{2}\times 2\times \frac{22}{7}\times 11 + \frac{1}{24}\times 2 \times \frac{22}{7} \times 22 \\ &= 22+\frac{242}{7}+\frac{121}{21} \\ &= 62\frac{1}{3} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=18^\circ$ and diameter $=$ 8 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 8+\frac{1}{2}\times 2 \times \pi \times r + \frac{18}{360}\times 2 \times \pi \times r_1 \\ &= 8+\frac{1}{2}\times 2\times \frac{22}{7}\times 4 + \frac{1}{20}\times 2 \times \frac{22}{7} \times 8 \\ &= 8+\frac{88}{7}+\frac{88}{35} \\ &= 23\frac{3}{35} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 8+\frac{1}{2}\times 2 \times \pi \times r + \frac{18}{360}\times 2 \times \pi \times r_1 \\ &= 8+\frac{1}{2}\times 2\times \frac{22}{7}\times 4 + \frac{1}{20}\times 2 \times \frac{22}{7} \times 8 \\ &= 8+\frac{88}{7}+\frac{88}{35} \\ &= 23\frac{3}{35} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=18^\circ$ and diameter $=$ 10 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 10+\frac{1}{2}\times 2 \times \pi \times r + \frac{18}{360}\times 2 \times \pi \times r_1 \\ &= 10+\frac{1}{2}\times 2\times \frac{22}{7}\times 5 + \frac{1}{20}\times 2 \times \frac{22}{7} \times 10 \\ &= 10+\frac{110}{7}+\frac{22}{7} \\ &= 28\frac{6}{7} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 10+\frac{1}{2}\times 2 \times \pi \times r + \frac{18}{360}\times 2 \times \pi \times r_1 \\ &= 10+\frac{1}{2}\times 2\times \frac{22}{7}\times 5 + \frac{1}{20}\times 2 \times \frac{22}{7} \times 10 \\ &= 10+\frac{110}{7}+\frac{22}{7} \\ &= 28\frac{6}{7} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=18^\circ$ and diameter $=$ 12 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 12+\frac{1}{2}\times 2 \times \pi \times r + \frac{18}{360}\times 2 \times \pi \times r_1 \\ &= 12+\frac{1}{2}\times 2\times \frac{22}{7}\times 6 + \frac{1}{20}\times 2 \times \frac{22}{7} \times 12 \\ &= 12+\frac{132}{7}+\frac{132}{35} \\ &= 34\frac{22}{35} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 12+\frac{1}{2}\times 2 \times \pi \times r + \frac{18}{360}\times 2 \times \pi \times r_1 \\ &= 12+\frac{1}{2}\times 2\times \frac{22}{7}\times 6 + \frac{1}{20}\times 2 \times \frac{22}{7} \times 12 \\ &= 12+\frac{132}{7}+\frac{132}{35} \\ &= 34\frac{22}{35} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=18^\circ$ and diameter $=$ 14 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 14+\frac{1}{2}\times 2 \times \pi \times r + \frac{18}{360}\times 2 \times \pi \times r_1 \\ &= 14+\frac{1}{2}\times 2\times \frac{22}{7}\times 7 + \frac{1}{20}\times 2 \times \frac{22}{7} \times 14 \\ &= 14+22+\frac{22}{5} \\ &= 40\frac{2}{5} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 14+\frac{1}{2}\times 2 \times \pi \times r + \frac{18}{360}\times 2 \times \pi \times r_1 \\ &= 14+\frac{1}{2}\times 2\times \frac{22}{7}\times 7 + \frac{1}{20}\times 2 \times \frac{22}{7} \times 14 \\ &= 14+22+\frac{22}{5} \\ &= 40\frac{2}{5} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=18^\circ$ and diameter $=$ 16 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 16+\frac{1}{2}\times 2 \times \pi \times r + \frac{18}{360}\times 2 \times \pi \times r_1 \\ &= 16+\frac{1}{2}\times 2\times \frac{22}{7}\times 8 + \frac{1}{20}\times 2 \times \frac{22}{7} \times 16 \\ &= 16+\frac{176}{7}+\frac{176}{35} \\ &= 46\frac{6}{35} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 16+\frac{1}{2}\times 2 \times \pi \times r + \frac{18}{360}\times 2 \times \pi \times r_1 \\ &= 16+\frac{1}{2}\times 2\times \frac{22}{7}\times 8 + \frac{1}{20}\times 2 \times \frac{22}{7} \times 16 \\ &= 16+\frac{176}{7}+\frac{176}{35} \\ &= 46\frac{6}{35} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=18^\circ$ and diameter $=$ 18 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 18+\frac{1}{2}\times 2 \times \pi \times r + \frac{18}{360}\times 2 \times \pi \times r_1 \\ &= 18+\frac{1}{2}\times 2\times \frac{22}{7}\times 9 + \frac{1}{20}\times 2 \times \frac{22}{7} \times 18 \\ &= 18+\frac{198}{7}+\frac{198}{35} \\ &= 51\frac{33}{35} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 18+\frac{1}{2}\times 2 \times \pi \times r + \frac{18}{360}\times 2 \times \pi \times r_1 \\ &= 18+\frac{1}{2}\times 2\times \frac{22}{7}\times 9 + \frac{1}{20}\times 2 \times \frac{22}{7} \times 18 \\ &= 18+\frac{198}{7}+\frac{198}{35} \\ &= 51\frac{33}{35} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=18^\circ$ and diameter $=$ 20 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 20+\frac{1}{2}\times 2 \times \pi \times r + \frac{18}{360}\times 2 \times \pi \times r_1 \\ &= 20+\frac{1}{2}\times 2\times \frac{22}{7}\times 10 + \frac{1}{20}\times 2 \times \frac{22}{7} \times 20 \\ &= 20+\frac{220}{7}+\frac{44}{7} \\ &= 57\frac{5}{7} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 20+\frac{1}{2}\times 2 \times \pi \times r + \frac{18}{360}\times 2 \times \pi \times r_1 \\ &= 20+\frac{1}{2}\times 2\times \frac{22}{7}\times 10 + \frac{1}{20}\times 2 \times \frac{22}{7} \times 20 \\ &= 20+\frac{220}{7}+\frac{44}{7} \\ &= 57\frac{5}{7} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=18^\circ$ and diameter $=$ 22 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 22+\frac{1}{2}\times 2 \times \pi \times r + \frac{18}{360}\times 2 \times \pi \times r_1 \\ &= 22+\frac{1}{2}\times 2\times \frac{22}{7}\times 11 + \frac{1}{20}\times 2 \times \frac{22}{7} \times 22 \\ &= 22+\frac{242}{7}+\frac{242}{35} \\ &= 63\frac{17}{35} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 22+\frac{1}{2}\times 2 \times \pi \times r + \frac{18}{360}\times 2 \times \pi \times r_1 \\ &= 22+\frac{1}{2}\times 2\times \frac{22}{7}\times 11 + \frac{1}{20}\times 2 \times \frac{22}{7} \times 22 \\ &= 22+\frac{242}{7}+\frac{242}{35} \\ &= 63\frac{17}{35} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=20^\circ$ and diameter $=$ 8 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 8+\frac{1}{2}\times 2 \times \pi \times r + \frac{20}{360}\times 2 \times \pi \times r_1 \\ &= 8+\frac{1}{2}\times 2\times \frac{22}{7}\times 4 + \frac{1}{18}\times 2 \times \frac{22}{7} \times 8 \\ &= 8+\frac{88}{7}+\frac{176}{63} \\ &= 23\frac{23}{63} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 8+\frac{1}{2}\times 2 \times \pi \times r + \frac{20}{360}\times 2 \times \pi \times r_1 \\ &= 8+\frac{1}{2}\times 2\times \frac{22}{7}\times 4 + \frac{1}{18}\times 2 \times \frac{22}{7} \times 8 \\ &= 8+\frac{88}{7}+\frac{176}{63} \\ &= 23\frac{23}{63} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=20^\circ$ and diameter $=$ 10 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 10+\frac{1}{2}\times 2 \times \pi \times r + \frac{20}{360}\times 2 \times \pi \times r_1 \\ &= 10+\frac{1}{2}\times 2\times \frac{22}{7}\times 5 + \frac{1}{18}\times 2 \times \frac{22}{7} \times 10 \\ &= 10+\frac{110}{7}+\frac{220}{63} \\ &= 29\frac{13}{63} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 10+\frac{1}{2}\times 2 \times \pi \times r + \frac{20}{360}\times 2 \times \pi \times r_1 \\ &= 10+\frac{1}{2}\times 2\times \frac{22}{7}\times 5 + \frac{1}{18}\times 2 \times \frac{22}{7} \times 10 \\ &= 10+\frac{110}{7}+\frac{220}{63} \\ &= 29\frac{13}{63} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=20^\circ$ and diameter $=$ 12 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 12+\frac{1}{2}\times 2 \times \pi \times r + \frac{20}{360}\times 2 \times \pi \times r_1 \\ &= 12+\frac{1}{2}\times 2\times \frac{22}{7}\times 6 + \frac{1}{18}\times 2 \times \frac{22}{7} \times 12 \\ &= 12+\frac{132}{7}+\frac{88}{21} \\ &= 35\frac{1}{21} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 12+\frac{1}{2}\times 2 \times \pi \times r + \frac{20}{360}\times 2 \times \pi \times r_1 \\ &= 12+\frac{1}{2}\times 2\times \frac{22}{7}\times 6 + \frac{1}{18}\times 2 \times \frac{22}{7} \times 12 \\ &= 12+\frac{132}{7}+\frac{88}{21} \\ &= 35\frac{1}{21} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=20^\circ$ and diameter $=$ 14 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 14+\frac{1}{2}\times 2 \times \pi \times r + \frac{20}{360}\times 2 \times \pi \times r_1 \\ &= 14+\frac{1}{2}\times 2\times \frac{22}{7}\times 7 + \frac{1}{18}\times 2 \times \frac{22}{7} \times 14 \\ &= 14+22+\frac{44}{9} \\ &= 40\frac{8}{9} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 14+\frac{1}{2}\times 2 \times \pi \times r + \frac{20}{360}\times 2 \times \pi \times r_1 \\ &= 14+\frac{1}{2}\times 2\times \frac{22}{7}\times 7 + \frac{1}{18}\times 2 \times \frac{22}{7} \times 14 \\ &= 14+22+\frac{44}{9} \\ &= 40\frac{8}{9} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=20^\circ$ and diameter $=$ 16 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 16+\frac{1}{2}\times 2 \times \pi \times r + \frac{20}{360}\times 2 \times \pi \times r_1 \\ &= 16+\frac{1}{2}\times 2\times \frac{22}{7}\times 8 + \frac{1}{18}\times 2 \times \frac{22}{7} \times 16 \\ &= 16+\frac{176}{7}+\frac{352}{63} \\ &= 46\frac{46}{63} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 16+\frac{1}{2}\times 2 \times \pi \times r + \frac{20}{360}\times 2 \times \pi \times r_1 \\ &= 16+\frac{1}{2}\times 2\times \frac{22}{7}\times 8 + \frac{1}{18}\times 2 \times \frac{22}{7} \times 16 \\ &= 16+\frac{176}{7}+\frac{352}{63} \\ &= 46\frac{46}{63} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=20^\circ$ and diameter $=$ 18 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 18+\frac{1}{2}\times 2 \times \pi \times r + \frac{20}{360}\times 2 \times \pi \times r_1 \\ &= 18+\frac{1}{2}\times 2\times \frac{22}{7}\times 9 + \frac{1}{18}\times 2 \times \frac{22}{7} \times 18 \\ &= 18+\frac{198}{7}+\frac{44}{7} \\ &= 52\frac{4}{7} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 18+\frac{1}{2}\times 2 \times \pi \times r + \frac{20}{360}\times 2 \times \pi \times r_1 \\ &= 18+\frac{1}{2}\times 2\times \frac{22}{7}\times 9 + \frac{1}{18}\times 2 \times \frac{22}{7} \times 18 \\ &= 18+\frac{198}{7}+\frac{44}{7} \\ &= 52\frac{4}{7} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=20^\circ$ and diameter $=$ 20 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 20+\frac{1}{2}\times 2 \times \pi \times r + \frac{20}{360}\times 2 \times \pi \times r_1 \\ &= 20+\frac{1}{2}\times 2\times \frac{22}{7}\times 10 + \frac{1}{18}\times 2 \times \frac{22}{7} \times 20 \\ &= 20+\frac{220}{7}+\frac{440}{63} \\ &= 58\frac{26}{63} \text{ cm} \end{align*}
Yay! Your are right.
\begin{align*} &\phantom{=} 20+\frac{1}{2}\times 2 \times \pi \times r + \frac{20}{360}\times 2 \times \pi \times r_1 \\ &= 20+\frac{1}{2}\times 2\times \frac{22}{7}\times 10 + \frac{1}{18}\times 2 \times \frac{22}{7} \times 20 \\ &= 20+\frac{220}{7}+\frac{440}{63} \\ &= 58\frac{26}{63} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=20^\circ$ and diameter $=$ 22 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 22+\frac{1}{2}\times 2 \times \pi \times r + \frac{20}{360}\times 2 \times \pi \times r_1 \\ &= 22+\frac{1}{2}\times 2\times \frac{22}{7}\times 11 + \frac{1}{18}\times 2 \times \frac{22}{7} \times 22 \\ &= 22+\frac{242}{7}+\frac{484}{63} \\ &= 64\frac{16}{63} \text{ cm} \end{align*}
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\begin{align*} &\phantom{=} 22+\frac{1}{2}\times 2 \times \pi \times r + \frac{20}{360}\times 2 \times \pi \times r_1 \\ &= 22+\frac{1}{2}\times 2\times \frac{22}{7}\times 11 + \frac{1}{18}\times 2 \times \frac{22}{7} \times 22 \\ &= 22+\frac{242}{7}+\frac{484}{63} \\ &= 64\frac{16}{63} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=24^\circ$ and diameter $=$ 8 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 8+\frac{1}{2}\times 2 \times \pi \times r + \frac{24}{360}\times 2 \times \pi \times r_1 \\ &= 8+\frac{1}{2}\times 2\times \frac{22}{7}\times 4 + \frac{1}{15}\times 2 \times \frac{22}{7} \times 8 \\ &= 8+\frac{88}{7}+\frac{352}{105} \\ &= 23\frac{97}{105} \text{ cm} \end{align*}
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\begin{align*} &\phantom{=} 8+\frac{1}{2}\times 2 \times \pi \times r + \frac{24}{360}\times 2 \times \pi \times r_1 \\ &= 8+\frac{1}{2}\times 2\times \frac{22}{7}\times 4 + \frac{1}{15}\times 2 \times \frac{22}{7} \times 8 \\ &= 8+\frac{88}{7}+\frac{352}{105} \\ &= 23\frac{97}{105} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=24^\circ$ and diameter $=$ 10 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 10+\frac{1}{2}\times 2 \times \pi \times r + \frac{24}{360}\times 2 \times \pi \times r_1 \\ &= 10+\frac{1}{2}\times 2\times \frac{22}{7}\times 5 + \frac{1}{15}\times 2 \times \frac{22}{7} \times 10 \\ &= 10+\frac{110}{7}+\frac{88}{21} \\ &= 29\frac{19}{21} \text{ cm} \end{align*}
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\begin{align*} &\phantom{=} 10+\frac{1}{2}\times 2 \times \pi \times r + \frac{24}{360}\times 2 \times \pi \times r_1 \\ &= 10+\frac{1}{2}\times 2\times \frac{22}{7}\times 5 + \frac{1}{15}\times 2 \times \frac{22}{7} \times 10 \\ &= 10+\frac{110}{7}+\frac{88}{21} \\ &= 29\frac{19}{21} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=24^\circ$ and diameter $=$ 12 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 12+\frac{1}{2}\times 2 \times \pi \times r + \frac{24}{360}\times 2 \times \pi \times r_1 \\ &= 12+\frac{1}{2}\times 2\times \frac{22}{7}\times 6 + \frac{1}{15}\times 2 \times \frac{22}{7} \times 12 \\ &= 12+\frac{132}{7}+\frac{176}{35} \\ &= 35\frac{31}{35} \text{ cm} \end{align*}
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\begin{align*} &\phantom{=} 12+\frac{1}{2}\times 2 \times \pi \times r + \frac{24}{360}\times 2 \times \pi \times r_1 \\ &= 12+\frac{1}{2}\times 2\times \frac{22}{7}\times 6 + \frac{1}{15}\times 2 \times \frac{22}{7} \times 12 \\ &= 12+\frac{132}{7}+\frac{176}{35} \\ &= 35\frac{31}{35} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=24^\circ$ and diameter $=$ 14 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 14+\frac{1}{2}\times 2 \times \pi \times r + \frac{24}{360}\times 2 \times \pi \times r_1 \\ &= 14+\frac{1}{2}\times 2\times \frac{22}{7}\times 7 + \frac{1}{15}\times 2 \times \frac{22}{7} \times 14 \\ &= 14+22+\frac{88}{15} \\ &= 41\frac{13}{15} \text{ cm} \end{align*}
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\begin{align*} &\phantom{=} 14+\frac{1}{2}\times 2 \times \pi \times r + \frac{24}{360}\times 2 \times \pi \times r_1 \\ &= 14+\frac{1}{2}\times 2\times \frac{22}{7}\times 7 + \frac{1}{15}\times 2 \times \frac{22}{7} \times 14 \\ &= 14+22+\frac{88}{15} \\ &= 41\frac{13}{15} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=24^\circ$ and diameter $=$ 16 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 16+\frac{1}{2}\times 2 \times \pi \times r + \frac{24}{360}\times 2 \times \pi \times r_1 \\ &= 16+\frac{1}{2}\times 2\times \frac{22}{7}\times 8 + \frac{1}{15}\times 2 \times \frac{22}{7} \times 16 \\ &= 16+\frac{176}{7}+\frac{704}{105} \\ &= 47\frac{89}{105} \text{ cm} \end{align*}
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\begin{align*} &\phantom{=} 16+\frac{1}{2}\times 2 \times \pi \times r + \frac{24}{360}\times 2 \times \pi \times r_1 \\ &= 16+\frac{1}{2}\times 2\times \frac{22}{7}\times 8 + \frac{1}{15}\times 2 \times \frac{22}{7} \times 16 \\ &= 16+\frac{176}{7}+\frac{704}{105} \\ &= 47\frac{89}{105} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=24^\circ$ and diameter $=$ 18 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 18+\frac{1}{2}\times 2 \times \pi \times r + \frac{24}{360}\times 2 \times \pi \times r_1 \\ &= 18+\frac{1}{2}\times 2\times \frac{22}{7}\times 9 + \frac{1}{15}\times 2 \times \frac{22}{7} \times 18 \\ &= 18+\frac{198}{7}+\frac{264}{35} \\ &= 53\frac{29}{35} \text{ cm} \end{align*}
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\begin{align*} &\phantom{=} 18+\frac{1}{2}\times 2 \times \pi \times r + \frac{24}{360}\times 2 \times \pi \times r_1 \\ &= 18+\frac{1}{2}\times 2\times \frac{22}{7}\times 9 + \frac{1}{15}\times 2 \times \frac{22}{7} \times 18 \\ &= 18+\frac{198}{7}+\frac{264}{35} \\ &= 53\frac{29}{35} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=24^\circ$ and diameter $=$ 20 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 20+\frac{1}{2}\times 2 \times \pi \times r + \frac{24}{360}\times 2 \times \pi \times r_1 \\ &= 20+\frac{1}{2}\times 2\times \frac{22}{7}\times 10 + \frac{1}{15}\times 2 \times \frac{22}{7} \times 20 \\ &= 20+\frac{220}{7}+\frac{176}{21} \\ &= 59\frac{17}{21} \text{ cm} \end{align*}
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\begin{align*} &\phantom{=} 20+\frac{1}{2}\times 2 \times \pi \times r + \frac{24}{360}\times 2 \times \pi \times r_1 \\ &= 20+\frac{1}{2}\times 2\times \frac{22}{7}\times 10 + \frac{1}{15}\times 2 \times \frac{22}{7} \times 20 \\ &= 20+\frac{220}{7}+\frac{176}{21} \\ &= 59\frac{17}{21} \text{ cm} \end{align*}
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region if $\theta=24^\circ$ and diameter $=$ 22 cm. Take $\pi=\frac{22}{7}$.
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\begin{align*} &\phantom{=} 22+\frac{1}{2}\times 2 \times \pi \times r + \frac{24}{360}\times 2 \times \pi \times r_1 \\ &= 22+\frac{1}{2}\times 2\times \frac{22}{7}\times 11 + \frac{1}{15}\times 2 \times \frac{22}{7} \times 22 \\ &= 22+\frac{242}{7}+\frac{968}{105} \\ &= 65\frac{83}{105} \text{ cm} \end{align*}
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\begin{align*} &\phantom{=} 22+\frac{1}{2}\times 2 \times \pi \times r + \frac{24}{360}\times 2 \times \pi \times r_1 \\ &= 22+\frac{1}{2}\times 2\times \frac{22}{7}\times 11 + \frac{1}{15}\times 2 \times \frac{22}{7} \times 22 \\ &= 22+\frac{242}{7}+\frac{968}{105} \\ &= 65\frac{83}{105} \text{ cm} \end{align*}
Three identical circles of radii 1 cm intersect through their centers as shown in the figure. Find the area of the shaded region. Take $\pi=\frac{22}{7}$.
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By cutting and pasting, we obtain a figure of a semi-circle \begin{equation*} \frac{1}{2}\times \frac{22}{7} \times 1 \times 1 = \frac{22}{2} \end{equation*} \begin{equation} \frac{11}{7}\text{ cm}^2 \end{equation}
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By cutting and pasting, we obtain a figure of a semi-circle \begin{equation*} \frac{1}{2}\times \frac{22}{7} \times 1 \times 1 = \frac{22}{2} \end{equation*} \begin{equation} \frac{11}{7}\text{ cm}^2 \end{equation}
Three identical circles of radii 2 cm intersect through their centers as shown in the figure. Find the area of the shaded region. Take $\pi=\frac{22}{7}$.
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By cutting and pasting, we obtain a figure of a semi-circle \begin{equation*} \frac{1}{2}\times \frac{22}{7} \times 2 \times 2 = \frac{88}{2} \end{equation*} \begin{equation} \frac{44}{7}\text{ cm}^2 \end{equation}
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By cutting and pasting, we obtain a figure of a semi-circle \begin{equation*} \frac{1}{2}\times \frac{22}{7} \times 2 \times 2 = \frac{88}{2} \end{equation*} \begin{equation} \frac{44}{7}\text{ cm}^2 \end{equation}
Three identical circles of radii 3 cm intersect through their centers as shown in the figure. Find the area of the shaded region. Take $\pi=\frac{22}{7}$.
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By cutting and pasting, we obtain a figure of a semi-circle \begin{equation*} \frac{1}{2}\times \frac{22}{7} \times 3 \times 3 = \frac{198}{2} \end{equation*} \begin{equation} \frac{99}{7}\text{ cm}^2 \end{equation}
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By cutting and pasting, we obtain a figure of a semi-circle \begin{equation*} \frac{1}{2}\times \frac{22}{7} \times 3 \times 3 = \frac{198}{2} \end{equation*} \begin{equation} \frac{99}{7}\text{ cm}^2 \end{equation}
Three identical circles of radii 4 cm intersect through their centers as shown in the figure. Find the area of the shaded region. Take $\pi=\frac{22}{7}$.
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By cutting and pasting, we obtain a figure of a semi-circle \begin{equation*} \frac{1}{2}\times \frac{22}{7} \times 4 \times 4 = \frac{352}{2} \end{equation*} \begin{equation} \frac{176}{7}\text{ cm}^2 \end{equation}
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By cutting and pasting, we obtain a figure of a semi-circle \begin{equation*} \frac{1}{2}\times \frac{22}{7} \times 4 \times 4 = \frac{352}{2} \end{equation*} \begin{equation} \frac{176}{7}\text{ cm}^2 \end{equation}
Three identical circles of radii 5 cm intersect through their centers as shown in the figure. Find the area of the shaded region. Take $\pi=\frac{22}{7}$.
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By cutting and pasting, we obtain a figure of a semi-circle \begin{equation*} \frac{1}{2}\times \frac{22}{7} \times 5 \times 5 = \frac{550}{2} \end{equation*} \begin{equation} \frac{275}{7}\text{ cm}^2 \end{equation}
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By cutting and pasting, we obtain a figure of a semi-circle \begin{equation*} \frac{1}{2}\times \frac{22}{7} \times 5 \times 5 = \frac{550}{2} \end{equation*} \begin{equation} \frac{275}{7}\text{ cm}^2 \end{equation}
Three identical circles of radii 6 cm intersect through their centers as shown in the figure. Find the area of the shaded region. Take $\pi=\frac{22}{7}$.
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By cutting and pasting, we obtain a figure of a semi-circle \begin{equation*} \frac{1}{2}\times \frac{22}{7} \times 6 \times 6 = \frac{792}{2} \end{equation*} \begin{equation} \frac{396}{7}\text{ cm}^2 \end{equation}
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By cutting and pasting, we obtain a figure of a semi-circle \begin{equation*} \frac{1}{2}\times \frac{22}{7} \times 6 \times 6 = \frac{792}{2} \end{equation*} \begin{equation} \frac{396}{7}\text{ cm}^2 \end{equation}
Three identical circles of radii 7 cm intersect through their centers as shown in the figure. Find the area of the shaded region. Take $\pi=\frac{22}{7}$.
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By cutting and pasting, we obtain a figure of a semi-circle \begin{equation*} \frac{1}{2}\times \frac{22}{7} \times 7 \times 7 = \frac{154}{2} \end{equation*} \begin{equation} 77\text{ cm}^2 \end{equation}
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By cutting and pasting, we obtain a figure of a semi-circle \begin{equation*} \frac{1}{2}\times \frac{22}{7} \times 7 \times 7 = \frac{154}{2} \end{equation*} \begin{equation} 77\text{ cm}^2 \end{equation}
Three identical circles of radii 8 cm intersect through their centers as shown in the figure. Find the area of the shaded region. Take $\pi=\frac{22}{7}$.
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By cutting and pasting, we obtain a figure of a semi-circle \begin{equation*} \frac{1}{2}\times \frac{22}{7} \times 8 \times 8 = \frac{1408}{2} \end{equation*} \begin{equation} \frac{704}{7}\text{ cm}^2 \end{equation}
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By cutting and pasting, we obtain a figure of a semi-circle \begin{equation*} \frac{1}{2}\times \frac{22}{7} \times 8 \times 8 = \frac{1408}{2} \end{equation*} \begin{equation} \frac{704}{7}\text{ cm}^2 \end{equation}
Three identical circles of radii 9 cm intersect through their centers as shown in the figure. Find the area of the shaded region. Take $\pi=\frac{22}{7}$.
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By cutting and pasting, we obtain a figure of a semi-circle \begin{equation*} \frac{1}{2}\times \frac{22}{7} \times 9 \times 9 = \frac{1782}{2} \end{equation*} \begin{equation} \frac{891}{7}\text{ cm}^2 \end{equation}
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By cutting and pasting, we obtain a figure of a semi-circle \begin{equation*} \frac{1}{2}\times \frac{22}{7} \times 9 \times 9 = \frac{1782}{2} \end{equation*} \begin{equation} \frac{891}{7}\text{ cm}^2 \end{equation}
Three identical circles of radii 10 cm intersect through their centers as shown in the figure. Find the area of the shaded region. Take $\pi=\frac{22}{7}$.
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By cutting and pasting, we obtain a figure of a semi-circle \begin{equation*} \frac{1}{2}\times \frac{22}{7} \times 10 \times 10 = \frac{2200}{2} \end{equation*} \begin{equation} \frac{1100}{7}\text{ cm}^2 \end{equation}
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By cutting and pasting, we obtain a figure of a semi-circle \begin{equation*} \frac{1}{2}\times \frac{22}{7} \times 10 \times 10 = \frac{2200}{2} \end{equation*} \begin{equation} \frac{1100}{7}\text{ cm}^2 \end{equation}
The area of a circle is given as $\pi r^2$, where $\pi=\frac{22}{7}$ or $3.14$. Find the area of the shaded regions in the figure, given that the area of square is 1 cm$^2$.
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The 4 radii are \begin{equation} 2\text{ cm},\ 3\text{ cm}, \ 4\text{ cm}, \ 5\text{ cm} \end{equation} Area shaded: \begin{align*} & \frac{1}{4}\pi\big[2^2+3^2+4^2+5^2+\big] \\ &= \frac{1}{4}\times \frac{22}{7} \times [4+9+16+25] \\ &= \frac{297}{7} \\ &= 42\frac{3}{7} \text{ cm}^2 \end{align*}
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The 4 radii are \begin{equation} 2\text{ cm},\ 3\text{ cm}, \ 4\text{ cm}, \ 5\text{ cm} \end{equation} Area shaded: \begin{align*} & \frac{1}{4}\pi\big[2^2+3^2+4^2+5^2+\big] \\ &= \frac{1}{4}\times \frac{22}{7} \times [4+9+16+25] \\ &= \frac{297}{7} \\ &= 42\frac{3}{7} \text{ cm}^2 \end{align*}
The area of a circle is given as $\pi r^2$, where $\pi=\frac{22}{7}$ or $3.14$. Find the area of the shaded regions in the figure, given that the area of square is 4 cm$^2$.
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The 4 radii are \begin{equation} 4\text{ cm},\ 6\text{ cm}, \ 8\text{ cm}, \ 10\text{ cm} \end{equation} Area shaded: \begin{align*} & \frac{1}{4}\pi\big[4^2+6^2+8^2+10^2+\big] \\ &= \frac{1}{4}\times \frac{22}{7} \times 2^2 \times [2^2+3^2+4^2+5^2] \\ &= \frac{4}{4}\times \frac{22}{7} \times [4+9+16+25] \\ &= \frac{1188}{7} \\ &= 169\frac{5}{7} \text{ cm}^2 \end{align*}
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The 4 radii are \begin{equation} 4\text{ cm},\ 6\text{ cm}, \ 8\text{ cm}, \ 10\text{ cm} \end{equation} Area shaded: \begin{align*} & \frac{1}{4}\pi\big[4^2+6^2+8^2+10^2+\big] \\ &= \frac{1}{4}\times \frac{22}{7} \times 2^2 \times [2^2+3^2+4^2+5^2] \\ &= \frac{4}{4}\times \frac{22}{7} \times [4+9+16+25] \\ &= \frac{1188}{7} \\ &= 169\frac{5}{7} \text{ cm}^2 \end{align*}
The area of a circle is given as $\pi r^2$, where $\pi=\frac{22}{7}$ or $3.14$. Find the area of the shaded regions in the figure, given that the area of square is 9 cm$^2$.
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The 4 radii are \begin{equation} 6\text{ cm},\ 9\text{ cm}, \ 12\text{ cm}, \ 15\text{ cm} \end{equation} Area shaded: \begin{align*} & \frac{1}{4}\pi\big[6^2+9^2+12^2+15^2+\big] \\ &= \frac{1}{4}\times \frac{22}{7} \times 3^2 \times [2^2+3^2+4^2+5^2] \\ &= \frac{9}{4}\times \frac{22}{7} \times [4+9+16+25] \\ &= \frac{2673}{7} \\ &= 381\frac{6}{7} \text{ cm}^2 \end{align*}
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The 4 radii are \begin{equation} 6\text{ cm},\ 9\text{ cm}, \ 12\text{ cm}, \ 15\text{ cm} \end{equation} Area shaded: \begin{align*} & \frac{1}{4}\pi\big[6^2+9^2+12^2+15^2+\big] \\ &= \frac{1}{4}\times \frac{22}{7} \times 3^2 \times [2^2+3^2+4^2+5^2] \\ &= \frac{9}{4}\times \frac{22}{7} \times [4+9+16+25] \\ &= \frac{2673}{7} \\ &= 381\frac{6}{7} \text{ cm}^2 \end{align*}
The area of a circle is given as $\pi r^2$, where $\pi=\frac{22}{7}$ or $3.14$. Find the area of the shaded regions in the figure, given that the area of square is 16 cm$^2$.
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The 4 radii are \begin{equation} 8\text{ cm},\ 12\text{ cm}, \ 16\text{ cm}, \ 20\text{ cm} \end{equation} Area shaded: \begin{align*} & \frac{1}{4}\pi\big[8^2+12^2+16^2+20^2+\big] \\ &= \frac{1}{4}\times \frac{22}{7} \times 4^2 \times [2^2+3^2+4^2+5^2] \\ &= \frac{16}{4}\times \frac{22}{7} \times [4+9+16+25] \\ &= \frac{4752}{7} \\ &= 678\frac{6}{7} \text{ cm}^2 \end{align*}
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The 4 radii are \begin{equation} 8\text{ cm},\ 12\text{ cm}, \ 16\text{ cm}, \ 20\text{ cm} \end{equation} Area shaded: \begin{align*} & \frac{1}{4}\pi\big[8^2+12^2+16^2+20^2+\big] \\ &= \frac{1}{4}\times \frac{22}{7} \times 4^2 \times [2^2+3^2+4^2+5^2] \\ &= \frac{16}{4}\times \frac{22}{7} \times [4+9+16+25] \\ &= \frac{4752}{7} \\ &= 678\frac{6}{7} \text{ cm}^2 \end{align*}
The area of a circle is given as $\pi r^2$, where $\pi=\frac{22}{7}$ or $3.14$. Find the area of the shaded regions in the figure, given that the area of square is 25 cm$^2$.
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The 4 radii are \begin{equation} 10\text{ cm},\ 15\text{ cm}, \ 20\text{ cm}, \ 25\text{ cm} \end{equation} Area shaded: \begin{align*} & \frac{1}{4}\pi\big[10^2+15^2+20^2+25^2+\big] \\ &= \frac{1}{4}\times \frac{22}{7} \times 5^2 \times [2^2+3^2+4^2+5^2] \\ &= \frac{25}{4}\times \frac{22}{7} \times [4+9+16+25] \\ &= \frac{7425}{7} \\ &= 1060\frac{5}{7} \text{ cm}^2 \end{align*}
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The 4 radii are \begin{equation} 10\text{ cm},\ 15\text{ cm}, \ 20\text{ cm}, \ 25\text{ cm} \end{equation} Area shaded: \begin{align*} & \frac{1}{4}\pi\big[10^2+15^2+20^2+25^2+\big] \\ &= \frac{1}{4}\times \frac{22}{7} \times 5^2 \times [2^2+3^2+4^2+5^2] \\ &= \frac{25}{4}\times \frac{22}{7} \times [4+9+16+25] \\ &= \frac{7425}{7} \\ &= 1060\frac{5}{7} \text{ cm}^2 \end{align*}
The area of a circle is given as $\pi r^2$, where $\pi=\frac{22}{7}$ or $3.14$. Find the area of the shaded regions in the figure, given that the area of square is 36 cm$^2$.
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The 4 radii are \begin{equation} 12\text{ cm},\ 18\text{ cm}, \ 24\text{ cm}, \ 30\text{ cm} \end{equation} Area shaded: \begin{align*} & \frac{1}{4}\pi\big[12^2+18^2+24^2+30^2+\big] \\ &= \frac{1}{4}\times \frac{22}{7} \times 6^2 \times [2^2+3^2+4^2+5^2] \\ &= \frac{36}{4}\times \frac{22}{7} \times [4+9+16+25] \\ &= \frac{10692}{7} \\ &= 1527\frac{3}{7} \text{ cm}^2 \end{align*}
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The 4 radii are \begin{equation} 12\text{ cm},\ 18\text{ cm}, \ 24\text{ cm}, \ 30\text{ cm} \end{equation} Area shaded: \begin{align*} & \frac{1}{4}\pi\big[12^2+18^2+24^2+30^2+\big] \\ &= \frac{1}{4}\times \frac{22}{7} \times 6^2 \times [2^2+3^2+4^2+5^2] \\ &= \frac{36}{4}\times \frac{22}{7} \times [4+9+16+25] \\ &= \frac{10692}{7} \\ &= 1527\frac{3}{7} \text{ cm}^2 \end{align*}
The area of a circle is given as $\pi r^2$, where $\pi=\frac{22}{7}$ or $3.14$. Find the area of the shaded regions in the figure, given that the area of square is 49 cm$^2$.
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The 4 radii are \begin{equation} 14\text{ cm},\ 21\text{ cm}, \ 28\text{ cm}, \ 35\text{ cm} \end{equation} Area shaded: \begin{align*} & \frac{1}{4}\pi\big[14^2+21^2+28^2+35^2+\big] \\ &= \frac{1}{4}\times \frac{22}{7} \times 7^2 \times [2^2+3^2+4^2+5^2] \\ &= \frac{49}{4}\times \frac{22}{7} \times [4+9+16+25] \\ &= 2079 \\ &= 2079 \text{ cm}^2 \end{align*}
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The 4 radii are \begin{equation} 14\text{ cm},\ 21\text{ cm}, \ 28\text{ cm}, \ 35\text{ cm} \end{equation} Area shaded: \begin{align*} & \frac{1}{4}\pi\big[14^2+21^2+28^2+35^2+\big] \\ &= \frac{1}{4}\times \frac{22}{7} \times 7^2 \times [2^2+3^2+4^2+5^2] \\ &= \frac{49}{4}\times \frac{22}{7} \times [4+9+16+25] \\ &= 2079 \\ &= 2079 \text{ cm}^2 \end{align*}
The area of a circle is given as $\pi r^2$, where $\pi=\frac{22}{7}$ or $3.14$. Find the area of the shaded regions in the figure, given that the area of square is 64 cm$^2$.
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The 4 radii are \begin{equation} 16\text{ cm},\ 24\text{ cm}, \ 32\text{ cm}, \ 40\text{ cm} \end{equation} Area shaded: \begin{align*} & \frac{1}{4}\pi\big[16^2+24^2+32^2+40^2+\big] \\ &= \frac{1}{4}\times \frac{22}{7} \times 8^2 \times [2^2+3^2+4^2+5^2] \\ &= \frac{64}{4}\times \frac{22}{7} \times [4+9+16+25] \\ &= \frac{19008}{7} \\ &= 2715\frac{3}{7} \text{ cm}^2 \end{align*}
Yay! Your are right.
The 4 radii are \begin{equation} 16\text{ cm},\ 24\text{ cm}, \ 32\text{ cm}, \ 40\text{ cm} \end{equation} Area shaded: \begin{align*} & \frac{1}{4}\pi\big[16^2+24^2+32^2+40^2+\big] \\ &= \frac{1}{4}\times \frac{22}{7} \times 8^2 \times [2^2+3^2+4^2+5^2] \\ &= \frac{64}{4}\times \frac{22}{7} \times [4+9+16+25] \\ &= \frac{19008}{7} \\ &= 2715\frac{3}{7} \text{ cm}^2 \end{align*}
The area of a circle is given as $\pi r^2$, where $\pi=\frac{22}{7}$ or $3.14$. Find the area of the shaded regions in the figure, given that the area of square is 81 cm$^2$.
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The 4 radii are \begin{equation} 18\text{ cm},\ 27\text{ cm}, \ 36\text{ cm}, \ 45\text{ cm} \end{equation} Area shaded: \begin{align*} & \frac{1}{4}\pi\big[18^2+27^2+36^2+45^2+\big] \\ &= \frac{1}{4}\times \frac{22}{7} \times 9^2 \times [2^2+3^2+4^2+5^2] \\ &= \frac{81}{4}\times \frac{22}{7} \times [4+9+16+25] \\ &= \frac{24057}{7} \\ &= 3436\frac{5}{7} \text{ cm}^2 \end{align*}
Yay! Your are right.
The 4 radii are \begin{equation} 18\text{ cm},\ 27\text{ cm}, \ 36\text{ cm}, \ 45\text{ cm} \end{equation} Area shaded: \begin{align*} & \frac{1}{4}\pi\big[18^2+27^2+36^2+45^2+\big] \\ &= \frac{1}{4}\times \frac{22}{7} \times 9^2 \times [2^2+3^2+4^2+5^2] \\ &= \frac{81}{4}\times \frac{22}{7} \times [4+9+16+25] \\ &= \frac{24057}{7} \\ &= 3436\frac{5}{7} \text{ cm}^2 \end{align*}
The area of a circle is given as $\pi r^2$, where $\pi=\frac{22}{7}$ or $3.14$. Find the area of the shaded regions in the figure, given that the area of square is 100 cm$^2$.
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The 4 radii are \begin{equation} 20\text{ cm},\ 30\text{ cm}, \ 40\text{ cm}, \ 50\text{ cm} \end{equation} Area shaded: \begin{align*} & \frac{1}{4}\pi\big[20^2+30^2+40^2+50^2+\big] \\ &= \frac{1}{4}\times \frac{22}{7} \times 10^2 \times [2^2+3^2+4^2+5^2] \\ &= \frac{100}{4}\times \frac{22}{7} \times [4+9+16+25] \\ &= \frac{29700}{7} \\ &= 4242\frac{6}{7} \text{ cm}^2 \end{align*}
Yay! Your are right.
The 4 radii are \begin{equation} 20\text{ cm},\ 30\text{ cm}, \ 40\text{ cm}, \ 50\text{ cm} \end{equation} Area shaded: \begin{align*} & \frac{1}{4}\pi\big[20^2+30^2+40^2+50^2+\big] \\ &= \frac{1}{4}\times \frac{22}{7} \times 10^2 \times [2^2+3^2+4^2+5^2] \\ &= \frac{100}{4}\times \frac{22}{7} \times [4+9+16+25] \\ &= \frac{29700}{7} \\ &= 4242\frac{6}{7} \text{ cm}^2 \end{align*}
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