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

byAmlesh Mahato •December 03, 2021

0

Solution:

⇒Let length of shorter side be x

∴ The Length of longer side = x + 30

⇒ From Pythagoras Theorem

$⇒ From Pythagoras Theorem$

$\large \Rightarrow h^{2}=x^{2}+\left ( x+30 \right )^{2}$

$\large \Rightarrow h^{2}=x^{2}+x^{2}+2\times x\times 30+30^{2}$

$\large \Rightarrow h^{2}=2x^{2}+60x+900$

$\large \Rightarrow h=\sqrt{2x^{2}+60x+900}$

A/Q

⇒ The diagonal is 60 meters more than the shorter side

h = x + 60

$⇒ The diagonal is 60 meters more than the shorter side$

⇒Squaring On Both Side

$\large \Rightarrow \left ( \sqrt{2x^{2}+60x+900} \right )^2=\left ( x+60 \right )^{2}$ $\large \Rightarrow 2x^{2}+60x+900=\left ( x+60 \right )^{2}$

$\large \Rightarrow 2x^{2}+60x+900=x^{2}+2\times x\times 60+60^{2}$

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

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

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

$\large \Rightarrow x^{2}-\left ( 90-30 \right )x-2700=0$

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

$\large \Rightarrow x\left ( x-90 \right )+30\left ( x-90 \right )=0$

$\large \Rightarrow \left ( x-90 \right )\left ( x+30 \right )=0$

$\large \Rightarrow x-90=0,\Rightarrow x+30=0$

$\large \Rightarrow x=90,\Rightarrow x=-30$

⇒ Length can not be negative , so -30 is rejected.

∴ Length of shorter side is = 90 m

⇒ Then Length of longer side = 90 + 30 = 120 m