Question

In triangle \(\mathrm{ODQ}, \angle \mathrm{Q}=\angle \mathrm{BPO}=90^{\circ}\)
\(\mathrm{AB}=2 \times \mathrm{OA}, \mathrm{BC}=3 \times \mathrm{OA}\) and \(\mathrm{CD}=4 \times \mathrm{OA}\).

Assertion (A):
\[\frac{\triangle \mathrm{OBP}}{\text { Trapezium BDQP }}=\frac{9}{100-9}\].

Reason (R): \[\triangle \mathrm{OBP} \sim \triangle \mathrm{ODQ}\]
and \[\frac{\triangle \mathrm{OBP}}{\triangle \mathrm{ODQ}}=\frac{(3 a)^{2}}{(10 \mathrm{a})^{2}}\]

पर्याय

• A is true, R is false.

• A is false, R is true.

• Both A and R are true and R is the correct reason for A.

• Both A and R are true and R is the incorrect reason for A .

Answer 1

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

Explanation:

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.

∴ \[\frac{\text{Area of } \triangle ODQ}{\text{Area of } \triangle OBP} = \frac{OD^2}{OB^2}\]

\[{} = \frac{(10a)^2}{(3a)^2}\]

\[{} = \frac{100a^2}{9a^2}\]

\[{} = \frac{100}{9}\]

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,

⇒ \[\frac{\triangle OBP}{\text{Trapezium } BDQP} = \frac{9m}{100m - 9m}\]

\[{} = \frac{9m}{(100 - 9)m}\]

\[{} = \frac{9}{100 - 9}\]

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