Q6.    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.

Answers (1)
D Divya Prakash Singh

Let the shorter side of the rectangle be x m.

Then, the larger side of the rectangle wil be = (x+30)\ m.

Diagonal of the rectangle:

= \sqrt{x^2+(x+30)^2}\ m

It is given that the diagonal of the rectangle is 60m more than the shorter side.

Therefore, 

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

\Rightarrow x^2+(x+30)^2 = (x+60)^2

\Rightarrow x^2+x^2+900+60x = x^2+3600+120x

\Rightarrow x^2-60x-2700 = 0

Solving by the factorizing method:

\Rightarrow x^2-90x+30x-2700 = 0

\Rightarrow x(x-90)+30(x-90)= 0

\Rightarrow (x+30)(x-90) = 0

Hence, the roots are: x = 90,\ -30

But the side cannot be negative.

Hence the length of the shorter side will be: 90 m 

and the length of the larger side will be (90+30)\ m =120\ m

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