Question

a right angled triangle has area 6 cm^2 and perimeter is 12cm. find its hypotenuse

Answer 1

Dear student
$$\begin{gathered} \mathrm{Given}:\mathrm{area}\mathrm{of}\mathrm{right}△=6{\mathrm{cm}}^2 \\ \frac{1}{2}\times \mathrm{Base}\left(\mathrm{B}\right)\times \mathrm{Perpendicular}\left(\mathrm{P}\right)=6c{m}^2 \\ \Rightarrow \mathrm{BP}=12...\left(1\right) \\ \mathrm{Given}\mathrm{perimeter}=12 \\ \mathrm{So},\mathrm{Perpendicular}\left(\mathrm{P}\right)+\mathrm{Base}\left(\mathrm{B}\right)+\mathrm{Hypotenuse}\left(\mathrm{H}\right)=12\mathrm{cm}. \\ \Rightarrow \mathrm{P}+\mathrm{B}=12-\mathrm{H} \\ \mathrm{Squaring}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ {\left(\mathrm{P}+\mathrm{B}\right)}^2={\left(12-\mathrm{H}\right)}^2 \\ {\mathrm{P}}^2+{\mathrm{B}}^2+2\mathrm{BP}=144+{\mathrm{H}}^2-24\mathrm{H} \\ \mathrm{Also}\mathrm{by}\mathrm{pythagoras}\mathrm{theorem}:{\mathrm{H}}^2={\mathrm{P}}^2+{\mathrm{B}}^2 \\ {\mathrm{H}}^2+2\mathrm{BP}=144+{\mathrm{H}}^2-24\mathrm{H} \\ 24=144-24\mathrm{H}\left[\mathrm{using}\left(1\right)\right] \\ 24\mathrm{H}=120 \\ \mathrm{H}=5\mathrm{cm} \\ \mathrm{So},\mathrm{hypotenuse}=5\mathrm{cm} \end{gathered}$$
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