Question

Class X Ex 12.2
7. a chord of a circle of radius 15cm subtends an angle of 60° at the centre.find the area of the corresponding minor and major segments of the circle
(π = 3.14 and √3 = 1.73)​

Answer 1

$\huge{ \tt {\underline{ \underline{ \green{Solution :}}}}}$

→ In the mentioned figure,

→ O is the centre of circle,

→ AB is a chord

→ AXB is a major arc,

→ OA=OB= radius = 15 cm

→ Arc AXB subtends an angle 60° at O.

$\tt{→ Area \: of \: sector \: AOB = \frac{360}{60}×π× {r}^{2} }$

$\tt{→ \frac{60}{360} \times 3.14 \times {(15)}^{2}}$

→ Area of minor segment (Area of Shaded region) = Area of sector AOB − Area of △ AOB

→ By trigonometry,

→ AC = 15 sin30

→ OC = 15 cos30

→ And,

→ AB = 2AC

$\tt{→ AB=2×15 \: sin \: 30=15 cm}$

$\tt{→ OC=15cos30}$

$\tt{→15 \frac{ \sqrt{3}}{2}} = 15 \times \frac{1.73}{2} = 12.975cm$

$\tt{→ Area \: of \: △ \: AOB=0.5×15×12.975</p><p> }$

$\tt{→ {97.3125cm}^{2}}$

→ Area of minor segment (Area of Shaded region)

$\tt{=117.75−97.3125={20.4375 cm}^{2} }$

→ Area of major segment = Area of circle − Area of minor segment

$\tt{=(3.14×15×15) \: − \: 20.4375</p><p></p><p> }$

$\tt{={686.0625cm}^{2}}$

Answer 2

Answer:

In the mentioned figure,

O is the centre of circle,

AB is a chord

AXB is a major arc,

OA=OB= radius = 15 cm

Arc AXB subtends an angle 60 °

at O.

Area of sector AOB=

=60/360 ×π×r ²

= 60/360 ×3.14×(15) ²

=117.75cm ²

Area of minor segment (Area of Shaded region) = Area of sector AOB− Area of △ AOB

By trigonometry,

AC=15sin30

OC=15cos30

And, AB=2AC

∴ AB=2×15sin30=15 cm

∴ OC=15cos30=15'2

=15× 2

1.73

=12.975 cm

∴ Area of △AOB=0.5×15×12.975=97.3125cm ²

∴ Area of minor segment (Area of Shaded region) =117.75−97.3125=20.4375 cm ²

Area of major segment = Area of circle − Area of minor segment

=(3.14×15×15)−20.4375

=686.0625cm²

@ananya