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If each side of a triangle is doubled, then find the ratio of area of the new triangle than formed and the given triangle.

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4: 1

1: 4

2: 1

1: 2

answer is A.

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Detailed Solution

Suppose, length of triangle = a, b and c So, area of triangle from heron’s formula

A_{1} = \sqrt{s (s - a) (s - b) (s - c)}

Here,
s=semi-perimeter of the triangle.

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

2 s = a + b + c

….(1)
When we doubled the sides of the triangle then the length we get = 2a, 2b, and 2c. So, the semiperimeter,

s^{'} = \frac{2 a + 2 b + 2 c}{2}

s^{'} = a + b + c

s' = 2 s

(from equation (1))
Now, Area of new triangle,

A_{2} = \sqrt{s^{'} (s^{'} - 2 a) (s^{'} - 2 b) (s^{'} - 2 c)}

= \sqrt{2 s (2 s - 2 a) (2 s - 2 b) (2 s - 2 c)}

= \sqrt{16 s (s - a) (s - b) (s - c)}

= 4 \sqrt{s (s - a) (s - b) (s - c)}

= 4 A_{1}

So, ratio of the areas

\frac{A_{2}}{A_{1}} = \frac{4 A_{1}}{A_{1}}

A_{2} : A_{1} = 4 : 1

Correct option is 1.

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