Question

Find the area of triangle two sides of which are 16cm and 22cm and perimeter is 64cm

Answer 1

semi perimeter =perimeter/2
=64cm/2
32cm
third side = 64cm -(16cm+22cm)
=64cm-38cm
=26cm
using heron formula
area of triangle √(s*(s-a)(s-b)(s-c)
=√(32*16*10*6)
=32√30
=32*5.47
=175.2cm²

Answer 2

$\huge\textsf{Answer:}$

Let a = 16cm, b = 22cm and c = c cm be the third side of ∆ABC, then

Perimeter of∆ABC = a + b + c

64 = 16 + 22 + c

=> c = 64 - 16 - 22

=> c = 64 - 38

=> c = 26cm.

$\therefore$ Area of ∆ABC = $\sf\sqrt{s(s-a)(s-b)(s-c)}$

Where a, b and c are the sides of the triangle

s = Semi-perimeter, i.e., half of the perimeter of the triangle

$\implies$$\huge\sf\frac{64}{2}$

= 32cm.

• s - a = 32 - 16 = 16cm
• s - b = 32 - 22 = 10cm
• s - c = 32 - 26 = 6cm.

$\therefore$ Area of ∆ABC = $\sf\sqrt{s(s-a)(s-b)(s-c)}$

$\implies$$\sf\sqrt{32×16×10×6}$$\sf{cm}^{2}$

$\implies$$\sf\sqrt{2×16×16×2</p><p>×5×2×3}$$\sf{cm}^{2}$

$\implies$2×16$\sf\sqrt{2×3×5}$$\sf{cm}^{2}$

$\implies$$\bf{32}$$\bf\sqrt{30}$$\sf{cm}^{2}$