Question:
In \( \triangle ABC \) and \( \triangle PQR \), which are similar triangles, \( AD \) is perpendicular to \( BC \) and \( PT \) is perpendicular to \( QR \). If \( AD = 9 \) cm and \( PT = 7 \) cm, then the ratio of the areas of triangles \( ABC \) and \( PQR \) is:
In \( \triangle ABC \) and \( \triangle PQR \), which are similar triangles, \( AD \) is perpendicular to \( BC \) and \( PT \) is perpendicular to \( QR \). If \( AD = 9 \) cm and \( PT = 7 \) cm, then the ratio of the areas of triangles \( ABC \) and \( PQR \) is:
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For similar triangles, the ratio of their areas is:
\[
\left(\frac{\text{Height}_1}{\text{Height}_2}\right)^2
\]
Updated On: Oct 27, 2025
- \( 9:7 \)
- \( 7:9 \)
- \( 16:25 \)
- \( 81:49 \)
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The Correct Option is D
Solution and Explanation
The ratio of the areas of similar triangles is the **square** of the ratio of their corresponding heights.
\[ \text{Area Ratio} = \left(\frac{\text{Height}_1}{\text{Height}_2}\right)^2 \] Substituting the values:
\[ \left(\frac{9}{7}\right)^2 = \frac{81}{49} \]
\[ \text{Area Ratio} = \left(\frac{\text{Height}_1}{\text{Height}_2}\right)^2 \] Substituting the values:
\[ \left(\frac{9}{7}\right)^2 = \frac{81}{49} \]
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