Question

(8)
Find the area of a triangle two sides of which are 18 cm and 10 cm and
perimeter is 42 cm​

Answer 1

$\Large{\underline{\underline{\tt{\purple{Hello !!}}}}}$

First side of triangle = a = 18 cm

Second side of triangle = b = 10 cm

Third side of triangle = c

a+b = 18 + 10 = 28 cm

Perimeter of triangle = a + b + c = 42 cm

Therefore ,

$= a = \sqrt{21(21 - 18)(21 - 10)(21 - 14)}$

c = 42 - 28 = 14 cm

Also ,

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

$\: \: \: \: \: = \frac{42}{2} = \frac{21}{1} = 21$

By heron's formula ,

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

$= \sqrt{21(21 - 18)(21 - 10)(21 - 14)}$

$a = \sqrt{21 \times 3 \times 11 \times 7}$

$a = \sqrt{21 \times 21 \times 11}$

$a = 21 \sqrt{11} \: \: {cm}^{2}$

Brainliest .. !! ❤⚠️

Answer 2

Answer:

hope it is helpful to you

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please mark the brainliest answer