Question

find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.​

Answer 1

Step-by-step explanation:

perimeter=42

18+10+x=42

28+x=42

x=42-28

x=14

area of triangle = 1/2* base*hight

=1/2*10*14

=70

Answer 2

To Find

• The area of triangle

Given

• The perimeter is 42 cm
• One side of triangle is 18 cm
• Second side of triangle is 10 cm

Suppose the third side of triangle is 'x'

Now Three sides of triangles are

• 18 cm 10 cm and x cm

$\implies\sf \: Now \: \: perimeter = 18+10+x \\ \implies \tt \: 42cm = 18 + 10 + x \\ \implies \tt \:42cm = 28 + x \\ \tt \:x = 42 - 28cm \\ \tt \:x = = 21cm$

Semi perimeter of triangle = 42/2 = 21cm

• Using Heron's Formula..

$\boxed {\underline{\sqrt{s(s - a)(s - b)(s - c)} }}$

• The area of triangle is :

$\sf {{\sqrt{s(s - a)(s - b)(s - c)} }} \\ \\ \sf \: \sqrt{21(21 - 18)(21 - 10)(21 - 14} \: {cm}^{2} \\ \\ \sf \: \sqrt{21 \times 3 \times 11 \times 7} \: {cm}^{2} \\ \\ 21\sqrt{11} \: \sf {cm}^{2}$

$\red{\sf \: Hence \: the \: area \: of \: triangle \: is \: 21 \sqrt{11 \: {cm}^{2} } }$