Question

the diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side. find the sides of the field​

Answer 1

The digital meter of a rectangular field is90 meters more than shorter side

Answer 2

$\huge\sf{Let,}$

$\large\sf{the\:shorter\:side=x}$

$\large\sf{the\:longer\:side=(x+30)m}$

$\large\sf{the\:diagonal\:side=(x+60)m}$

━━━━━━━━━━━━━━━

$\huge\sf{In\:∆ABC,}$

$\large\sf{By\:Pythagoras\:theorem,}$

$\small\sf{{(AB)}^{2} + {(BC)}^{2} = {(AC)}^{2}}$

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

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

$\small\sf{{(2x)}^{2}+60x+900={x}^{2}+120x+3600}$

$\small\sf{{2x}^{2}-{x}^{2}+60x-120x+900-3600}$

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

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

$\small\sf{x(x-90)+30(x-90)=0}$

$\small\sf{(x+30)(x-90)=0}$

$\small\sf{x+30=0}$ $\small\sf{x-90=0}$

$\small\sf{x=-30}$ $\small\sf{x=90}$

━━━━━━━━━━━━━━━

Since, side cannot be negative,

Therefore,shorter side = 90m

longer side = 120m

& diagonal = 150m