Question

The area of an isosceles triangle is 8 root 15 cm2

. If the base is 8 cm, find the length of each of its equal sides.​

Answer 1

Step-by-step explanation:

Area of the triangle = 8√15 cm² ; base = 8cm

Area of the triangle = 1/2 * base * height

8√15 = 1/2 * 8 * height

8√15 = 4 * height

Height = 8√15 / 4

Height = 2√15 cm

In triangle ABC, AB =AC ; AD = DC = 8/2 = 4cm

Triangle ADC is a right angled triangle.

Using Pythagoras Theorem,

AC² = AD² + DC²

AC² = (2√15)² + 4²

AC² = (4 * 15) + 16

AC² = 60 + 16

AC² = 76

AC² = 4 * 19

AC = √4 * √19

AC = 2√19 cm

∴The length of equal sides is 2√19 cm

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Answer 2

${ \sf{ \huge{ \underline{solution}}}}$

${ \rm{ \large \bold{ {given \colon}}}}$

${ \rm{ \large{ {area \: of \: isosceles \: triangle = 8 \sqrt{15} {cm}^{2} }}}}$

${ \rm{ \large{ {base =8cm }}}}$

${ \rm{ \large \bold{ {area \: of \: it}}}} = {{{{\dfrac{1}{2} \times b \times h}}}}$

${ \rm{ \large { { \implies8 \sqrt{15} = \dfrac{1}{2} \times 8 \times h}}}}$

${ \rm{ \large { { \implies8 \sqrt{15} = \dfrac{1}{ \not2} \times { \not8}^{4} \times h}}}}$

${ \rm{ \large { { \implies h = \dfrac{8 \sqrt{15} }{4} }}}}$

${ \rm{ \large { { \implies h = \dfrac{ { \not8}^{2} \sqrt{15} }{ \not4} }}}}$

${ \rm{ \large { { \implies h = 2 \sqrt{15} }}}}$

${ \rm{ \large { { \therefore \: the \: value \: of \: h = 2 \sqrt{15} }}}}$