Question

The perimeter of the right angled triangle is five times the length of the shortest side. The numerical value of the area of the triangle is 15 times the numerical value of the length of the shortest side. Find the lengths of the three sides of the triangle.

Answer 1

Step 1: Determine the quadratic equation:

Let the base and altitude of a right-angled triangle be $x\mathrm{and}y$ units respectively.

Let $x$ unit be the shortest side of the right angled triangle. and $y$ unit be other side of this triangle.

Then the hypotenuse of the triangle is given by the Pythagoras theorem.

$\begin{array}{rcl}{\left(\mathrm{Hypotenuse}\right)}^2& =& {\left(\mathrm{Base}\right)}^2+{\left(\mathrm{Altitude}\right)}^2\\ & \Rightarrow & {\left(AC\right)}^2={\left(BC\right)}^2+{\left(AB\right)}^2\\ & \Rightarrow & {\left(AC\right)}^2=x^2+y^2\\ \therefore AC& =& \sqrt{x^2+y^2}\end{array}$

Now, perimeter of the triangle is the sum of all its sides and the perimeter is five times the length of its shortest side.

$\begin{array}{rcl}P& =& x+y+\sqrt{x^2+y^2}\\ P& =& 5\times x\dots \left(1\right)\end{array}$

And, the area of the right angled triangle is,

$A=\frac{1}{2}\times \mathrm{base}\times \mathrm{height}$

It is given that the numerical value of the area of triangle is 15 times the numerical value of the length of the shortest side.

$$\begin{gathered} \frac{1}{2}\times x\times y=15\times x\dots \left(2\right) \\ \Rightarrow x\times y=2\times 15\times x \\ \Rightarrow y=2\times 15\left[\mathrm{Divide}\mathrm{both}\mathrm{sides}\mathrm{by}x\right] \\ \therefore y=30...\left(3\right) \end{gathered}$$

Therefore, the altitude of the right-angled triangle is $30\mathrm{units}$.

From equation (1),

$$\begin{gathered} \begin{array}{rcl}x+y+\sqrt{x^2+y^2}& =& 5\times x\\ & \Rightarrow & \sqrt{x^2+y^2}=5x-x-y\\ & \Rightarrow & \sqrt{x^2+y^2}=4x-y\\ & \Rightarrow & x^2+y^2={\left(4x-y\right)}^2\left[\mathrm{Squaring}\mathrm{both}\mathrm{sides}\right]\end{array} \\ \Rightarrow x^2+y^2=16x^2+y^2-8xy \\ \Rightarrow 16x^2-x^2+y^2-y^2-8xy=0 \\ \Rightarrow 15x^2-8xy=0 \\ \Rightarrow 15x^2-\left(8x\times 30\right)=0\left[\mathrm{Substitute}y=30\right] \\ \therefore 15x^2-240x=0 \end{gathered}$$

Step 2: Determine the three sides of the triangle:

Use the factorization method to solve the obtained quadratic equation.

$\begin{array}{rcl}15x^2-240x& =& 0\\ & \Rightarrow & 15x\left(x-16\right)=0\\ & \Rightarrow & x=0\mathrm{or}x=16\end{array}$

As length cannot be zero, thus, reject $x=0$.

$\therefore x=16$

Then, the third sides of the right-angled triangle is calculated as,

$\begin{array}{rcl}AC& =& \sqrt{x^2+y^2}\\ & =& \sqrt{{\left(16\right)}^2+{\left(30\right)}^2}\\ & =& \sqrt{1156}\\ \therefore AC& =& 34\mathrm{unit}\end{array}$

Therefore, the three sides of the triangle are$16\mathrm{units},30\mathrm{units}\mathrm{and}34\mathrm{units}$.