Question

the side of a triangle are 8 CM 11 cm and 13 cm what is its area ​

Answer 1

Let the sides of a triangle be a b and c

a = 8cm

b = 11cm

c= 13 cm

S=(a+b+c)/2=(8cm+11cm+13cm)/2=32cm/2=16cm

$Area = \sqrt{s(s - a)(s - b)(s - c)} \\ \: \: \: \: \: \: \: = \sqrt{16(16 - 8)(16 - 11)(16 - 13)} \\ \: \: \: \: \: \: \: = \sqrt{16 \times 8 \times 5 \times 3} \\ \: \: \: \: \: \: \: \: = \sqrt{2 \times 8 \times 8 \times 5 \times 3} \\ \: \: \: \: \: \: \: \: = 8 \sqrt{30}$

Therefore, Area of triangle =8√30 cm^2

Answer 2

Answer:

Area of triangle = $8\sqrt{30}$ $cm^{2}$

Step-by-step explanation:

Given:- Sides of the triangle are 8 cm, 11 cm, 13 cm.

To Find:- Area of the triangle

Solution:-

Let the sides of the triangle be a, b and c.

∴ a = 8cm, b = 11cm, c= 13 cm

Now,

⇒ S = $\frac{a +b+c }{2}$

       = $\frac{8+11+13}{2}$

       = $\frac{32}{2}$

       = 16

Now,

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

        = $\sqrt{16(16-8)(16-11)(16-13)}$

        = $\sqrt{16(8)(5)(3)}$

        = $\sqrt{1920}$

        = $8\sqrt{30}$

∴ Area of triangle = $8\sqrt{30}$ $cm^{2}$

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