Question

the diagonal of a rectangular field is 60 M more than the shorter side if the longer side is 30 M more than the shorter sides find the sides of the field

Answer 1

Answer :

The sides of the rectangle are 90 m and 120 m.

Step-by-step explanation :

Let ABCD be the required rectangular field where AC is the diagonal.

Let us take the shorter side to be $x$ metres.

Given, diagonal is 60 m more than the shorter side.

$\therefore AC=x+60$ metres

Also it's given that longer side is 30 m more than the shorter side.

$\therefore BC=x+30$ metres

Since, in $\triangle$ ABC,

$\angle$ B = 90°

Therefore, by Pythagoras theorem,

$Hypotenuse^{2}=Base^{2}+Height^{2}$

$AC^{2}=AB^{2}+BC^{2}$

$\implies (x+60)^{2}=x^{2}+(x+30)^{2}$

$\implies x^{2}+3600+120x=x^{2}+x^{2}+900+60x$

$\implies x^{2}-x^{2}-x^{2}+120x-60x+3600-900=0$

$\implies -x^{2}+60x+2700=0$

$\implies x^{2}-60x-2700=0$

$\implies x^{2}-90x+30x-2700=0$

$\implies x(x-90)+30(x-90)=0$

$\implies (x+30)(x-90)=0$

$\implies x=-30\:OR\:x=90$

Since, $x$ is length, therefore it can't be negative.

So, $x=90$

Shorter side = $x$ = $90$ metres.

Longest side = $x+30=90+30=120$ metres

Answer 2

Answer -

According to question-

AC=x+60

BC=x+30

by Pythagoras theorem,

》$Hypotenuse^{2}=Base^{2}+Height^{2}$

》$(x+60)^{2}=x^{2}+(x+30)^{2}$

》$x^{2}+3600+120x=x^{2}+x^{2}+900+60x$

》$x^{2}-x^{2}-x^{2}+120x-60x+3600-900=0$

》$-x^{2}+60x+2700=0$

》$x^{2}-60x-2700=0$

》$x^{2}-90x+30x-2700=0$

》$x(x-90)+30(x-90)=0$

》$(x+30)(x-90)=0$

》$x=-30\:and\:x=90$

x is length so it can't be negative.

Breadth = 90 m

Length = 120 m