Question

(i) \( \triangle ABC \sim \triangle PQR , A (\triangle ABC )=81 cm ^{2} \). If \( AB =6, PQ =12 cm \), then find \( A (\triangle PQR ) \).

Answer 1

shreyash1144, hello

Nice fun problem.

Say AB is the base of triangle ABC.  Then one-half the base is 3. If the area is 81cm², then the height of triangle ABC must be 27cm (3x27=81).

Since triangle PQR its similar to triangle ABC, let the base be PQ=12cm, and the height must be 54cm (twice the height of triangle ABC).  The area of triangle PQR is then:

1/2 x 12cm x 54cm = 324cm²

Answer 2

Given:

\[ \triangle ABC \sim \triangle PQR \]

\[ \text{Ar}(\triangle ABC) = 81 \, \text{cm}^2 \]

\[ AB = 6 \, \text{cm}, \quad PQ = 12 \, \text{cm} \]

To Find:

\[ \text{Ar}(\triangle PQR) \]

Solution:

The required area is \( 324 \, \text{cm}^2 \).

We can obtain the area using the formula:

\[ \frac{\text{Ar}(\triangle ABC)}{\text{Ar}(\triangle PQR)} = \left( \frac{AB}{PQ} \right)^2 \]

Substituting the given values:

\[ \frac{81}{\text{Ar}(\triangle PQR)} = \frac{6^2}{12^2} \]

\[ \frac{81}{\text{Ar}(\triangle PQR)} = \frac{1}{2^2} \]

\[ 81 \times 4 = \text{Ar}(\triangle PQR) \]

\[ \text{Ar}(\triangle PQR) = 324 \]

Therefore, the required area is \( 324 \, \text{cm}^2 \).