Question

If each side of a triangle is doubled, then find the percentage increase in the
area of the triangle​

Answer 1

Answer:

There is a 300% increase in area of triangle when sides are doubled.

Step-by-step explanation:

Let S1, S2 and S3 be the lengths of the sides of the triangle

Let S = (S1 + S2 + S3)/2

By Heron's formula, we know:

Area of triangle = √[S(S - S1)(S - S2)(S - S3)]

Let T1, T2, and T3 represent the sides of the "doubled" triangle.

=> T1 = 2S1, T2 = 2S2 and T3 = 3S3

T = (T1 + T2 + T3)/2

Area of new triangle = √[T(T - T1)(T - T2)(T - T3)]

Now,

T = (T1 + T2 + T3)/2

   = (2S1 + 2S2 + 2S3)/2

   = S1 + S2 + S3 = 2S

Area of new triangle  

= √[T(T - T1)(T - T2)(T - T3)]

= √[2S(2S - 2S1)(2S - 2S2)(2S - 2S3)]

= √[2S*2*(S - S1)*2*(S - S2)*2*(S - S3)]

= √[16*S(S - S1)(S - S2)(S - S3)]

= 4√[S(S - S1)(S - S2)(S - S3)]

= 4*Area of original triangle

=> Area of new triangle = 4 * Area of old triangle

=> The area increases by 300%

More explanation

Why 300%?

Let NEW = 4*OLD

=> NEW = OLD + 3*OLD

=> NEW = OLD + 300/100*OLD

=> NEW = OLD + 300% of OLD

=> (NEW - OLD) = 300% of OLD

=> Increase = 300%

Remember: % increase is the increase from old to new, expressed as a percentage of old.

Answer 2

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \boxed{\boxed { \huge \mathcal\red{ solution}}}}$

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$\underline{\red{\textbf{FROM Heron's Formula}}}$

$\blue{\textbf{For a triangle with the sides x,y and z}}$

$\boxed{\purple{\textbf{Area=}\sqrt{s(s-x)(s-y)(s-z)}}}$

$\red{\textbf{where, s=}\bf\frac{x+y+z}{2}}$

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For a Triangle with sides a, b, c

$\textbf{Area_1}=\sqrt{s_1(s_1-a)(s_1-b)(s_1-c)}$

$\textbf{where, s_1=}\frac{a+b+c}{2}$

$\therefore Area_1=\sqrt{s_1(s_1-a)(s_1-b)(s_1-c)}\\</p><p>\bf\implies \footnotesize{Area_1=\sqrt{\frac{a+b+c}{2}(\frac{a+b+c}{2}-a)(\frac{a+b+c}{2}-b)(\frac{a+b+c}{2}-c)}}\\</p><p>\bf\implies \footnotesize{Area_1=\sqrt{\frac{a+b+c}{2}(\frac{a+b+c-2a}{2})(\frac{a+b+c-2b}{2})(\frac{a+b+c-2c}{2})}}\\</p><p></p><p>\bf\implies\footnotesize{ Area_1=\sqrt{\frac{a+b+c}{2}(\frac{b+c-a}{2})(\frac{a-b}{2})(\frac{a+b-c}{2})}}\\</p><p></p><p>\bf\implies\footnotesize{ Area_1=\sqrt{\frac{a+b+c}{2}(\frac{b+c-a}{2})(\frac{a-b+c}{2})(\frac{a+b-c}{2})}}\\</p><p>\bf\implies \footnotesize{Area_1=\sqrt{\frac{(a+b+c)(b+c-a)(a-b+c)(a+b-c)}{2\times2\times2\times2}}}\\</p><p>\bf\implies \footnotesize{Area_1=\frac{\sqrt{(a+b+c)(b+c-a)(a-b+c)(a+b-c)}}{2\times2}}\\</p><p>\bf\implies \footnotesize{Area_1=\frac{\sqrt{(a+b+c)(b+c-a)(a-b+c)(a+b-c)}}{4}}\\</p><p>\bf\implies \footnotesize{4\times Area_1=\sqrt{(a+b+c)(b+c-a)(a-b+c)(a+b-c)}}</p><p>$

For a Triangle with sides 2a, 2b,2c :-

$\textbf{Area_2=}\sqrt{s_2(s_2-2a)(s_2-2b)(s_2-2c)}$

$\textbf{where, s_2=}\frac{2a+2b+2c}{2}=(a+b+c)$

$\therefore Area_2=\sqrt{s_2(s_2-2a)(s_2-2)(s_2-2c)}\\</p><p>\bf\implies\footnotesize{ Area_2=\sqrt{(a+b+c)(a+b+c-2a)(a+b+c-2b)(a+b+c-2c)}}\\</p><p>\bf\implies \footnotesize{Area_2=\sqrt{(a+b+c)(b+c-a)(a-b+c)(a+b-c)}}\\</p><p>\bf\implies\boxed{\green{ \bf Area_2=4\times Area_1}}$

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$\therefore$% Of Increase in The area of the of the Triangle:-

$\bf\implies percentage \:increase=\frac{Area_2}{Area_1}\times100\%\\</p><p></p><p>\bf\implies \% \:increase=\frac{\cancel{Area_1}\times4}{\cancel{Area_1}}\times100\%\\</p><p>\bf\implies \% \:increase=4\times100\%\\</p><p>\boxed{\red{\bf\implies \% \:increase=400\%}}$

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$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

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