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In triangle ODQ, ∠Q = ∠BPO = 90° AB = 2 x OA, BC = 3 x OA and CD = 4 x OA.

In triangle ODQ, ∠Q = ∠BPO = 90° AB = 2 x OA, BC = 3 x OA and CD = 4 x OA. Concise Mathematics Solutions ICSE Class 10.

Assertion (A) : Δ OBPTrapezium BDQP=9100−9\dfrac{\text{Δ OBP}}{\text{Trapezium BDQP}} = \dfrac{9}{100 - 9}.

Reason (R) : Δ OBP - ODQ and Δ OBPΔ ODQ=(3a)2(10a)2\dfrac{\text{Δ OBP}}{\text{Δ ODQ}} = \dfrac{(3a)^2}{(10a)^2}.

  1. A is true, R is false.

  2. A is false, R is true.

  3. Both A and R are true and R is correct reason for A.

  4. Both A and R are true and R is incorrect reason for A.

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Answer

In Δ ODQ and Δ OBP

⇒ ∠DOQ = ∠BOP (Common angle)

⇒ ∠OQD = ∠OPB (Both equal to 90°)

∴ Δ ODQ ∼ Δ OBP (By A.A. postulate)

From figure,

OB = a + 2a = 3a

OD = a + 2a + 3a + 4a = 10a

We know that,

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

∴Area of Δ ODQArea of Δ OBP=OD2OB2=(10a)2(3a)2=100a29a2=1009.\therefore \dfrac{\text{Area of Δ ODQ}}{\text{Area of Δ OBP}} = \dfrac{OD^2}{OB^2}\\[1em] = \dfrac{(10a)^2}{(3a)^2}\\[1em] = \dfrac{100a^2}{9a^2}\\[1em] = \dfrac{100}{9}.

Let the area of Δ OBP be 9m and that of area of Δ ODQ be 100m.

From figure,

Area of trapezium BPQD = Area of Δ ODQ - Area of Δ OBP = 100m - 9m

Now,

⇒Δ OBPTrapezium BDQP=9m100m−9m=9m(100−9)m=9100−9\Rightarrow \dfrac{\text{Δ OBP}}{\text{Trapezium BDQP}} = \dfrac{9m}{100m - 9m}\\[1em] = \dfrac{9m}{(100 - 9)m}\\[1em] = \dfrac{9}{100 - 9}

∴ Both A and R are true and R is correct reason for A.

Hence, option 3 is the correct option.

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