Question

In the adjoining figure ,if AB+AC=5AD and AC-AD=8, then the area of the rectangle ABCD is

Answer 1

Answer:

Step-by-step explanation:

 suppose AB=x=CD,   AD  =y=BC,  

       z=AC

z^2=x^2+y^2......equation (1)

given (AB+AC=5AD,)     x+z=5y  ⇒z=5y-x   equation (2)

both side saquarting equation 2

z^2=25y^2+x^2-10xy ....(3)

subtracting  from 1 in 3

24y^2 -10xy =0

x/y=24/10=12/5

AC-AD=8

z-y=8 ,z=8+5=13

we get area of rectangle ABCD  =length*breadth

                                                     =x*y

                                                      =12*5=60

hope it helped

         

Answer 2

Solution:

If ABCD is a rectangle than AB,BC,CD,AD are the sides of rectangle.

Let AB =DC = x

BC = AD = y

and AC is diagonal ,we can compute AC by Pythagoras Theorem,as
$= \sqrt{ {x}^{2} + {y}^{2} }$

$x + \sqrt{ {x}^{2} + {y}^{2} } = 5y \\ \\ \sqrt{ {x}^{2} + {y}^{2} } - y = 8 \\ \\ put \: the \: value \: of \: \sqrt{ {x}^{2} + {y}^{2} } \: in \: second \: eq \\ \\ - x + 4y = 8 \: \: eq1$

With the given data only one equation can be formed. For second equation it's perimeter should be given.

No image attached .