A rectangular community garden has a straight gravel path connecting one corner of the garden to the opposite corner. The path is meters long. At one end of the path, the angle between the path and one side of the garden is . Which choice is the area of the garden, in square meters?

Solution available with step-by-step explanation

SAT Right Triangles & Trigonometry Question 47 | Free SAT Prep | Aniko

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Question 47·Medium·Right Triangles and Trigonometry

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A rectangular community garden has a straight gravel path connecting one corner of the garden to the opposite corner. The path is $24$ meters long. At one end of the path, the angle between the path and one side of the garden is $60^\circ$ .

Which choice is the area of the garden, in square meters?

When a diagonal and an angle in a rectangle are given, draw the right triangle formed by the diagonal (hypotenuse) and the side lengths (legs). Use $\cos(\theta)$ and $\sin(\theta)$ to find the adjacent and opposite legs from the hypotenuse, then multiply the two side lengths to get the area.

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Use the diagonal as a hypotenuse

A rectangle’s diagonal and its two side lengths form a right triangle.

Assign adjacent and opposite sides

Relative to the $60^\circ$ angle, one side length is adjacent and the other is opposite. Use cosine for adjacent and sine for opposite.

Multiply to get area

Once you have both side lengths, the rectangle’s area is their product.

Desmos Guide

Compute the side lengths from the diagonal

In Desmos, enter:

$a=24\cos(60^\circ)$
$b=24\sin(60^\circ)$

Compute the area

Enter $A=a\,b$ to calculate the rectangle’s area.

Match to an answer choice

Compare the computed value of $A$ to the choices; it should match one exactly.

Step-by-step Explanation

Model the situation as a right triangle

The garden’s diagonal (the path) is the hypotenuse of a right triangle whose legs are the side lengths of the rectangle. Let the side adjacent to the $60^\circ$ angle be $a$ and the opposite side be $b$ .

Use cosine and sine to find the side lengths

Using the hypotenuse $24$ :

\cos 60^\circ=\frac{a}{24}\Rightarrow a=24\cos 60^\circ=24\cdot\frac{1}{2}=12.

\sin 60^\circ=\frac{b}{24}\Rightarrow b=24\sin 60^\circ=24\cdot\frac{\sqrt{3}}{2}=12\sqrt{3}.

Compute the area of the rectangle

The area is

A=ab=(12)(12\sqrt{3})=144\sqrt{3}.

Answer: $144\sqrt{3}$

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation