Question

In the figure given below, AC is a diameter of the circle.

AP = 3 cm and PB = 4 cm and QP ⊥ AB. 

If the area of ΔAPQ is 18 cm², then the area of shaded portion QPBC is:

Options

• 32 cm²

• 49 cm²

• 80 cm²

• 98 cm²

Answer 1

80 cm² 

Explanation:

We know that, angle in a semicircle is a right angle.

∴ ∠ABC = 90°

In △APQ and △ABC,

⇒ ∠APQ = ∠ABC   ...(Both equal to 90°)

⇒ ∠PAQ = ∠BAC   ...(Common angles)

∴ △APQ ~ △ABC   ...(By A.A. axiom)

We know that the ratio of the area of similar triangles is equal to the ratio of the square of the corresponding sides.

∴ `("Area of" △APQ)/("Area of" △ABC) = (AP^2)/(AB^2)`

⇒ `18/("Area of" △ABC) = (AP^2)/(AB^2)`

⇒ Area of △ABC = `(AB^2)/(AP^2) xx 18`

⇒ Area of △ABC = `(3 + 4)^2/3^2 xx 18`

⇒ Area of △ABC = `7^2/3^2 xx 18`

⇒ Area of △ABC = `49/9 xx 18`

⇒ Area of △ABC = 98 cm²

From figure,

Area of QPBC = Area of △ABC – Area of △APQ 

= 98 – 18

= 80 cm²