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.

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.

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.