Question

The diagonals of a rectangular field is 60m more than the shorter side. if the longest side is 30 m more than the shorter side.find the side of the field.​

Answer 1

Answer:

Step-by-step explanation:

∴Let shorter side(AC) be x m.

∴longer side(AB) =x+30 m.

∴diagonal (BC ) =x+60 m.

∴In right ΔABC:-

∴(AB)²+(AC)²=(BC)²

∴(x+30)²+(x)²=(x+60)²

∴(x²+900+60x)+(x²)=(x²+3600+120x)

∴x²+900+x²+60x=x²+120x+3600

∴2x²+60x+900=x²+120x+3600

∴2x²-x²=120x-60x+3600-900

∴x²=60x+2700

∴x²-60x-2700=0

∴x²+30x-90x-2700=0

∴x(x+30)-90(x+30)=0

∴(x-90) (x+30)=0

∴x=90 or x= -30

∴Length can't be negative. So,x=90 m.

∴Shorter side = 90 m.

∴Longer side =90+30

                       =120 m.

∴diagonal =90+60

                 =150 m.    

Answer 2

Let the shorter side be x.

Diagonal = x+60

Longer side = x+30

By applying Pythagoras theorem,

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

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

x^2-60x-2700=0

D= 3600-4(1)(-2700) = 3600+10800=14400

=120

But, x cannot be negative. So, x = 90

Thus, shorter side = 90m, longer side= 120 m

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