Question

In the figure , P is the centre of the sector. m ∠APB = m ∠DPE =60° . PA = 7 cm , PD = 21 cm .Find the area of shaded portion​

Answer 1

Area of shaded portion is 205.34 cm².

Step-by-step explanation:

Given:-

• P is center of sector.
• ∠APB = ∠DPE = 60°.
• AP = 7 cm.
• DP = 21 cm.

To find:-

• Area of shaded portion.

Solution:-

Let,

Area of sector PED be A₁

Area of sector PBA be A₂

We know that,

Area of sector = θ/360° × πr²

___________________________________

For A₁

Angle of sector is 60°.

Radius of circle is 21 cm.

$\longrightarrow \sf \dfrac{60}{360} \times \dfrac{22}{7} \times 21 \times 21$

$\longrightarrow \sf \cfrac{27720}{120}$

$\longrightarrow \sf 231$

Area of A₁ is 231 cm².

___________________________________

For A₂

Angle of sector is 60°.

Radius of circle is 7 cm.

$\longrightarrow \sf \cfrac{60}{360} \times \cfrac{22}{ \cancel{7}} \times \cancel{7} \times 7$

$\longrightarrow \sf \cfrac{9240}{360}$

$\longrightarrow \sf 25.66$

Area of A₂ is 25.666 cm².

___________________________________

Area of shaded portion = A₁ - A₂

= 231 - 25.66

= 205.34 cm²

Therefore,

Area of shaded portion is 205.34 cm².