Geometry Help for Diagonals and Area

Question

In a certain rectangle, a diagonal has a length of  The perimeter of the rectangle is  Find the area of the rectangle.

Answer 1

We can use L and W for the length and width. Since there are 2 unknowns, we need 2 equations. We get these from the information that is supplied.

"The diagonal has a length of 41": Use this with the Pythagorean Theorem to get:

L² + W² = 41²

"The perimeter is 98":

2L + 2W = 98, So: L + W = 49

Use substitution to solve the pair of equations......from the second one we can write:

L = 49 - W

Use this to replace L in the 1st equation:

(49 - W)² + W² = 41²

Simplify this and solve for W. Then use the second equation to find L

Answer 2

draw a diagram

P=2l+2w

98=2l+2w

49=l+w

l=49-w

use the Pythagorean Theorem

w^2+l^2=41^2

w^2+(49-w)^2=1681

w^2+2401-98w+w^2=1681

2w^2-98w+720=0

w^2-49w+360=0

factror or use the quadratic formula

360=2*2*2*3*3*5=(2*2*2*5)*(3*3)=40*9 and 40+9=49, the coefficient of the middle term

(w-40)(w-9)=0

w-40=0 and w-9=0

w=40 or w=9

length=40 and width=9 or vice-versa

A=l*w

A=40*9

A=360 square units

note: 40^2+9^2=41^2, 1600+81=1681 and 41^2=1681

Answer 3

Given: Hypotenuse = W^2 + L^2 = 41

Given: Perimeter = 2W + 2L = 98

Then W+L = 49, W = 49 - L

(49-L)^2 + L^2 = 41

49^2 - 98L + 2L^2 = 41

2L^2 -98L = 41 - 49^2

L^2 - 49L + 1180 = 0

Using the Quadratic Formula, finding the Discriminant:

B^2 - 4AC = 49^2 - 4*1*1180 = 2401 - 4720 = -2319

So the Discriminant says that the roots are imaginary and thus Area cannot be found in the real sense.