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