Question

the percentage increases in the area of a triangle if each of its sides is double each

Answer 1

$\textbf{ Hello Mate! }$

Area of triangle by heron's formula

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

$s= \frac{a + b + c}{2}$
As per the sides are doubled,

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

s' / s = (a+b+c)/[(a+b+c)/2]

s' = 2s

$= \sqrt{s'(s' - 2a)(s' - 2b)(s' - 2c)} \\ = \sqrt{2s(2s - 2a)(2s - 2b)(2s - 2c)} \\ = \sqrt{2s \times 2(s - a) \times 2(s - b) \times 2(s - c)} \\ = \sqrt{16s(s - a)(s - b)(s - c)} \\ = 4 \times \: area \: of \: triangle$

Chane in area = [( 4 - 1 ) area of triangle / area of triangle ] × 100

= 300%

$\textsf{ \red{ Percentage increase is 300}}$

Have great future ahead!

Answer 2

$<b><u>Hii !<u><b>$

Here is your answer !!

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

Semiperimeter = s

EQUATION ( 1 )

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

When sides doubled ,

Sides are 2a , 2b and 2c

Semiperimeter = 2s

$area \: = \sqrt{2s(2s - 2a)(2s - 2b)(2s - 2c)} \\ \\ \: \: \: \: \: \: \: \: \: \: \: = \sqrt{2s \times 2(s - a) \times 2(s - b) \times 2(s - c)} \\ \\ \: \: \: \: \: \: \: \: \: \: \: = 4 \sqrt{s(s - a)(s - b)(s - c)}$
FROM EQUATION ( 1 ) ,

$\: \: \: \: \: \: \: \: \: \: \: = 4a$

Increase in area =

4a - a = 3a

Percentage increase =

$\frac{3a}{a} \times 100 = 300 \: \% \: \: \: \: ans$

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THANKS !!!!