Question

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

Answer 1

Answer:

$\sf \huge \underline{solution}$

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

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

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

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

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

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

$\sf{ \therefore \: {area \: of \triangle}} \: = \sqrt{s(s - a)(s - b)(s - c) } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sqrt{46 \times 8 \times 5 \times 3} \sf{ \: \: \: \: {cm}^{2} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 8 \sqrt{30} \sf{ {cm}^{2} }$

Answer 2

Answer:

area of triangle = s(s﹣a)(s﹣b)(s﹣c)

= P=a+b+c

s =a+b+c/2

Solving forA

A=1

4﹣P4+4P3a+4P3b﹣4(Pa)2﹣12P2ab﹣4(Pb)2+8Pa2b+8Pab2=1

4·﹣324+4·323·8+4·323·11﹣4·(32·8)2﹣12·322·8·11﹣4·(32·11)2+8·32·82·11+8·32·8·112

≈43.8178cm²

area of triangle is 43.817 sq.cm