Question

Find the area of a triangle, two sides of which are 8 cm and 11 cm and the perimeter is 32

cm.​

Answer 1

☘️ Solution :

Here we have perimeter of triangle = 32 cm, a = 8 cm and b = 11 cm.

Third side c = 32 cm - (8 + 11)cm = 13 cm

So, 2s = 32, i.e., s = 16 cm,

s - a = (16 - 8) cm = 8 cm,

s - b = (16 - 11) cm = 5 cm,

s - c = (16 - 13) cm = 3 cm.

Therefore, area of the triangle

$\sf \: = \sqrt{s(s - a) \: (s - a) \: (s - c) } \\ \\ \sf \: = \sqrt{16 \times 8 \times 5 \times 3} \: cm {}^{2} \\ \\ \sf \: = \large \purple{ 8 \sqrt{30} \: cm {}^{2} }$

☘️ Answer :

• 8√30 cm²

Answer 2

$\huge\mathfrak\purple{Answer} \:$

$8 \sqrt{30}$

Solution:-

s(semi-perimeter) and a,b and c are sides of a triangle

using Heron's formula,

$\sqrt{s(s - a)(s - b)(s - c) } \\ a = 8cm \\ b = 11cm \\ c = 32 - (8 + 11) \sqrt{(16(8)(5)(3)} \\ c = 32 - 19 \\ c = 13cm \\ s = 8 + 11 + \frac{13}{2} = 16 \\ \sqrt{16(16 - 8)(16 - 11)(16 - 13)} \\ \sqrt{(16(8)(5)(3)} \\ \sqrt{2* \: 2*2*2*2*2*2*5* 3} \\ 2* 2* 2* \sqrt{30} \\ \\ = 8 \sqrt{30 \:} (is \: \: the \: \: correct \: \: answer \: )$