The area of a rectangle is 108m2 and its diagonal is 15m. find the perimeter of the rectangle if the sides are of integer lengths.

Considering the figure,

In ΔABC, right angled at B, we can use Pythagoras theorem :

..................(1)

Also we are given area of rectangle as 108 m²

Area of rectangle = length * breadth = x * y

.............(2)

........(3)

plugging the value of x from equation (3) in (1),

We can use quadratic formula to solve this.

On solving this we get four values of y as :

y=12, y=-12, y=9 and y=-9

since length cannot be negative we have two y values as :

y=12 and y=9

plugging these in equation (3) to get x as

x=108/y = 108/12 = 9

x = 108/9 =12

so we have two answers:

length =9 m and breadth = 12m

length =12 m and breadth =9m

Perimeter = 2(l+b)

Perimeter = 2(9+12) = 42m

We want to find the perimeter of a rectangle given that we know the area and length of the diagonal of the rectangle.

We will see that the perimeter is equal to 42 meters.

We know that for a rectangle of length L and width W, the area is given by:

A = L*W

And the length of the diagonal is given by:

d = √(L^2 + W^2)

From the given information we have that:

A = 108m^2 = L*W

d = 15m = √(L^2 + W^2)

From the first equation we can write:

L = (108m^2)/W

Now we can replace this in the other equation to get:

15m = √(((108m^2)/W)^2 + W^2)

(15m)^2 =  ((108m^2)/W)^2 + W^2

Multiplying both sides by W^2 we get:

(15m)^2*W^2 =  (108m^2)^2 + W^4

Replacing W^2 = x we get:

(15m)^2*x =  (108m^2)^2 + x^2

Now we can rewrite this as:

x^2 +  11,664 m^4 - (225m^2)*x = 0

And solve it using Bhaskara's formula:

Because X is the width squared, we take the positive solution:

X = (225m^2 + 63m^2)/2 = 144m^2

W = √(144m^2) = 12m

Then we can replace this in the area equation:

A = 108m^2 = L*12m

108m^2/12m = L = 9m

Now that we know the length and width, we can write the perimeter as:

P = 2*(L + W) = 2*(12m + 9m) = 42m

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