Question : The area of a sector of a circle is 88 sq. cm., and the angle of the sector is 45°. Find the radius of the circle. (Use $\pi=\frac{22}{7}$)
Option 1: $3 \sqrt{ 11} \mathrm{~cm}$
Option 2: $4 \sqrt{ 14} \mathrm{~cm}$
Option 3: $6 \sqrt{ 13} \mathrm{~cm}$
Option 4: $5 \sqrt{ 14} \mathrm{~cm}$
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Correct Answer: $4 \sqrt{ 14} \mathrm{~cm}$
Solution : Given, that the angle of the sector is 45°. The area of a sector of a circle is 88 sq. cm. Let the radius of the circle be $r$ cm. We know that Area of sector = $\frac{\theta}{360^{\circ}}\times \pi r^2$ According to the question, $\frac{\theta}{360^{\circ}}\times \pi r^2=88$ ⇒ $\frac{22}{7}\times r^2\times \frac{45}{360}=88$ ⇒ $r^2=224$ $\therefore r=4\sqrt{14} \mathrm{~cm}$ Hence, the correct answer is $4 \sqrt{ 14} \mathrm{~cm}$.
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Question : The area of a sector of a circle is 66 cm2 and the angle of the sector is 60°. Find the radius of the circle.
Option 1: $5 \sqrt{15} \mathrm{~cm}$
Option 2: $6 \sqrt{14} \mathrm{~cm}$
Option 3: $7 \sqrt{19} \mathrm{~cm} $
Option 4: $3 \sqrt{14} \mathrm{~cm} $
Question : The area of the sector of a circle of radius 12 cm is $32 \pi \;\mathrm{cm}^2$. Find the length of the corresponding arc of the sector.
Option 1: $\frac{16}{3} \pi$ cm
Option 2: $\frac{13}{3} \pi$ cm
Option 3: $\frac{10}{3} \pi$ cm
Option 4: $\frac{8}{3} \pi$ cm
Question : The radius of a circle with centre at O is 6 cm and the central angle of a sector is 40°. Find the area of the sector.
Option 1: $6 \pi ~\mathrm{cm}^2$
Option 2: $5 \pi ~\mathrm{cm}^2$
Option 3: $4 \pi ~\mathrm{cm}^2$
Option 4: $8 \pi ~\mathrm{cm}^2$
Question : A circular arc whose radius is 4 cm makes an angle of 45º at the centre. Find the perimeter of the sector formed. (Take $\pi=\frac{22}{7}$)
Option 1: $\frac{78}{7} \mathrm{~cm}$
Option 2: $\frac{72}{7} \mathrm{~cm}$
Option 3: $\frac{74}{7} \mathrm{~cm}$
Option 4: $\frac{76}{7} \mathrm{~cm}$
Question : The volume of a cone with a height equal to the radius and slant height of 5 cm is:
Option 1: $\frac{125 \pi}{12 \sqrt{3}} \mathrm{~cm}^3$
Option 2: $\frac{125 \pi}{6 \sqrt{3}} \mathrm{~cm}^3$
Option 3: $\frac{125 \pi}{12 \sqrt{2}} \mathrm{~cm}^3$
Option 4: $\frac{125 \pi}{6 \sqrt{2}} \mathrm{~cm}^3$
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