Question

Find the length of the hypotenuse of an isosceles triangle with area 72cm2.
please help.... ​

Answer 1

       $\underline{\bold{Solution:}}$

       $\text{It is given that a right angled isosceles triangle has an area of 72 cm}^{2}$

       $\text{We know that, Area of triangle} = \dfrac12(\text{base})(\text{height)}$

       $\text{Since in a right angled isoscleles triangle base = height}$

$\implies \text{Area of the triangle} = \dfrac12(\text{base)}^{2}$

$\implies 72 = \dfrac12(\text{base})^{2}$

       $\text{Transposing 2 to L.H.S}$

$\implies 72(2) = (\text{base})^{2}$

       $\text{Simplifying...}$

$\implies 144 = (\text{base})^{2}$

       $\text{Rooting both sides}$

$\implies \sqrt{144} = \text{base}$

       $\text{Simplifying...}$

$\implies +12,-12 = \text{base}$

       $\text{Since, length cannot be negative}$

$\implies \text{12 cm = base = height}$

       $\text{Applying Pythagorus theorem}$

$\implies (\text{base})^{2} + (\text{height})^{2} = (\text{hypotenuse})^{2}$

       $\text{Substituting the values}$

$\implies (12)^{2} + (12)^{2} = (\text{hypotenuse})^{2}$

       $\text{Simplifying...}$

$\implies 144 + 144 = (\text{hypotenuse})^{2}$

       $\text{Simplifying...}$

$\implies 288 = (\text{hypotenuse})^{2}$

       $\text{Rooting both sides}$

$\implies \sqrt{288} = \text{hypotenuse}$

       $\text{Simplifying...}$

$\implies +12\sqrt2,-12\sqrt2 = \text{hypotenuse}$

       $\text{Since, length cannot be negative}$

$\implies \boxed{12\sqrt{2} \text{ cm}= \text{hypotenuse}}$