Question

If you can solve, I will be very impressed.

The area of a rectangle is 108m^2 and its diagonal is 15m. Find the perimeter of the rectangle.

Answer 1

42m

Explanation:

Given: Area of rectangle= 108
`m^(2)`

Diagonal= 15m

Lets assume the value of width of rectangle be `w` and length be `l`.

We know that Area of rectangle=
`w* l`

∴
`108= w* l`

cross multiplying both side

⇒
`w= (108)/(l)`

∴ w=
`(108)/(l)`

Now solving to get value for l and w.

Remember,
`Diagonal^(2) =l^(2) +w^(2)`

⇒
`15^(2) = l^(2) +w^(2)`

Next substituting the value of w in the equation.

⇒
`225= l^(2) +((108)/(l)) ^(2)`

Opening parenthesis

⇒
`225= l^(2) + (11664)/(l^(2) )`

Now taking LCD and cross multiplying both side.

⇒
`225l^(2) = l^(4) + 11664`

Use quadratic equation to solve.

∴
`l^(4) -225l^(2) +11664= 0`

⇒
`l^(4) -144l^(2) -81l^(2) -11664=0`

Solving it we get two value of l, which is 9 and 12.

We can use any these value of l in substituting.

∴ w=
`(108)/(9) = 12m`

l= 9m and w= 12m

Now, solving to get perimeter of the rectangle.

perimeter=
`2(w+l)`

⇒ Perimeter=
`2* (12+9)= 2* 21`

∴ Perimeter of the rectangle is 42m.