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The length of the diagonal of a certain rectangle is 8. If

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by deloitte247 » Tue Apr 09, 2019 7:26 am
$$Diagonal\ of\ rec\tan gle\ =\sqrt{l^2+w^2}=8$$
Square both sides, we have
$$\left(\sqrt{l^2+w^2}\right)^2=8^2$$
$$l^2+w^2=8^2$$
$$Sum\ of\ the\ lengths\ of\ the\ sides=10$$
$$\left(l+w\right)=10$$
$$Note\ that\ the\ area\ of\ a\ rec\tan gle\ is\ equal\ to\ l\cdot w$$
$$So,\ Area=l\cdot w$$
$$\left(l+w\right)^2=\left(l^2+w^2\right)+2lw$$
$$\left(10\right)^2=\left(8^2\right)+2lw$$
$$100-64=2lw$$
$$lw=\frac{36}{2}=18\ \ \ \ \ \ \ \ \left(Option\ A\right)$$
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by Scott@TargetTestPrep » Wed Apr 10, 2019 4:49 pm
swerve wrote:The length of the diagonal of a certain rectangle is 8. If the sum of the lengths of the sides is 10, what is the area of the rectangle?

A. 18
B. 36
C. 64
D. 80
E. It cannot be determined.

The OA is A

Source: Veritas Prep
We can use the Pythagorean theorem to create the following equation:

L^2 + W^2 = 8^2

L^2 + W^2 = 64

We also know the sum of the length and the width:

L + W = 10

Squaring both sides of the equation, we have:

(L + W)^2 = 10^2

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

Since we already know that L^2 + W^2 = 64, we have:

64 + 2LW = 100

2LW = 36

LW = 18, which is the area of the rectangle.

Answer: A

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