Question

Abcd is a rectangle such that ac+ab=5ad and ac-ad=8,then the area of rectangle is

Answer 1

Buddy the question is perfectly WRONG................

Answer 2

Answer: 60

Step-by-step explanation:

Let

AB = x = CD ,

AD = y = BC ,

AC = z,

By the Pythagoras theorem,

z^2 = x^2+y^2 ........ eqn 1

Given:- AC + AB = 5 AD

z + x = 5y

z = 5y - x .......... eqn 2

Squaring both sides in equation 2

we get

z^2 = 25y^2+x^2−10xy ........ eqn 3

Now Subtracting 1 from 3 we get

0 = 24y^2−10xy

0 = 24y−10x

x / y = 24 / 10

= 12 / 5

AC - AD = 8

z - y = 8

z = 8 + 5

= 13

we will get the area of rectangle ABCD = length × breadth

= 12×5

= 60