if the diagonal and area of a rectangle are 25m and 168 sq.m, what is the length and breadth of rectangle?

Question

please give the working steps also.

Answer 1

the diagonal is (a^2+b^2) ^0.5 =25 (1)
the area ab =168(2) b=168/a (2')
lets square (1)

a^2+b^2= 25*25 =625 replace b by 2'

a^2+ (168)^2/a^2 = 625

a^4 +(168)^2 = 625a^2

a^4-625a^2 +28224=0 Put a^2 =A

A^2-625A +28224=0

you find A^2 =576 so
a= 24 and b=7

length 24 m , width 7m

I tried to solve a similar problem and isaw i made a blunder for the quadrat of hypothenuse

Answer 2

lets side of rectrangle is a, b
digonlos(d) ^2= a^2+b^2
(25)^2= a^2+b^2
a^2+b^2= 625 --------------------(I)
Area = a*b
ab = 168 ----------------------------(II)
According to (1) & (ii)
(a+b)^2= a^2+b^2+2*a*b
a+b = 31 --------------------(iii)
a-b = 17 ---------------------(iv)
we solve the equetion (iii) & (iv)
we get value
a= 24
b= 7
lenght of trangle is 24
breath of trangle is 7