# Let Δ be the area of a triangle. Find the area of a triangle whose each side is twice the side of the given triangle. - Mathematics

Let Δ be the area of a triangle. Find the area of a triangle whose each side is twice the side of the given triangle.

#### Solution

We are given assumed value Δ   is the area of a given triangle ABC

We assume the sides of the given triangle ABC be a, b, c

The area of a triangle having sides aband s as semi-perimeter is given by,

A = sqrt(s(s-a)(s-b)(s-c))

Δ  = sqrt( s(s-a)(s-b)(s-c))                              ...................(1)

Where,

s =(a+b+c)/2

2s = a +b+c                                                  ....................(2)

We take the sides of a new triangle as 2a, 2b, 2c that is twice the sides of previous one

Now, the area of a triangle having sides 2a, 2b, and 2c and s1 as semi-perimeter is given by,

A_1 = sqrt(s_1(s_1-2a)(s_1-2b)(s_1-2c)), where

s_1 = ( 2a+2b+2c)/2

s_1 = (2(a+b+c))/2

s1 = a+ b+ c

s1 =2s                                                       ..................(using (2))

Now,

A_1 = sqrt(2s(2s-2a)(2s-2b)(2s-2c)

A_1= sqrt(2x xx 2 (s-a) xx 2 (s-b) xx 2(s-c))

A_1 = 4sqrt(4(s-a)(s-b)(s-c))

A_1 = 4Δ                                          ...................(using (1))

Concept: Application of Heron’s Formula in Finding Areas of Quadrilaterals
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#### APPEARS IN

RD Sharma Mathematics for Class 9
Chapter 17 Heron’s Formula
Exercise 17.3 | Q 8 | Page 24

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