The diagonal of a rectangular field is $25$ metres more than the shorter side. If longer side is $23$ metres more than the shorter side, find the sides of the field.

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Solution

Let length of shorter side $= x$
Diagonal $= d = x + 25$
Longer side $= l = x + 23$
Applying pythagoras theorem
$x^{2} + l^{2} = d^{2} x^{2} + {(x + 23)}^{2} = {(x + 25)}^{2} x^{2} + x^{2} + 529 + 46 x = x^{2} + 625 + 50 x x^{2} - 4 x - 96 = 0 x^{2} - 12 x + 8 x - 96 = 0 x (x - 12) + 8 (x - 12) = 0 (x - 12) (x + 8) = 0 x = 12, - 8$
But length can not be negative
$∴ x = 12$
So shorter side $= 12 m$
Longer side $= 12 + 23 = 35 m$

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The diagonal of a rectangular field is 25 metres more than the shorter side. If longer side is 23 metres more than the shorter side, find the sides of the field.

Let length of shorter side =xDiagonal =d=x+25Longer side =l=x+23Applying pythagoras theoremx2+l2=d2x2+(x+23)2=(x+25)2x2+x2+529+46x=x2+625+50xx2−4x−96=0x2−12x+8x−96=0x(x−12)+8(x−12)=0(x−12)(x+8)=0x=12,−8But length can not be negative∴x=12So shorter side =12mLonger side =12+23=35m

The diagonal of a rectangular field is $25$ metres more than the shorter side. If longer side is $23$ metres more than the shorter side, find the sides of the field.