A diagonal of a rectangle has length 41, and the perimeter is 98. Find the area of the rectangle.

Thanks guys,     Hi Mellie    

A diagonal of a rectangle has length 41, and the perimeter is 98. Find the area of the rectangle.

This is a really neat completing the squares question !

2(L+B)=98

so

L+B=49

$L^2+B^2=41^2\\ \mbox{complete the square}\\ L^2+B^2+2BL=41^2+2BL\\ (L+B)^2=41^2+2BL\\ sub\;\;L+B=49\\ 49^2=41^2+2BL\\ 49^2-41^2=2BL\\ BL=\frac{49^2-41^2}{2}\\ AREA=\frac{49^2-41^2}{2}\\ AREA=360 \;units^2$

Let length=L

Let width =W

L^2 + W^2=41^2

2[L + W] =98

L = 40 and W = 9

Area= L X W=40 X 9=360

Let length=L

Let width =W

L^2 + W^2=41^2

2[L + W] =98

2L + 2W=98

2L=98 - 2W

2L=2[49 - W] divide both sides by 2

L=49 - W Substitute

[49 - W]^2 + W^2=41^2=1,681

Solve for W:
(49-W)^2+W^2 = 1681

Expand out terms of the left hand side:
2 W^2-98 W+2401 = 1681

Divide both sides by 2:
W^2-49 W+2401/2 = 1681/2

Subtract 2401/2 from both sides:
W^2-49 W = -360

Add 2401/4 to both sides:
W^2-49 W+2401/4 = 961/4

Write the left hand side as a square:
(W-49/2)^2 = 961/4

Take the square root of both sides:
W-49/2 = 31/2 or W-49/2 = -31/2

Add 49/2 to both sides:
W = 40 or W-49/2 = -31/2

Add 49/2 to both sides:
Answer: | W = 40                      or                 W = 9

2(L + W)  = 98

L + W  = 49

L = 49 - W

L^2 + W^2  = 41^2

(49 - W)^2 + W^2  = 1681

2W^2 - 98W + 2401 = 1681

2W^2 - 98W + 720  = 0

W^2 - 49W + 360  = 0

(W - 40) (W - 9)  = 0

W  = 40 and L = 9     or W = 9 and L = 40        [ depending upon your orientation ]

And the area = L * W   = 40 * 9   = 360  units^2

Let x and y be the dimensions of the rectangle. Then from the given information $\sqrt{x^2 + y^2} = 41$ and , so x + y = 49.

Squaring the equation $\sqrt{x^2 + y^2} = 41$, we get $x^2 + y^2 = 1681.$

Squaring the equation x + y= 49, we get $x^2 + 2xy + y^2 = 2401$

Subtracting these equations, we get 2xy = 720, so the area of the rectangle is 360 degrees.