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If a rectangle has perimeter of 20 and a diagonal with length 9, what is the area of the rectangle?

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Re: If a rectangle has perimeter of 20 and a diagonal with length 9, what is the area of the rectangle?

by ravimani » Tue Oct 27, 2020 12:14 pm
let sides be a and b

perimeter = 2(a+b) = 20
so, a+b = 10

next, diagonal length = 9. So by right angled triangle theorem.
$$a^2\ +\ b^2\ =\ c^2$$

so,
$$a^2\ +\ b^2\ = 81$$

Now, we know that
$$\left(a+b\right)^2\ =\ a^2\ +\ b^2\ +\ 2ab$$

Let's replace the value with given values
$$10^2\ =\ a^2\ +\ b^2\ +\ 2ab$$

100 = 81 + 2ab
2ab = 19
ab = 9.5

So, area of rectable => a*b = 9.5
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Re: If a rectangle has perimeter of 20 and a diagonal with length 9, what is the area of the rectangle?

by Scott@TargetTestPrep » Sat Oct 31, 2020 2:13 pm
AAPL wrote: ↑
Mon Oct 26, 2020 4:21 am
 Magoosh

If a rectangle has perimeter of 20 and a diagonal of 9, what is the area of the recatngle?

A. 7.5
B. 8
C. 8.5
D. 9
E. 9.5

OA E
Solution:

We can let L = the length and W = the width of the rectangle. Our goal is to determine the value of LW since that is the area of the rectangle. We can create two equations:

By the Pythagorean theorem:

L^2 + W^2 = 9^2 = 81

and

by the formula for the perimeter of a rectangle:

2L + 2W = 20

L + W = 10

Squaring both sides of L + W = 10, we have:

L^2 + W^2 + 2LW = 100

Substituting, we have:

81 + 2LW = 100

2LW = 19

LW = 9.5

Answer: E

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