Question

4. A chord subtends an angle of 90°at the centre of a circle whose radius is 20 cm. Compute the area of the corresponding major segment of the circle.

5. A square is inscribed in a circle. Calculate the ratio of the area of the circle and the square.

6. Find the area of the sector of a circle with radius 4cm and of angle 30°. Also, find the area of the corresponding major sector.​

Answer 1

4. $\downarrow$

Given:-

• A chord subtends an angle of 90° at the centre.
• Radius of the circle is of 20 cm.

To Find:-

• Area of the major segment of the circle ?

Solution:-

Here, To find area of the major segment we use certain formula

$\sf{So,\ Area\ of\ the\ major\ segment\ =\ \dfrac{\pi r^{2} (360^{\circ} - \theta)}{360^{\circ}} + \dfrac{1}{2}r^{2} sin\theta}$

Now, Putting all the values in the formula we get,

$\sf{\Longrightarrow \dfrac{\dfrac{22}{7} \times 20^{2} (360^{\circ} - 90^{\circ})}{360^{\circ}} + \dfrac{1}{2} \times 20^{2} \times sin90^{\circ}}$

$\sf{\Longrightarrow \dfrac{\dfrac{22}{7} \times 400 \times 270^{\circ}}{360^{\circ}} + \dfrac{1}{2} \times 400 \times 1}$

$\sf{\Longrightarrow \dfrac{\dfrac{22}{7} \times 400 \times 3}{4} + 200}$

$\sf{\Longrightarrow \dfrac{22}{7} \times 100 \times 3 + 200}$

$\sf{\Longrightarrow 3.14 \times 100 \times 3 + 200}$

$\sf{\Longrightarrow 942 + 200}$

$\sf{\Longrightarrow 1142\ cm^{2}}$

$\sf{Hence,\ Area\ of\ the\ major\ segment\ is\ 1142\ cm^{2}.}$

5. $\downarrow$

Given:-

• A Square is inscribed in a circle.

To Find:-

• The ratio of the area of the circle and the square.

Solution:-

Here, Let the side of the square be of $\sf{x\ units.}$

$\sf{So,\ Diagonal\ of\ the\ square\ =\ \sqrt{2}x}$

$\sf{We\ Know\ that,\ Radius\ of\ the\ circle\ =\ \dfrac{Diagonal}{2}}$

                                                       $\sf{= \dfrac{\sqrt{2}x}{2}}$

                                                       $\sf{= \dfrac{x}{\sqrt{2}}}$

Now, Area of square = $\sf{x^{2}}$

And, Area of circle = $\sf{\pi r^{2}}$

                               $\sf{= \pi \bigg(\dfrac{x}{\sqrt{2}}\bigg)}$

$\sf{Now,\ Ratio\ =\ \dfrac{Area\ of\ circle}{Area\ of\ square}}$

                   $\sf{= \dfrac{\pi \bigg(\dfrac{x}{\sqrt{2}}\bigg)}{x^{2}}}$

                   $\sf{= \dfrac{\pi \times x^{2}}{x^{2} \times 2}}$

                   $\sf{= \dfrac{\pi}{2}}$

$\sf{Hence,\ Ratio\ of\ the\ area\ of\ the\ circle\ and\ the\ square\ is\ \pi : 2.}$

6. $\downarrow$

Given:-

• Radius of the circle is 4 cm.
• Angle at the centre is of 30°.

To Find:-

• Area of the minor sector ?
• Area of major sector ?

Solution:-

Here, To find area of the sector we use certain formula,

$\sf{Area\ of\ the\ sector\ =\ \dfrac{\theta}{360^{\circ}} \times \pi r^{2}}$

Now, Putting all the values we get,

$\sf{\Longrightarrow \dfrac{30^{\circ}}{360^{\circ}} \times \dfrac{22}{7} \times 4^{2}}$

$\sf{\Longrightarrow \dfrac{1}{12} \times \dfrac{22}{7} \times 16}$

$\sf{\Longrightarrow \dfrac{88}{21}}$

$\sf{\Longrightarrow 4\dfrac{4}{21}\ cm^{2}}$    $\sf{\Longrightarrow 4.19\ cm^{2}}$

$\sf{Hence,\ Area\ of\ the\ minor\ sector\ is\ \sf{4\dfrac{4}{21}\ cm^{2}}\ or\ \sf{4.19\ cm^{2}}.}$

Now, Angle of major segment = 360° - 30°

                                                  ⇒ 330°

$\sf{Area\ of\ the\ sector\ =\ \dfrac{\theta}{360^{\circ}} \times \pi r^{2}}$

Putting all the values in the above formula we get,

$\sf{\Longrightarrow \dfrac{330^{\circ}}{360^{\circ}} \times \dfrac{22}{7} \times 4^{2}}$

$\sf{\Longrightarrow \dfrac{11}{12} \times \dfrac{22}{7} \times 16}$

$\sf{\Longrightarrow \dfrac{11}{3} \times \dfrac{22}{7} \times 4}$

$\sf{\Longrightarrow \dfrac{968}{21}}$

$\sf{\Longrightarrow 46\dfrac{2}{21}\ cm^{2}}$    $\sf{\Longrightarrow 46.09\ cm^{2}}$

$\sf{Hence,\ Area\ of\ the\ major\ sector\ is\ \sf{46\dfrac{2}{21}\ cm^{2}}\ or\ \sf{46.09\ cm^{2}}.}$