If the area of two similar triangles ABC and PQR are 49 $c m^{2}$ and 36 $c m^{2}$ respectively and altitude of the first triangle is 2.1 cm then altitude of the second triangle is

2.95 cm

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2.7 cm

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2.4 cm

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1.8 cm

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Solution

The correct option is D 1.8 cm
In similar triangles, $\sqrt{\frac{a r e a o f Δ A B C}{a r e a o f Δ P Q R}} = \frac{a l t i t u d e A D}{a l t i t u d e P S}$
$⟹ \frac{7}{6} = \frac{2.1}{a l t i t u d e P S} ⟹ a l t i t u d e P S = 1.8 c m$

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