Question

The hypotenuse of a right triangle is 2 centimeters more than the longer of the side containing the right angle. The shorter side containing the right angle is 7 centimeters less than the longer side containing the right angle. Find the length of the hypotenuse.

Answer 1

Hint: We start solving this question by assuming shorter of the side to be $x$, so the longer of the side be $x+7$ as given in the question the shorter of the side containing the right angle is 7 centimeters less than the longer side containing the right angle. Also, the hypotenuse will be $x+9$ as given in the question the hypotenuse is 2 centimeters more than the longer of the side. Then, we apply the Pythagoras theorem to get the answer.

Complete step-by-step answer:
We have given that the hypotenuse of a right triangle is 2 centimeters more than the longer of the side containing the right angle and the shorter of the side containing the right angle is 7 centimeters less than the longer side containing the right angle.
Let us assume that the length of the shorter side is $x$.
So, the length of the longer side will be $x+7$ and the length of the hypotenuse will be $x+9$.
We have to find the length of the hypotenuse.

We have given that it is a right triangle, so we use the Pythagoras theorem.
According to the Pythagora's theorem, square of the hypotenuse of a right angled triangle is equal to the sum of the square of other two sides (base and perpendicular).
${{\left( \text{Hypotanuse} \right)}^{\text{2}}}\text{=}{{\left( \text{Base} \right)}^{\text{2}}}\text{+}{{\left( \text{Perpendicular} \right)}^{\text{2}}}$
When we put the values, we get
${{\left( x+9 \right)}^{2}}={{\left( x+7 \right)}^{2}}+{{x}^{2}}$
Now, we know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
So, when we solve further, we get
$\begin{align}
  & \Rightarrow {{x}^{2}}+{{9}^{2}}+\left( 2\times 9\times x \right)={{x}^{2}}+{{7}^{2}}+\left( 2\times 7\times x \right)+{{x}^{2}} \\
 & \Rightarrow {{x}^{2}}+81+18x=2{{x}^{2}}+49+14x \\
 & \Rightarrow 2{{x}^{2}}-{{x}^{2}}+14x-18x=81-49 \\
 & \Rightarrow {{x}^{2}}-4x=32 \\
 & \Rightarrow {{x}^{2}}-4x-32=0 \\
\end{align}$
Now, we solve this equation by using the factorization method.
$\begin{align}
  & \Rightarrow {{x}^{2}}-8x+4x-32=0 \\
 & \Rightarrow x\left( x-8 \right)+4\left( x-8 \right)=0 \\
 & \Rightarrow \left( x-8 \right)\left( x+4 \right)=0 \\
 & \Rightarrow \left( x-8 \right)=0\text{ or }\left( x+4 \right)=0 \\
 & \Rightarrow x=8\text{ or }x=-4 \\
\end{align}$
As $x$ is the length of the side of a triangle, so it has only positive value. We leave the negative value.
So, the length of the shorter of the side is $x=8\text{ cm}$
Length of the longer of the side is $x+7=8+7=15\text{ cm}$
Length of the hypotenuse is $x+9=8+9=17\text{ cm}$

Note: The point to be noted is that in the right angle triangle hypotenuse is the longest side. Also, we can cross verify our answer by using Pythagoras theorem.
${{\left( \text{Hypotenuse} \right)}^{\text{2}}}\text{=}{{\left( \text{Base} \right)}^{\text{2}}}\text{+}{{\left( \text{Perpendicular} \right)}^{\text{2}}}$
We have Hypotenuse $=17\text{ cm}$
Base $=8\text{ cm}$
Perpendicular $=15\text{ cm}$
Now, put this values, we get
$\begin{align}
  & {{17}^{2}}={{8}^{2}}+{{15}^{2}} \\
 & 289=64+225 \\
 & 289=289 \\
\end{align}$
L.H.S=R.H.S
So, our answer is correct.