Question

Using Heron's formula, find the area of the triangle whose sides measure
10 cm, 24 cm and 26 cm.

Answer 1

$Let \: a,b \:and\: c \:are \: sides \: measures \\of \: a \: triangle$

$Given \: a = 10 \:cm, b = 24 \:cm \: and \\ c = 26 \:cm$

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

$\implies s = \frac{10+24+26}{2} \\= \frac{60}{2} \\= 30$

$\boxed { \pink { Area \:of \: a \: triangle = \sqrt{s(s-a)(s-b)(s-c)} }}$

$\triangle = \sqrt{ 30(30-10)(30-24)(30-26)} \\= \sqrt{30\times 20 \times 6 \times 4 } \\= \sqrt{10^{2} \times 2^{2} \times 6^{2} } \\= 10 \times 2 \times 6 \\= 120$

Therefore.,

$\red { Area \:of \: the \: triangle} \green {= 120 \: cm^{2}}$

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Answer 2

Answer:

120

Step-by-step explanation:

a = 10, b = 24, c = 26

s = a + b + c / 2

s = 10 + 24 + 26/ 2

s= 30

A = √ 30 (30 - 10) (30- 24) (30- 26)

= √30*20*6*4

=√ 2*2*5*2*3*2*2*30

= 2*2 √ 5*2*3*30

= 4√ 30*30

= 4*30

= 120