Question

A semicircle is divided into two sectors whose angles are in ratio 4:5 find the ratio of their areas

Answer 1

Answer

Ratio of angles = 4:5

Let, angles = 4x and 5x

Ratio of areas:

$( \frac{4x}{360} \times \pi {r}^{2} ) \div ( \frac{5x}{360} \times \pi {r}^{2} ) \\ \\ = > \frac{4x}{360} \times \pi {r}^{2} \times \frac{360}{5x} \times \frac{1}{\pi {r}^{2} } \\ \\ = > \frac{4}{5}$

=> 4:5

Answer 2

Hey Dear,

◆ Answer -

A1 : A2 = 4 : 5

● Explanation -

Let r be the radius of semicircle.

Let x be some common multiple such that small sector has angle θ1=4x and large sector has angle θ2=5x.

We know that semicircle measures 180°.

4x + 5x = 180°

9x = 180°

x = 20°

Measure of angles -

θ1 = 4x = 4×20 = 80°

θ2 = 5x = 5×20 = 100°

Area of small sector -

A1 = πr^2.θ1/360°

A1 = πr^2 × 80/360

Area of large sector -

A2 = πr^2.θ2/360°

A2 = πr^2 × 100/360

Ratio of areas is -

A1/A2 = (πr^2×80/360) / (πr^2×100/360)

A1/A2 = 4/5

Therefore, ratio of areas of two sectors is 4 : 5.

Thanks dear...