Question

If the sides of the triangle are doubled then the area
$\begin{align}
  & \text{a) Remains the same} \\
 & \text{b) Becomes doubled} \\
 & \text{c) becomes three times} \\
 & \text{d) becomes four times} \\
\end{align}$

Answer 1

Hint: Now we know that the area of triangle with sides a, b and c is given by the formula $\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$ where s is half of the perimeter of triangle. Now we will form a new triangle with double the lengths of the previous triangle and hence find the area by substituting the sides of the triangle. Hence we will find the relation between the original area of the triangle and the new area of the triangle obtained.

Complete step by step solution:
Now consider a triangle with sides a, b and c.
We know that the perimeter of the triangle is nothing but the sum of all the sides of the triangle.
Hence we get perimeter is $a+b+c$
Let $p=a+b+c$
Now let s be such that $s=\dfrac{p}{2}.............\left( 1 \right)$
Then we know that the area of triangle is given by \[\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}\]
Now let us say we have formed another triangle which has all the sides of the triangle double the original side.
Hence the sides of the triangle are 2a, 2b and 2c.
Now the perimeter of the triangle will be $2a+2a+2c$
Now $s'=\dfrac{2a+2b+2b}{2}$
Hence we get
\[\begin{align}
  & \Rightarrow s'=\dfrac{2\left( a+b+c \right)}{2}=a+b+c \\
 & \Rightarrow s'=p \\
\end{align}\]
Now from equation (1) we get,
\[\Rightarrow s'=2s\]
Hence we get the new s which will be 2 times the original s.
Hence the formula for area will be $\sqrt{2s\left( 2s-2a \right)\left( 2s-2b \right)\left( 2s-2c \right)}$
Now taking 2 out from the whole expression we get
$\begin{align}
  & \Rightarrow A=\sqrt{2s\left( 2 \right)\left( s-a \right)\left( 2 \right)\left( s-b \right)\left( 2 \right)\left( s-c \right)} \\
 & \Rightarrow A=\sqrt{16s\left( s-a \right)\left( s-b \right)\left( s-c \right)} \\
\end{align}$
Now we can take 16 out from the root as $16={{4}^{2}}$ . Hence we get,
$\Rightarrow A=4\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$

So, the correct answer is “Option d”.

Note: Now note that we can also solve this example by taking an example of a right angle triangle. Now we know that in the right angle triangle the base and height are the sides of the triangle itself and the area of the triangle is given by $A=\dfrac{1}{2}\times b\times h$ . Now if we double the sides we get the new base and height of the triangle as $A=\dfrac{1}{2}\times \left( 2b \right)\times \left( 2h \right)$ . Hence we have $A=4\times \left( \dfrac{1}{2}\times b\times h \right)$ . Hence we can easily say that the area of the triangle is 4 times the original triangle.