Question

The base of a right triangle is 8 cm and hypotenuse is 10 cm. Then its area is _________.

Answer 1

Given:
The base of a right triangle is 8 cm
hypotenuse is 10 cm


$$\begin{gathered} \mathrm{Using}\mathrm{pythagoras}\mathrm{theorem}, \\ {\left(\mathrm{Hypotenuse}\right)}^2={\left(\mathrm{Base}\right)}^2+{\left(\mathrm{Height}\right)}^2 \\ \Rightarrow {10}^2=8^2+{\left(\mathrm{Height}\right)}^2 \\ \Rightarrow 100=64+{\left(\mathrm{Height}\right)}^2 \\ \Rightarrow {\left(\mathrm{Height}\right)}^2=100-64 \\ \Rightarrow {\left(\mathrm{Height}\right)}^2=36 \\ \Rightarrow \mathrm{Height}=6 \end{gathered}$$


$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}=\frac{1}{2}\times \left(\mathrm{base}\right)\times \left(\mathrm{height}\right) \\ =\frac{1}{2}\times 8\times 6 \\ =4\times 6 \\ =24{\mathrm{cm}}^2 \end{gathered}$$


Hence, the area is $\underline{24{\mathrm{cm}}^2}$.