Question

The areas of two similar triangles are 16cm² and 25cm² respectively.If the different of their corresponding altitudes is 10 cm find the lengths of altitudes

Answer 1

$\frac{area \: of \: first \: triangle}{area \: of \: second \: triangle} = { (\frac{altitude \: 1}{altitude \: 2}) }^{2}$
Thus,
Let one altitude be x and other be y
x-y=10
y=x-10
$\frac{4}{5} = \frac{y}{x} \\ \frac{4}{5} = \frac{x - 10}{x} \\ 4x = 5x - 50 \\ x = 50 cm \\thus \: y = 40cm$