Question

If every side of triangle is doubled, then increase in area of the triangle is . . . . . . . . . . . . . . .
a). \[100\sqrt{2}%\]
b). \[200%\]
c). \[300%\]
d). \[400%\]

Answer 1

Hint: We have to know the heron’s formula and parameters, in that formulae there are s which is known as semi perimeter of the triangle and is given by \[s=\dfrac{a+b+c}{2}\] and a, b, c are the sides of triangle, the area of a triangle is given by A.

Complete step-by-step solution -
Heron’s formula states that for a triangle with sides of lengths a, b, c the area of the triangle is given by
\[A=\sqrt{s*\left( s-a \right)\left( s-b \right)\left( s-c \right)}\]. . . . . . . . . . . . . . . . . . . . . . . . 1
Where s is given by\[s=\dfrac{a+b+c}{2}\] . . . . . . . . . . . . . . . . . . . . . . . .2
Now if you double all the sides of triangle then a becomes 2a, b becomes 2b, c becomes 2c
Now s is given by after doubling all the sides of triangle
\[s=\dfrac{2a+2b+2c}{2}=2s\]. . . . . . . . . . . . . . . . . . . . . . . . 3
Now substitute the value 2s in the places of s in the formulae of area of triangle then the area of triangle is given by
\[A=\sqrt{2s*\left( 2s-2a \right)\left( 2s-2b \right)\left( 2s-2c \right)}\]
Taking the total value of 16 from inside square root to outside we will get
4\[\sqrt{s*\left( s-a \right)\left( s-b \right)\left( s-c \right)}\]
Which is equivalent to 4A
So the increase in area is 4A-A=3A
So, the increase in percentage in the area is \[\dfrac{3A}{A}\times 100\]
=\[300%\]
So the correct option is (C)

Note: The heron’s formula is applicable for all types of triangles whether it is equilateral triangle or isosceles triangle or scalene triangle. But there are also simple derived formulas for areas for equilateral triangles. But the area of the scalene triangle is mostly derived only using heron’ s formula.