The perimeter of a rectangle is 98, and the length of one of its diagonals is 41. Find the area of the rectangle.

Question

can i get help on how to solve this?

Answer 1

Draw and label a diagram!

The short side of the triangle is "x"

The lone side of the triangle is "49 - x"

By Pythagoras

41² = x² + (49 - x)²

Can you solve for x and answer?

Answer 2

580 Square units

Answer 3

If x and y sides of rectangle then we have 2 equations:

x + y = 49 and x² + y² = 41².

If we square first equation we get: x² + 2xy + y² = 49².

From this 2 equations we have 41² + 2xy = 49².

From here xy = (49² - 41²/2 = 360. But xy is area of rectangle.

Answer: area of rectangle equel 360 square units.

Answer 4

2(L+W)=98 => L+W=49 => W=49-L

sqrt(L²+W²)=41 => L²+W² = 1681

Substitute: L²+(49-L)²=1681

Expand and collect terms and then divide by 2 to get: L²-49L+720=0

The quadratic formula give L=40 or L=9, which means W=9 or W=40.

Answer 5

P = 2 L + 2 w

98 = 2 L + 2 w

49 = L + w

49 - w = L

L^2 + w^2 = 41^2

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

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

w^2 - 98w + 720 = 0

(w - 90)(w - 8) = 0

w=90 results in negative measures

w=8 ---> L = 41

Area is 328

Answer 6

Let a and b be the sides of the rectangle, and the diagonal - 41.

Using Pythagorean theorem

a² + b² = 41²

This is the first equation. We need the second to find out both A and B.

It is the perimeter:

2a + 2b = 98

Next you solve this system of equations (substitution method)

And find the area as a*b

Answer 7

draw a diagram, a rectangle with a diagonal labeled 41, l and w

P=2l+2w

98=2l+2w

49=l+w

l=49-w

use the Pythagorean Theorem

the hypotenuse is 41, one leg is w and the other leg is 49-w

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

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

1681=2401-98w+2w^2

2w^2-98w+720=0

w^2-49w+360=0

factor

360=2*2*2*3*3*5

group the factors such that you have two numbers whose product is 360 and whose sum is 49

2*2*2*3*3*5=(2*2*2*5)*(3*3)=40*9, 40*9=360 and 40+9=49

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

w-40=0

w=40

w-9=0

w=9

the length and width are 40 and 9 or 9 and 40

the area of a rectangle is A=lw

A=40*9

A=360 sq units

notice that you could have stopped at the equation above the word "factor" but you wouldn't know the length and width

Answer 8

Hi Emma,

The Pythagorean Theorem can be used to relate the length of the diagonal to the measure of the rectangle's length and width:

Length^2 + Width^2 = Diagonal^2 = 41^2

The bit about the perimeter gives you another equation:

Perimeter = 2 (Length + Width) = 98

But this can be rearranged to obtain

Length = 49 - Width

... which we can substitute into the first equation:

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

Width^2 - 49 Width + 360 = 0

(Width - 40) (Width - 9) = 0

So the rectangle is either 40 x 9 or 9 x 40.

Answer 9

In case it's not obvious....

Once you have x, use the perimeter to find y. Then--once you have y, you can find the area