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 cm^2 2. 49 cm^2 3. 80 cm^2 4. 98 cm^2

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. From figure, Area of QPBC = Area of △ ABC - Area of △ APQ = 98 - 18 = 80 cm2. Hence, Option 3 is the correct option.

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Mathematics

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 :
32 cm²
49 cm²
80 cm²
98 cm²

Circles

ICSE Sp 2025

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Mathematics

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 :
32 cm²
49 cm²
80 cm²
98 cm²

Circles

ICSE Sp 2025

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Mathematics

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 :32 cm249 cm280 cm298 cm2

Circles

ICSE Sp 2025

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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 :
32 cm²
49 cm²
80 cm²
98 cm²

A polynomial in 'x' is divided by (x - a) and for (x - a) to be a factor of this polynomial, the remainder should be :
-a
0
a
2a

Radha deposited ₹ 400 per month in a recurring deposit account for 18 months. The qualifying sum of money for the calculation of interest is :
₹ 3600
₹ 7200
₹ 68,400
₹ 1,36,800

In the adjoining diagram, O is the center of the circle and PT is a tangent. The value of x is :
20°
40°
55°
70°