Question

find the area of triangle whose sides are 25cm,24cm and 7cm

Answer 1

S=$\frac{a+b+c}{2}$= (25+24+7)/2= 28

A= $\sqrt{s(s-a)(s-b)(s-c)} =\sqrt{28(28-25)(28-24)(28-7)} =\sqrt{7056} =84 cm^{2}$

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Answer 2

The area of triangle is 84 square centimeter

Solution:

Given that,

We have to find the area of triangle

Given sides are 25 cm, 24 cm and 7 cm

The area of triangle when sides are given is as:

$A = \sqrt{s(s-a)(s-b)(s-c)}$

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

s is the half of the triangles perimeter

$s = \frac{a+b+c}{2}$

From given,

a = 25

b = 24

c = 7

Therefore,

$s = \frac{25+24+7}{2}\\\\s = \frac{56}{2}\\\\s = 28$

Therefore, area is given as:

$A = \sqrt{28 \times (28-25) \times (28-24) \times (28-7)}\\\\A = \sqrt{28 \times 3 \times 4 \times 21}\\\\A = \sqrt{7056}\\\\ \boxed{A=84}$

Thus area of triangle is 84 square centimeter

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