Question & Answer
QUESTION

The hypotenuse of a right triangle is 25 cm. The difference between the lengths of the other two sides of the triangle is 5 cm. Find the perimeter of the triangle.

ANSWER Verified Verified
Hint: In this question, let's assume the other two sides as some variables then by using the pythagoras theorem we can get a relation between the two sides. Now, by solving the two equations we can get the sides. Then, as we know all the sides we can get the perimeter.

Complete step-by-step answer:
Let us assume the other two sides of the right triangle as a and b

By the pythagoras theorem, in a right triangle it follows that the hypotenuse of this triangle has the length of the square root of the sum of the squares of the other two sides.

\[c=\sqrt{{{a}^{2}}+{{b}^{2}}}\]

Where, a and b are the sides of the right triangle and c is the hypotenuse

Now, as given in the question that hypotenuse is 25 cm we get,

\[\Rightarrow 25=\sqrt{{{a}^{2}}+{{b}^{2}}}\]

Now, by squaring on both sides we get,

\[\Rightarrow {{a}^{2}}+{{b}^{2}}={{\left( 25 \right)}^{2}}.........\left( 1 \right)\]

Given, in the question that the difference between the two sides is 5 which can be written as:

\[\Rightarrow a-b=5........\left( 2 \right)\]

Now, let us square this above equation on both sides.

\[\Rightarrow {{a}^{2}}+{{b}^{2}}-2ab=25\]

Now, let use substitute the value we got in the equation (1) in this above equation.

\[\begin{align}

  & \Rightarrow {{\left( 25 \right)}^{2}}-2ab=25 \\

 & \Rightarrow 625-2ab=25 \\

\end{align}\]

Let us now rearrange the terms on the both sides and then simplify it further.

\[\begin{align}

  & \Rightarrow 2ab=625-25 \\

 & \Rightarrow 2ab=600 \\

\end{align}\]

Now, divide with 2 on both sides.

\[\Rightarrow ab=300\]

Now, let us write a in terms of b

\[\Rightarrow a=\dfrac{300}{b}\]

Let us now substitute this relation in the equation (2).

\[\begin{align}

  & \Rightarrow \dfrac{300}{b}-b=5 \\

 & \Rightarrow 300-{{b}^{2}}=5b \\

\end{align}\]

Now, on rearranging the terms we get a quadratic equation in b.

\[\Rightarrow {{b}^{2}}+5b-300=0\]

Now, by using the direct formula we can get the value of b.

\[\Rightarrow b=\dfrac{-5\pm \sqrt{{{5}^{2}}-4\times 1\times \left( -300 \right)}}{2\times 1}\]

Now, on simplifying the terms under the square root we get,

\[\begin{align}

  & \Rightarrow b=\dfrac{-5\pm \sqrt{25+1200}}{2} \\

 & \Rightarrow b=\dfrac{-5\pm 35}{2} \\

\end{align}\]

Now, we neglect the - case here as the side of a triangle cannot be negative.

\[\begin{align}

  & \Rightarrow b=\dfrac{35-5}{2} \\

 & \therefore b=15cm \\

\end{align}\]

Now, by substituting this value of b in the relation between a and b we get,

\[\begin{align}

  & \Rightarrow a=\dfrac{300}{b} \\

 & \Rightarrow a=\dfrac{300}{15} \\

 & \therefore a=20cm \\

\end{align}\]

Now, let us assume the perimeter of this triangle as P.

\[\Rightarrow P=a+b+25\]

Now, on substituting the respective values of a and b we get,

\[\begin{align}

  & \Rightarrow P=20+15+25 \\

 & \therefore P=60cm \\

\end{align}\]

Note: Instead of writing a in terms of b we can also write b in terms of a and then substitute this value of b in any of the two equations and then solve for the value of a. Then by substituting this value of a in the relation between a and b we get the value of b. Both the methods give the same value of a and b.
While getting the value of b from the quadratic equation formed instead of using the direct formula method we can also use factorisation method. But this may take a little more time as we need to check for the possible values.
Here, from the direct formula we consider only the case which gives the positive value of b but there is possibility of another value of b as it is a quadratic equation. We can neglect that other value here as the side of a triangle cannot be negative.