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ICSE Class 10 Mathematics Question 1 of 21

Solved 2025 Specimen Paper ICSE Class 10 Mathematics — Question 3

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Question

Question 1(iii)

In the adjoining figure, AC is a diameter of the circle, AP = 3 cm and PB = 4 cm and QP ⊥ AB. If the area of △ APQ is 18 cm2, then the area of shaded portion QPBC is :

  1. 32 cm2

  2. 49 cm2

  3. 80 cm2

  4. 98 cm2

In the adjoining figure, AC is a diameter of the circle, AP = 3 cm and PB = 4 cm and QP ⊥ AB. If the area of △ APQ is 18 cm2, then the area of shaded portion QPBC is : ICSE 2025 Maths Solved Question Paper.
Answer

We know that,

Angle in a semi-circle 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 area of similar triangles is equal to the ratio of the square of the corresponding sides.

∴Area of △ APQArea of △ ABC=AP2AB2⇒18Area of △ ABC=AP2AB2⇒Area of △ ABC=AB2AP2×18⇒Area of △ ABC=(3+4)232×18⇒Area of △ ABC=7232×18⇒Area of △ ABC=499×18=98 cm2.\therefore \dfrac{\text{Area of △ APQ}}{\text{Area of △ ABC}} = \dfrac{AP^2}{AB^2} \\[1em] \Rightarrow \dfrac{18}{\text{Area of △ ABC}} = \dfrac{AP^2}{AB^2} \\[1em] \Rightarrow \text{Area of △ ABC} = \dfrac{AB^2}{AP^2} \times 18 \\[1em] \Rightarrow \text{Area of △ ABC} = \dfrac{(3 + 4)^2}{3^2} \times 18 \\[1em] \Rightarrow \text{Area of △ ABC} = \dfrac{7^2}{3^2} \times 18 \\[1em] \Rightarrow \text{Area of △ ABC} = \dfrac{49}{9} \times 18 = 98 \text{ cm}^2.

From figure,

Area of QPBC = Area of △ ABC - Area of △ APQ = 98 - 18 = 80 cm2.

Hence, Option 3 is the correct option.