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The length of a rectangle's diagonal is
2sqrt29, and the length of the longer side is 10 . What is the area of the rectangle?
A) 116
B)
20sqrt29
C)
10sqrt29
D) 40

The length of a rectangle's diagonal is $2 \sqrt{29}$ , and the length of the longer side is $10$ . What is the area of the rectangle? $\newline$ A) $116$ $\newline$ B) $20 \sqrt{29}$ $\newline$ C) $10 \sqrt{29}$ $\newline$ D) $40$

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Transformations of functions

Full solution

Q. The length of a rectangle's diagonal is

2 \sqrt{29}

, and the length of the longer side is

10

. What is the area of the rectangle?

\newline

A)

116

\newline

B)

20 \sqrt{29}

\newline

C)

10 \sqrt{29}

\newline

D)

40

Identify Rectangle Dimensions: To find the area of the rectangle, we need to know both the length and the width. We already know the length is $10$ . We can use the Pythagorean theorem to find the width, since the diagonal ( $d$ ), length ( $l$ ), and width ( $w$ ) of a rectangle form a right triangle: $d^2 = l^2 + w^2$ .
Calculate Diagonal Squared: We are given that the diagonal $d$ is $2\sqrt{29}$ . Let's square this to find $d^2$ : $(2\sqrt{29})^2 = 4 \times 29 = 116$ .
Calculate Length Squared: We know the length $l$ is $10$ , so let's square this to find $l^2$ : $10^2 = 100$ .
Substitute into Pythagorean Theorem: Now we can substitute $d^2$ and $l^2$ into the Pythagorean theorem to find $w^2$ : $116 = 100 + w^2$ .
Solve for Width: Subtract $100$ from both sides to solve for $w^2$ : $116 - 100 = w^2$ , so $w^2 = 16$ .
Find Width: Take the square root of both sides to find $w$ : $\sqrt{w^2} = \sqrt{16}$ , so $w = 4$ .
Calculate Area: Now that we have both the length ( $10$ ) and the width ( $4$ ), we can find the area of the rectangle by multiplying the length by the width: $\text{Area} = \text{length} \times \text{width} = 10 \times 4$ .
Final Area Calculation: Calculate the area: $10 \times 4 = 40$ .

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