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SAT II Math I : Right Triangles and Similar Triangles

Study concepts, example questions & explanations for SAT II Math I

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Example Questions

Example Question #1 : Right Triangles And Similar Triangles

Aas

Find the area of the triangle.

Possible Answers:

$12\sqrt{3}$ $\displaystyle 12\sqrt{3}$

$6\sqrt{3} + 6$ $\displaystyle 6\sqrt{3} + 6$

$6\sqrt{3} + 12$ $\displaystyle 6\sqrt{3} + 12$

$12\sqrt{3} + 12$ $\displaystyle 12\sqrt{3} + 12$

$18\sqrt{3} + 18$ $\displaystyle 18\sqrt{3} + 18$

Correct answer:

$18\sqrt{3} + 18$ $\displaystyle 18\sqrt{3} + 18$

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

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

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Example Question #1 : Right Triangles And Similar Triangles

Fire tower A is $\displaystyle 2$ miles due west of fire tower B. Fire tower A sees a fire in the direction $\displaystyle 40$ degrees west of north. Fire tower B sees the same fire in the direction $\displaystyle 50$ degrees east of north. Which tower is closer to the fire and by how much?

Possible Answers:

Fire tower B; 0.24 miles

Fire tower A; 1.53 miles

Fire tower A; 0.24 miles

Fire tower B; 1.29 miles

The two fire towers are equidistant to the fire.

Correct answer:

Fire tower B; 0.24 miles

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.

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

$AF \approx 1.53 miles$ $\displaystyle AF \approx 1.53 miles$

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

$BF \approx 1.29 miles$ $\displaystyle BF \approx 1.29 miles$

Fire tower B is $\displaystyle 1.53 - 1.29 = 0.24$ miles closer to the fire.

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Example Question #2 : Right Triangles And Similar Triangles

Los

Find the length of side $\displaystyle AB$.

Possible Answers:

5.0 $\displaystyle 5.0$

5.8 $\displaystyle 5.8$

5.3 $\displaystyle 5.3$

5.2 $\displaystyle 5.2$

4.9 $\displaystyle 4.9$

Correct answer:

5.0 $\displaystyle 5.0$

Explanation:

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

First find angle A:

$\displaystyle 180-(83+44)=53$

Then use Law of Sines.

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

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Example Question #1 : Right Triangles And Similar Triangles

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

Possible Answers:

10cm $\displaystyle 10cm$

6.40cm $\displaystyle 6.40cm$

9cm $\displaystyle 9cm$

3cm $\displaystyle 3cm$

4cm $\displaystyle 4cm$

Correct answer:

3cm $\displaystyle 3cm$

Explanation:

$\displaystyle Pythagorean Theorem: a^2+b^2=c^2$

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

$\displaystyle a^2+4^2=5^2$

$\displaystyle a^2+16=25$

Subtracting 16 from either side of the equation gives us:

a^2=9 $\displaystyle a^2=9$

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

a=3cm $\displaystyle a=3cm$

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SAT II Math I : Right Triangles and Similar Triangles

Study concepts, example questions & explanations for SAT II Math I

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Example Question #1 : Right Triangles And Similar Triangles

Example Question #1 : Right Triangles And Similar Triangles

Example Question #2 : Right Triangles And Similar Triangles

Example Question #1 : Right Triangles And Similar Triangles

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