Question

area of an isosceles triangle is
$8 \sqrt{15 \: cm {}^{2} }$
if the base is 8 cm ,find the length of each of its equal sides​

Answer 1

in the given triangle base and the area is given

$base = 8cm \\ area = 8 \sqrt{15} {cm}^{2} \\ we \: know \\ = > area = \frac{1}{2} \times b \times h \\ = > 8 \sqrt{15} = \frac{1}{2} \times 8 \times h \\ = > h = 2 \sqrt{15}$

we know that the half of the base

height and the side of triangle makes an equilateral triangle

hence using phythogorus thereom we get

$= > l {}^{2} = 4 {}^{2} + {(2 \sqrt{15} )}^{2} \\ = > {l}^{2} = 16 + 4 \times 15 \\ = > l = \sqrt{16 + 60} = \sqrt{76} = 2 \sqrt{19}$

side is equal to

$2 \sqrt{19}$