Question

ABCD is a rectangle. AC + AB= 5AD and AC- AD =8. Find the area of the rectangle.

Answer 1

Answer:

Area of ABCD = 4 AD² + 8 AD

Step-by-step explanation:

Given :

ABCD is a rectangle.

AC + AB = 5AD

AC - AD = 8

To find: Area of the rectangle.

Solution:

Area of rectangle = Length * Breadth

Area of rectangle ABCD = AB * AD

AC = AD - 8

Substituting this in first equation

AD - 8 + AB = 5 AD

AB - 8 = 4 AD

AB = 4 AD + 8

Now area of rectangle ABCD = (4 AD + 8) * AD = 4 AD² + 8 AD

The above equation has area calculated using the width.

Similarly the area of the rectangle in terms of the length and diagonal can be calculated.

Answer 2

Answer:

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 ..........2

squaring both sides in equation 2,

we get,

z^2=(5y-x)^2

z^2 = 25y^2+x^2−10xy.......eqn 3

Now Subtracting eqn1 from eqn 3

we get ,

z^2 = 25y^2+x^2-10xy

-z^2 = - y^2 - x^2 - 0

0 = 24y^2 − 10xy

x / y = 24 / 10

= 12 / 5

AC - AD = 8

z - y = 8

z = 8 + 5

z = 13

we will get the area of rectangle ABCD,

Area of rectangle = length × breadth

= 12×5

= 60

Area of rectangle ABCD = 60