Question

Find the area of a the triangle whose sides are 10 cm, 10 cm and 8 cm using Hero’s formula.​

Answer 1

Answer:

A SIDE =10 CM B= BASE =10 CM C = SIDE = 8CM

Step-by-step explanation:

A=s(s-a )(s-b) (s-c) s=a+b+c_2 A1_ 4_ a4 +2(ab)+2 (ac)2_b4 1_4 -10 4 +2 (10.10)2 + 2.(10) ans is = 36.66061

Answer 2

$\large\underline{\sf{Solution-}}$

Given that sides of a triangle are 10 cm, 10 cm and 8 cm respectively.

Let assume that sides of triangle be represented as

↝ a = 10 cm

↝ b = 10 cm

↝ c = 8 cm

We know,

$\underline{\boxed{\sf \: \: Semi \ perimeter, \: s \: = \: \dfrac{a + b + c}{2} \: \: }}$

So,

$\rm :\longmapsto\:s = \dfrac{10 + 10 + 8}{2}$

$\rm :\longmapsto\:s = \dfrac{28}{2}$

$\rm \implies\:\boxed{ \tt{ \: s \: = \: 14 \: cm \: \: }}$

We know, By Heron's Formula

$\underline{\boxed{\sf Area \ of \ triangle=\sqrt{s(s-a)(s-b)(s-c)} }}$

So, on substituting the values, we get

$\rm :\longmapsto\:Area \: of \: \triangle \: = \sqrt{14(14 - 10)(14 - 10)(14 - 8)}$

$\rm :\longmapsto\:Area \: of \: \triangle \: = \sqrt{14(4)(4)(6)}$

$\rm :\longmapsto\:Area \: of \: \triangle \: = \sqrt{7 \times 2 \times 4 \times 4 \times 2 \times 3}$

$\rm :\longmapsto\:Area \: of \: \triangle \: = 2 \times 4\sqrt{7 \times 3}$

$\rm :\longmapsto\:Area \: of \: \triangle \: = 8 \sqrt{21} \: {cm}^{2}$

Hence,

$\rm :\longmapsto\:\boxed{ \tt{ \: \: Area \: of \: \triangle \: = 8 \sqrt{21} \: {cm}^{2} \: \: }}$