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Mathematics · Areas Related to Circles

NCERT Examples

All 2 worked examples from Areas Related to Circles, with independently verified solutions and direct navigation.

Mathematics · Chapter 11 · NCERT Examples

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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.

Circle with centre O, radius 4 centimetres and a 30 degree minor sector OAPB.
Figure 11-5 from the supplied NCERT chapter.

Solution

Board-exam working:

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

Area of the minor sector:

Aminor=θ360∘×πr2A_{minor}=\frac{\theta}{360^\circ}\times\pi r^2
=30∘360∘×3.14×42=\frac{30^\circ}{360^\circ}\times3.14\times4^2
=112×3.14×16=\frac{1}{12}\times3.14\times16
=12.563=4.186… cm2=\frac{12.56}{3}=4.186\ldots\text{ cm}^2
Aminor≈4.19 cm2A_{minor}\approx4.19\text{ cm}^2

Area of the whole circle:

πr2=3.14×42=50.24 cm2\pi r^2=3.14\times4^2=50.24\text{ cm}^2

Therefore, area of the corresponding major sector:

Amajor=50.24−4.19=46.05 cm2A_{major}=50.24-4.19=46.05\text{ cm}^2
Amajor≈46.1 cm2A_{major}\approx46.1\text{ cm}^2

Final answer

Aminor≈4.19 cm2,Amajor≈46.1 cm2A_{minor}\approx4.19\text{ cm}^2,\qquad A_{major}\approx46.1\text{ cm}^2

NCERT Example

Example 2

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.

Circle of radius 21 centimetres with a 120 degree central angle and shaded segment AYB.
Figure 11-6 from the supplied NCERT chapter.
Triangle AOB split into two right triangles by perpendicular OM, with OA and OB each 21 centimetres.
Figure 11-7 from the supplied NCERT chapter.

Solution

Board-exam working:

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

First, find the area of sector OAYB:

Asector=120∘360∘×227×212A_{sector}=\frac{120^\circ}{360^\circ}\times\frac{22}{7}\times21^2
=13×227×21×21=462 cm2=\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.

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

In right triangle OMA:

OMOA=cos⁡60∘=12\frac{OM}{OA}=\cos60^\circ=\frac12
OM=21×12=212 cmOM=21\times\frac12=\frac{21}{2}\text{ cm}
AMOA=sin⁡60∘=32\frac{AM}{OA}=\sin60^\circ=\frac{\sqrt3}{2}
AM=21×32=2132 cmAM=21\times\frac{\sqrt3}{2}=\frac{21\sqrt3}{2}\text{ cm}
AB=2AM=213 cmAB=2AM=21\sqrt3\text{ cm}

Now find the area of triangle OAB:

A△OAB=12×AB×OMA_{\triangle OAB}=\frac12\times AB\times OM
=12×213×212=44134 cm2=\frac12\times21\sqrt3\times\frac{21}{2}=\frac{441\sqrt3}{4}\text{ cm}^2

Therefore:

Asegment=462−44134A_{segment}=462-\frac{441\sqrt3}{4}
=214(88−213) cm2=\frac{21}{4}(88-21\sqrt3)\text{ cm}^2

Final answer

214(88−213) cm2\boxed{\frac{21}{4}(88-21\sqrt3)\text{ cm}^2}