Question

Find the area of a triangle of side 50 cm ,50 cm and 20 cm using Heron's Formula.​

Answer 1

Answer :-

${\underline{\boxed{\sf Area = 200\sqrt{6}}}$

Step-by-step explanation :-

Heron's Formula :-

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

-----------------------------------

S = a + b + c /2

= 50 + 50 + 20/2

= 120/2

= 60

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Substituting values in formula :-

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

= $\sqrt{60(60-50)(60-50)(60-20)$

= $\sqrt{60*10*10*40}$

= $\sqrt{2*2*3*5*2*5*2*5*2*2*2*5}$

= $200\sqrt{6}$

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Therefore the area of triangle is 200√6

Answer 2

SOLUTION:-

Let a, b & c are the sides of triangle and S is the semi- perimeter, then its area is given by:

$a = \sqrt{s(s - a)(s - b)(s - c)} \: where \: \\ s = \frac{a + b + c}{2} \: \: \: </u></strong><strong><u>[</u></strong><strong><u>H</u></strong><strong><u>eron</u></strong><strong><u>'</u></strong><strong><u>s</u></strong><strong><u> \: formula</u></strong><strong><u>]</u></strong><strong><u>$

Given,

a= 50cm

b= 50cm

c= 20cm

$s = \frac{a + b + c}{2} \\ \\ = > \frac{50 + 50 + 20}{2} = \frac{120}{2} \\ \\ = > 60cm$

Now, Area of ∆

$a = \sqrt{s(s - a)(s - b)(s - c)} \\ \\ = > \sqrt{60(60 - 50)(60 - 50)(60 - 20)} \\ \\ = > \sqrt{60(10)(10)(40)} \\ \\ = > \sqrt{2 \times 2 \times 3 \times 5 \times 10 \times 10 \times 2 \times 2 \times 2 \times 5} \\ \\ = > 2 \times 2 \times 5 \times 10 \sqrt{6} \\ \\ = > 200 \sqrt{6} {cm}^{2}$

Hope it helps ☺️