Right Triangles and Similar Triangles

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SAT Math › Right Triangles and Similar Triangles

Questions 1 - 4

Aas

Find the area of the triangle.

$18\sqrt{3} + 18$

$6\sqrt{3} + 6$

$12\sqrt{3}$

$12\sqrt{3} + 12$

$6\sqrt{3} + 12$

Explanation

Aas_key

Dropping the altitude creates two special right triangles as shown in the diagram. Use the area formula of a triangle to get

$A = \frac {1}{2}bh = \frac {1}{2} \cdot (6 \sqrt{3} + 6) \cdot 6 = 18\sqrt{3} + 18$

What is the length of the leg of a right triangle whose hypotenuse is 5cm and other leg is 4cm?

Explanation

One leg is 4cm and the hypotenuse is 5cm. Plug in 4 for one of the legs and 5 for the hypotenuse (c).

Subtracting 16 from either side of the equation gives us:

The last step is to take the square root both sides resulting in:

Los

Find the length of side .

Explanation

In an angle-side-angle problem, Law of Sines will solve the triangle.

First find angle A:

Then use Law of Sines.

$\frac {AB}{\sin 83^\circ} = \frac {4}{\sin 53^\circ} \rightarrow AB = \frac {4 \sin 83^\circ}{\sin 53^\circ} \approx 5.0$

Fire tower A is miles due west of fire tower B. Fire tower A sees a fire in the direction degrees west of north. Fire tower B sees the same fire in the direction degrees east of north. Which tower is closer to the fire and by how much?

Fire tower B; 0.24 miles

Fire tower A; 0.24 miles

Fire tower B; 1.29 miles

Fire tower A; 1.53 miles

The two fire towers are equidistant to the fire.

Explanation

Fire

First, realize that the angles given are from due north, which means you need to find the complements to find the interior angles of the triangle. This triangle happens to be a right triangle, so the fast way to compute the distances is using trigonometry.

$\sin 50^\circ = \frac{AF}{2}$

$AF \approx 1.53 miles$

$\sin 40^\circ = \frac{BF}{2}$

$BF \approx 1.29 miles$

Fire tower B is miles closer to the fire.

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