Question

The diagonal of a rectangular field is $60m$ more than shorter side. If the longer side is $30m$ more than the shorter side, find the sides of the field.

Answer 1

Step 1. Explaining the diagram.

Let the $ABCD$ be a rectangle in which shorter side of rectangle be $BC=am$.

According to given condition The diagonal of a rectangular field is $AC=60m$ more than shorter side

Therefore diagonal $=(a+60)m$

Also longer side is $30m$ more than the shorter side

Therefore longer side $AB=(a+30)m$

Step 2.Finding the sides

In $∆ABC$, $\angle B=90°$ [As each angle of rectangle is of $90°$]

Using Pythagoras theorem In $∆ABC$ we have

$$\begin{gathered} A{C}^2=A{B}^2+B{C}^2 \\ \Rightarrow {(a+60)}^2={(a+30)}^2+a^2 \\ \Rightarrow a^2+{60}^2+2xax60=a^2+{30}^2+2xax30+a^2 \\ \Rightarrow a^2+3600+120a=a^2+900+60a+a^2 \\ \Rightarrow a^2-a^2-a^2+120a-60a+3600-900=0 \\ \Rightarrow -a^2+60a+2700=0 \end{gathered}$$

$\Rightarrow a^2-60a-2700=0$ [ multiply by $(-)$ on both sides]

Step 3: Solving quadratic equation.

$$\begin{gathered} a^2-60a-2700=0 \\ \Rightarrow a^2-(90-30)a-2700=0[\because 90x30=2700and90-30=60] \\ \Rightarrow a^2-90a+30a-2700=0 \\ \Rightarrow a(a-90)+30(a-90)=0 \\ \Rightarrow (a+30)(a-90)=0 \end{gathered}$$

$\Rightarrow$ either $a+30=0$ or $a-90=0$

$\Rightarrow$ either $a=-30$ or $a=90$

Since, length can not be negative, therefore $a\ne -30$

Hence, shorter side $a=90m$ and longer side $=(a+30)=90+30=120m$