Question

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

Answer 1

Perimeter = 32cm
1st side = 8cm
2nd side =11cm
3rd side = 32 - (8 +11)
=32 - 19
= 13 cm
Area of the triangle
$= \sqrt{s\times (s - a)(s - b)(s - c)}$
where s is the semi perimeter
ie , 32 ÷ 2 = 16cm
a is the first side
b is the second side
c is the third side
So,
$\sqrt{16 \times (16 - 8) \times (16 -11 ) \times (16 - 13)} \\ = \sqrt{16 \times 8 \times 5 \times 3} \\ = \sqrt{4 \times 4 \times 2 \times 2 \times 2 \times 5 \times 3} \\ = 4 \times 2 \sqrt{2 \times 3 \times 5} \\ = 8 \sqrt{30} \: {cm}^{2}$
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Answer 2

Answer:

8√30 cm^2

Step-by-step explanation:

Here we have perimeter of the triangle = 32cm , a =

8cm and b = 11cm.

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

so,

2s = 32, I. e.,s = 16cm,

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

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

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

Therefore,

area of the triangle = √s(s-a)(s-b)(s-c)

= √16×8×5×3 cm^2

= 8√30 cm^2.