Question

Find the altitudes of a trapezium whose area is $65c{m}^2$ and whose bases are $13cm$ and $26cm$.

Answer 1

Given area of traperium $=65c{m}^2$

bases of trapezium are $13cm$ and $26cm$

we know

Area of trapezium $=\frac{1}{2}\times h\times$ (sum of bases)

Where h is height of trapezium

According to question

$65=\frac{1}{2}\times h\times (13+26)$

$h=\frac{130}{39}cm$

$h=\frac{10}{3}cm$

Hence altitude of this trapezium is $\frac{10}{3}cm$