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
The length of the diagonal of a certain rectangle is 8. If
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$$Diagonal\ of\ rec\tan gle\ =\sqrt{l^2+w^2}=8$$
Square both sides, we have
$$\left(\sqrt{l^2+w^2}\right)^2=8^2$$
$$l^2+w^2=8^2$$
$$Sum\ of\ the\ lengths\ of\ the\ sides=10$$
$$\left(l+w\right)=10$$
$$Note\ that\ the\ area\ of\ a\ rec\tan gle\ is\ equal\ to\ l\cdot w$$
$$So,\ Area=l\cdot w$$
$$\left(l+w\right)^2=\left(l^2+w^2\right)+2lw$$
$$\left(10\right)^2=\left(8^2\right)+2lw$$
$$100-64=2lw$$
$$lw=\frac{36}{2}=18\ \ \ \ \ \ \ \ \left(Option\ A\right)$$
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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We can use the Pythagorean theorem to create the following equation: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
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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