Find the percentage increase in size of the triangle if its side is doubled

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

`A = sqrt(s(s-a)(s-b)(s-c))`  

Where,

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

`2s = a+b+c`

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

Now,

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

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

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

`A_1 = 4A`

Therefore, increase in the area of the triangle

=A1 -A

=4A-A

=3A

Percentage increase in area 

`=(3A)/A xx 100 `

= 300%

Solution : 1) Area of triangle is `A_1=sqrt(s(s-a)(s-b)(s-c)`
2) If each side is doubled
`A_2=sqrt(2s(2s-2a)(2s-2a)(2s-2c)`
`=4sqrt(s(s-a)(s-b)(s-c)`
`A_2=4A_1`
Percentage change`=(A_2-A_1)/A_1*100`
`=(3A_1)/A_1*100`
`=300%`.

Hint: In this question use the direct formula for area of triangle in terms of perimeter s and side a, b, c which is $\sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} $, where the perimeter will be the sum of sides. Now as the sides are doubled so using the mentioned concept relation between the previous area and the new area could be calculated and thus the overall change in percentage.

Complete step-by-step answer:
As we know that the area of triangle is given as
$ \Rightarrow A = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} $……………. (1)
Where, s = perimeter of the triangle.
And a, b and c are the sides of the triangle respectively.
Now it is given that each side of the triangle is doubled.
So the sides of the new triangle become 2a, 2b and 2c.
Now as we know that the perimeter of the triangle is the sum of all sides.
Therefore (s = a + b + c)
And the new perimeter becomes,
S’ = (2a + 2b + 2c) = 2(a + b + c) = 2s.
So the new area (A’) of the triangle becomes,
$ \Rightarrow A' = \sqrt {2s\left( {2s - 2a} \right)\left( {2s - 2b} \right)\left( {2s - 2c} \right)} $
Now simplify it we have,
$ \Rightarrow A' = \sqrt {16s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} = 4\sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} $
Now from equation (1) new area of the triangle becomes,
$ \Rightarrow A' = 4A$
So the increase in area is the subtraction of new area and previous area.
Therefore increase in area = A’ – A
                                                = 4A – A = 3A.
Now we have to calculate the percentage increase in area, so the percentage increase in the area is the ratio of increase in area to the old area multiplied by 100.
Therefore the percentage increase in area $ = \dfrac{{3A}}{A} \times 100 = 300$ %.
So the percentage increase in the area of a triangle if each side is doubled is 300%.
So this is the required answer.

Note: Whenever we face such type of problems the key concept is the area of the triangle in terms of its perimeter and sides which is stated above now according to question the sides of the triangle is doubled so the perimeter of the triangle is doubled and the area of the triangle becomes four times the calculation is shown above so the area increase three times and the percentage increase in area is 300 % which is the required answer.

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