NCERT Examples

NCERT Example

Example 1

area of sector · major sector

Find the area of the sector of a circle with radius 4 cm and angle 30°. Also find the area of the corresponding major sector. Use π = 3.14.

Board-exam working:

Given: radius r = 4 cm and central angle θ = 30°.

Area of the minor sector:

A_{minor}=\frac{\theta}{360^\circ}\times\pi r^2

=\frac{30^\circ}{360^\circ}\times3.14\times4^2

=\frac{1}{12}\times3.14\times16

=\frac{12.56}{3}=4.186\ldots\text{ cm}^2

A_{minor}\approx4.19\text{ cm}^2

Area of the whole circle:

\pi r^2=3.14\times4^2=50.24\text{ cm}^2

Therefore, area of the corresponding major sector:

A_{major}=50.24-4.19=46.05\text{ cm}^2

A_{major}\approx46.1\text{ cm}^2

A_{minor}\approx4.19\text{ cm}^2,\qquad A_{major}\approx46.1\text{ cm}^2

NCERT Example

area of segment · sector and triangle

Find the area of segment AYB shown in Figure 11.6 if the radius of the circle is 21 cm and ∠AOB = 120°. Use π = 22/7.

Board-exam working:

Area of segment AYB = area of sector OAYB − area of triangle OAB.

First, find the area of sector OAYB:

A_{sector}=\frac{120^\circ}{360^\circ}\times\frac{22}{7}\times21^2

=\frac13\times\frac{22}{7}\times21\times21=462\text{ cm}^2

Draw OM perpendicular to AB. Since OA = OB, OM bisects both AB and ∠AOB.

\angle AOM=\angle BOM=\frac{120^\circ}{2}=60^\circ

In right triangle OMA:

\frac{OM}{OA}=\cos60^\circ=\frac12

OM=21\times\frac12=\frac{21}{2}\text{ cm}

\frac{AM}{OA}=\sin60^\circ=\frac{\sqrt3}{2}

AM=21\times\frac{\sqrt3}{2}=\frac{21\sqrt3}{2}\text{ cm}

AB=2AM=21\sqrt3\text{ cm}

Now find the area of triangle OAB:

A_{\triangle OAB}=\frac12\times AB\times OM

=\frac12\times21\sqrt3\times\frac{21}{2}=\frac{441\sqrt3}{4}\text{ cm}^2

Therefore:

A_{segment}=462-\frac{441\sqrt3}{4}

=\frac{21}{4}(88-21\sqrt3)\text{ cm}^2

\boxed{\frac{21}{4}(88-21\sqrt3)\text{ cm}^2}