Question

The ratio of the sides in a triangle are $5:12:13$ and its area is $270$ square cm. The sides of the triangle in cm are
A. $5,12,13$
B. $10,24,26$
C. $15,36,39$
D. $20,48,52$

Answer 1

Hint: To solve this question, we will assume the value of the sides from the given ratio $5:12:13$ to some variable and determine its value. We will use the formula of the area of triangle and substitute all the given value of area and value of sides assumed from the ratio. After simplifying we will determine the value of the variable assumed.
We will then substitute this value of variable in the values of the sides and determine the value of sides of the triangle in cm.
Formula used:
$Area=\frac{1}{2}\times b\times h$, where $b,h$ are base and height of a right-angled triangle.
Pythagoras theorem: ${{H}^{2}}={{P}^{2}}+{{B}^{2}}$
Complete step-by-step solution:
We are given a triangle in which the ratio of all the three sides are $5:12:13$ and its area is $270$ square cm. We have to calculate the value of the sides of the triangle in cm.
Let us assume the values of the sides from the ratio be $5x,12x$ and $13x$.
We will check if the triangle is right angled triangle or not by using Pythagoras theorem ${{H}^{2}}={{P}^{2}}+{{B}^{2}}$.
$\begin{align}
  & {{(13x)}^{2}}={{(12x)}^{2}}+{{(5x)}^{2}} \\
 & 169{{x}^{2}}=144{{x}^{2}}+25{{x}^{2}} \\
 & 169{{x}^{2}}=169{{x}^{2}}
\end{align}$
Hence we can say that the triangle is right angled triangle. Now we will use the area of the right angled triangle,
\[\begin{align}
  & 270=\frac{1}{2}\times 5x\times 12x \\
 & 270=30{{x}^{2}} \\
 & 9={{x}^{2}} \\
 & x=3
\end{align}\]
We will now substitute this value $x=3$in the value of the sides $5x,12x$ and $13x$.
$\begin{align}
  & 5x=5\times 3 \\
 & =15cm \\
 & 12x=12\times 3 \\
 & =36cm \\
 & 13x=13\times 3 \\
 & =39cm
\end{align}$
The sides will be$15,36,39$.
The sides of the triangle in cm are $15,36,39$ when the ratio of the sides is $5:12:13$ and its area $270$square cm. Hence the correct option is (C).
Note:
We could have directly used Heron’s formula \[Area=\sqrt{s(s-a)(s-b)(s-c)}\,\,\,\,,s=\frac{a+b+c}{2}\] instead of checking the triangle being right angled and then using its formula of area.