Question

If each side of any triangle is doubled then find the percentage of increase in its area
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Answer 1

$\huge{\text{\underline{Question:-}}}$

If each side of any triangle is doubled then, find the percentage increase in the area of a triangle if its each side is doubled.

$\huge{\text{\underline{Solution:-}}}$

Let a, b and c be the sides of the original triangle and s be its semi perimeter.

$\large{\boxed{\text{S = (a + b + c) / 2}}}$

2s = a + b + c $\implies$(1)

★ The sides of a new triangle are 2a, 2b and 2c.

★ Point to remember:-

• Side is doubled [Given]

★ Let s' be the new semi perimeter.

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

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

s' = a + b + c

s' = 2s ( From eq 1)$\implies$(2)

★ Let ∆ = area of original triangle

∆ = √s(s-a)(s-b)(s-c)$\implies$(3)

★ And,

∆' = area of new Triangle

∆' = √s'(s' - 2a)(s' - 2b)(s' - 2c)

∆' = √ 2s(2s - 2a)(2s - 2b)(2s - 2c)[From eq 2]

∆' = √ 2s × 2(s-a) × 2(s - b) × 2(s - c)

= √16s(s - a)(s - b)(s - c)

∆' = 4 √s(s - a)(s - b)(s - c)

∆' = 4 ∆ [From eq (3)]

★ Increase in the area of the triangle

$\implies$ ∆' - ∆

$\implies$ 4 ∆ - 1 ∆

$\implies$ 3 ∆

Percentage increase in area = (increase in the area of the triangle / original area of the triangle) × 100

$\implies$(3∆ /∆) × 100

$\implies$3 × 100 = 300 %

$\large{\boxed{\text{= 300}}}$

Hence, the percentage increase in the area of a triangle is 300%

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Answer 2

SOLUTION:-

Given:

If each side of any triangle is doubled.

To find:

The percentage of increase in its area.

Explanation:

Let the sides of ∆ are a, b, c.

We can use Heron's Formula to determine the area of a ∆.

We have,

• First side=a
• First side=aSecond side=b
• First side=aSecond side=bThird side=c

[Heron's Formula]

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

Or

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

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

&

When each side of ∆ is doubled.

• First side= 2a
• Second side= 2b
• Third side= 2c

$S' = \frac{2a + 2b + 2c}{2} \\ \\ S'= \frac{2(a + b + c)}{2} \\ \\ S' = (a + b + c) \\ \\ S' = 2s \: \: \: \: \: \: [Semi \: perimeter]$

Now,

Assume the original area be R

$Area \: of \triangle' = \sqrt{s(s - a)(s - b)(s - c)} \\ \\ Are a \: of \triangle '= \sqrt{2s(2s - 2a)(2s - 2b)(2s - 2c)} \\ \\ Are a \: of \triangle' = \sqrt[4]{s(s - a)(s - b)(s - c)} \\ \\ Area \: of \triangle' = 4R$

Increase in area = 4R -R

Increase in area= 3R

&

Percentage increase in area:

$\frac{Increase \: area}{Original \: area} \times 100 \\ \\ (\frac{3R}{R} \times 100)\% \\ \\ ( \frac{300R}{R} )\% \\ \\ 300\%$

Thus,

The percentage increase in the area of ∆ is 300%.