Question

find the are of an isosceles triangle each of whose equal sides measures 13 cm and whose base measures 20cm.

Answer 1

$\textbf {Area of triangle = }\sqrt{p(p-a)(p-b)9p-c)}$

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STEP 1: Find p:

$\text{p} = \dfrac{13 + 13 + 20}{2} = \dfrac{46}{2} = 23$

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STEP 2: Find area:

$\text {Area of triangle = } \sqrt{23(23-13)(23-13)(23-20)}$

$\text {Area of triangle = } \sqrt{6900}$

$\text {Area of triangle = } 83 \text { units}^2$

Answer 2

hey user!!!

given the equal side of the isosceles triangle is 13cm and base is 20cm.

semi perimeter of the triangle = (13 + 13 + 20)/2
= 46/2 = 23cm

let the sides of the triangle be :-

a = 13cm, b = 13cm and c = 20cm
$area \: of \: the \: triangle \: by \: herons \: \\ formula = \sqrt{s(s - a)(s - b)(s - c)} \\ = \sqrt{23(23 - 13)(23 - 13)(23 - 20)} \\ = \sqrt{23 \times 10 \times 10 \times 3} \\ = \sqrt{6900} \\ = 10 \sqrt{69} {cm}^{2} \\ = 83.06 {cm}^{2} (approximately)$

u can also solve it by Pythagoras theorem..

for that plz refer to the attachment..

cheers!!!