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SAT Math : Triangles

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Example Questions

Example Question #3 : How To Find The Length Of The Side Of A Right Triangle

Bob the Helicopter is at 30,000 ft. above sea level, and as viewed on a map his airport is 40,000 ft. away. If Bob travels in a straight line to his airport at 250 feet per second, how many minutes will it take him to arrive?

Possible Answers:

2 hours and 30 minutes

4 hours and 0 minutes

3 minutes and 20 seconds

3 minutes and 50 seconds

1 hour and 45 minutes

Correct answer:

3 minutes and 20 seconds

Explanation:

Draw a right triangle with a height of 30,000 ft. and a base of 40,000 ft. The hypotenuse, or distance travelled, is then 50,000ft using the Pythagorean Theorem. Then dividing distance by speed will give us time, which is 200 seconds, or 3 minutes and 20 seconds.

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Example Question #4 : How To Find The Length Of The Side Of A Right Triangle

A right triangle has two sides, 9 and x, and a hypotenuse of 15. What is x?

Possible Answers:

Correct answer:

Explanation:

We can use the Pythagorean Theorem to solve for x.

9² + x² = 15²

81 + x² = 225

x² = 144

x = 12

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Example Question #3 : How To Find The Length Of The Side Of A Right Triangle

The area of a right traingle is 42. One of the legs has a length of 12. What is the length of the other leg?

Possible Answers:

$7$

$9$

$11$

$6$

$5$

Correct answer:

$7$

Explanation:

$Area= \frac{1}{2}\times base\times height$

$42=\frac{1}{2}\times base\times 12$

$42=6\times base$

$base=7$

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Example Question #81 : Triangles

Solve for x.

Possible Answers:

Correct answer:

Explanation:

Use the Pythagorean Theorem. Let a = 8 and c = 10 (because it is the hypotenuse)

$\small a^2+x^2=c^2$

$\small 8^2+x^2=10^2$

$\small 64+x^2=100$

$\small x^2=100-64=36$

$\small x=6$

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Example Question #2 : How To Find The Length Of The Side Of A Right Triangle

Solve each problem and decide which is the best of the choices given.

If $sin x =\frac{12}{13}$ , what is tan (x) ?

Possible Answers:

$tan(x)=\frac{12}{5}$

$tan(x)=\frac{5}{12}$

$tan(x)=\frac{5}{13}$

$tan(x)=\frac{12}{13}$

$tan(x)=\frac{13}{12}$

Correct answer:

$tan(x)=\frac{12}{5}$

Explanation:

This is a 5-12-13 triangle. We can find the value of the other leg by using the Pythagorean Theorem.

Remembering that

$sin=\frac{opposite}{hypotenuse}$ .

Thus,

a^2+12^2=13^2

a^2+144=169

a^2=25

a=5 .

If $sin(x)=\frac{12}{13}$ , you know the adjacent side is .

Thus, making

$tan(x)=\frac{opposite}{adjacent}=\frac{12}{5}$ because tangent is opposite/adjacent.

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Example Question #11 : How To Find The Length Of The Side Of A Right Triangle

Given $\bigtriangleup ABC$ with $m \angle A > m \angle B$ and $m \angle C = 90^{\circ }$ .

Which of the following could be the correct ordering of the lengths of the sides of the triangle?

I) AB > BC > AC

II) BC > AB > AC

III) BC > AC > AB

Possible Answers:

I or II only

II only

III only

I only

II or III only

Correct answer:

I only

Explanation:

Given two angles of unequal measure in a triangle, the side opposite the greater angle is longer than the side opposite the other angle. Since $m \angle A > m \angle B$ , it follows that BC > AC . Also, in a right triangle, the hypotenuse must be the longest side; in $\bigtriangleup ABC$ , since $\angle C$ is the right side, this hypotenuse is $\overline{AB}$ . It follows that AB > BC > AC , and that (I) is the only statement that can possibly be true.

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Example Question #82 : Triangles

Triangle

$\left ( \bigtriangleup ABC \textup{ is a right triangle where } \angle B \textup{ is a right angle}\right)$

If $\angle A=30^{\circ}$ and $\overline{AB}=4$ , what is the length of $\overline{BC}$ ?

Possible Answers:

$\frac{4}{3}$

$\frac{4\sqrt{3}}{3}$

Correct answer:

$\frac{4\sqrt{3}}{3}$

Explanation:

AB is the leg adjacent to Angle A and BC is the leg opposite Angle A.

Since we have a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle, the opposites sides of those angles will be in the ratio $x-x\sqrt{3}-2x$ .

Here, we know the side opposite the sixty degree angle. Thus, we can set that value equal to $x\sqrt{3}$ .

$4=x\sqrt{3}$

$\frac{4}{\sqrt{3}}=\frac{x\sqrt{3}}{\sqrt{3}}$

$\frac{4}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}=x$

$x=\frac{4\sqrt{3}}{3}$

which also means

$\overline{BC}=\frac{4\sqrt{3}}{3}$

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Example Question #12 : How To Find The Length Of The Side Of A Right Triangle

A single-sided ladder is leaning against a wall. The angle between the end of the ladder that is on the ground and the ground itself is represented by $x^{\circ}$ . The ladder is sliding down the wall at a rate of 6 feet per second. If

$\sin x^{\circ}=\frac{3}{5}$

how many seconds does it take for the ladder to fall all the way to the ground? (The wall is a right angle to the ground.)

Possible Answers:

$1 \text{ second}$

$6 \text{ seconds}$

$2.2 \text{ seconds}$

$0.5 \text{ seconds}$

$1.1 \text{ seconds}$

Correct answer:

$0.5 \text{ seconds}$

Explanation:

The ladder leaning against the wall forms a right triangle. The hypotenuse of the triangle is 5 ft., the length of the ladder.

Because sin x= opposite/hypotenuse, sine of the angle is equal to the length of the side opposite the angle divided by the hypotenuse. In this case, the length of the side opposite the angle is h, the height of the end of the ladder that is touching the wall. Thus,

$\sin x^{\circ} = \frac{h}{5}$

Because we are told that

$\sin x^{\circ}=\frac{3}{5}$

we know that h=3. Therefore, 3 feet is the height of the ladder. If the ladder is falling at a rate of 6 feet per second, we can find the number of seconds it will take the ladder to hit the ground with the equation

h=6s

where h represents the height the ladder is falling from, and s represents the number of seconds it takes the ladder to fall. We can now solve for s:

$s=\frac{h}{6}$

$s=\frac{3}{6}=\frac{1}{2}=0.5$

It takes the ladder 0.5 seconds to fall to the ground.

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Example Question #13 : How To Find The Length Of The Side Of A Right Triangle

Right triangle 7

Refer to the provided figure. Give the length of $\overline{BC}$ .

Possible Answers:

$\frac{20}{3} \sqrt{3}$

$\frac{20}{3} \sqrt{2}$

$10\sqrt{2}$

$10\sqrt{3}$

Correct answer:

$10\sqrt{2}$

Explanation:

The figure shows a right triangle. The acute angles of a right triangle have measures whose sum is $90^{\circ }$ , so

$m \angle A+ m \angle C = 90^{\circ }$

Substituting $45^{\circ }$ for $m \angle C$ :

$m \angle A+ 45^{\circ } = 90^{\circ }$

$m \angle A= 45^{\circ }$

This makes $\bigtriangleup ABC$ a 45-45-90 triangle. By the 45-45-90 Triangle Theorem, the length of leg $\overline{BC}$ is equal to that of hypotenuse $\overline{AC}$ , the length of which is 20. Therefore,

$BC = \frac{AC}{\sqrt{2}} = \frac{20}{\sqrt{2}}$

Rationalize the denominator by multiplying both halves of the fraction by $\sqrt{2}$ :

$BC = \frac{20 \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \frac{20 \sqrt{2}}{2} = 10 \sqrt{2}$

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Example Question #1 : How To Find If Right Triangles Are Similar

Triangles

In the figure above, line segments DC and AB are parallel. What is the perimeter of quadrilateral ABCD?

Possible Answers:

Correct answer:

Explanation:

Because DC and AB are parallel, this means that angles CDB and ABD are equal. When two parallel lines are cut by a transversal line, alternate interior angles (such as CDB and ABD) are congruent.

Now, we can show that triangles ABD and BDC are similar. Both ABD and BDC are right triangles. This means that they have one angle that is the same—their right angle. Also, we just established that angles CDB and ABD are congruent. By the angle-angle similarity theorem, if two triangles have two angles that are congruent, they are similar. Thus triangles ABD and BDC are similar triangles.

We can use the similarity between triangles ABD and BDC to find the lengths of BC and CD. The length of BC is proportional to the length of AD, and the length of CD is proportional to the length of DB, because these sides correspond.

We don’t know the length of DB, but we can find it using the Pythagorean Theorem. Let a, b, and c represent the lengths of AD, AB, and BD respectively. According to the Pythagorean Theorem:

a² + b²= c²

15² + 20²= c²

625 = c²

c = 25

The length of BD is 25.

Similar_triangles

We now have what we need to find the perimeter of the quadrilateral.

Perimeter = sum of the lengths of AB, BC, CD, and DA.

Perimeter = 20 + 18.75 + 31.25 + 15 = 85

The answer is 85.

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SAT Math : Triangles

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #3 : How To Find The Length Of The Side Of A Right Triangle

Example Question #4 : How To Find The Length Of The Side Of A Right Triangle

Example Question #3 : How To Find The Length Of The Side Of A Right Triangle

Example Question #81 : Triangles

Example Question #2 : How To Find The Length Of The Side Of A Right Triangle

Example Question #11 : How To Find The Length Of The Side Of A Right Triangle

Example Question #82 : Triangles

Example Question #12 : How To Find The Length Of The Side Of A Right Triangle

Example Question #13 : How To Find The Length Of The Side Of A Right Triangle

Example Question #1 : How To Find If Right Triangles Are Similar

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