Question

The areas of two similar triangles are 81cm² and 49 cm² respectively. If the altitude of the bigger triangle is 4.5 cm. Find the corresponding altitude of the smaller triangle.

Answer 1

Hey there!

$\textbf{If two triangles are equal,}$

Then,

$\frac{ \textbf{Area of bigger triangle }} { \textbf{Area of smaller triangle} }= ( \frac{ \textbf{altitude of bigger \: triangle}}{ \textbf{altitude of smaller \: triangle}} )^{2}$


Given,
Area of larger triangle = 81cm²
Area of smaller triangle = 49cm²

The altitude of bigger triangle = 4.5cm .
Let the altitude of smaller triangle be x

$\frac{81}{49} = (\frac{4.5}{x})^{2} \\ \\ \frac{9}{7} = \frac{4 . 5}{x} \\ \\ 9x = 4.5 \times 7 \\ \\ x = \frac{4 .5 \times 7}{9} \\ \\ x = 0.5 \times 7 = 3.5$

Therefore, The altitude of smaller triangle is 3 5 cm.

Answer 2

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