Question

Two sides of a triangle are 10 cm and 20 cm . The length of the altitude to the longer side is 5 cm . Find the length of the altitude to the shorter side . Also find the area of the triangle .

Answer 1

area $△ABC=\frac{1}{2}\times BC\times AD=\frac{1}{2}\times 20\times 5=50{cm}^2$

Again, area $△ABC=\frac{1}{2}\times AB\times CE$

$50=\frac{1}{2}\times 10\times h$

$\Rightarrow h=\frac{100}{10}=10cm$

$\therefore$ Length of altitude on shorter side $=10cm$

Area of $△=50{cm}^2$