Math 2 Quiz: Special Right Triangles

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Special Right TrianglesQuestion 1 of 9

Triangle $ABC$ has vertices at $A(0, 0)$ , $B(12, 0)$ , and $C(6, 6\sqrt{3})$ . Point $P$ is inside the triangle such that triangles $PAB$ , $PBC$ , and $PCA$ all have equal areas. If $P$ is at coordinates $(6, 2\sqrt{3})$ , what is the distance from $P$ to the nearest vertex?

4\sqrt{3}

6

2\sqrt{21}

8

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Math 2 Quiz

Math 2 Quiz: Special Right Triangles

Practice Special Right Triangles in Math 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Special Right Triangles, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$4\sqrt{3}$ (correct answer)
$6$
$2\sqrt{21}$
$8$

Explanation: Point P(6, 2√3) is the centroid of triangle ABC. Triangle ABC is equilateral with side length 12 (can be verified: AB = 12, BC = √((12-6)² + (0-6√3)²) = √(36 + 108) = 12, CA = √((6-0)² + (6√3-0)²) = 12). The distance from P to each vertex: PA = √(6² + (2√3)²) = √(36 + 12) = √48 = 4√3. PB = √((6-12)² + (2√3)²) = √(36 + 12) = 4√3. PC = √((6-6)² + (2√3-6√3)²) = √(0 + 48) = 4√3. All distances are equal to 4√3. Choice B would be for a different triangle. Choice C uses incorrect calculation. Choice D is too large.

Question 2

A ladder leans against a wall at a $60°$ angle with the ground. If the ladder is $20$ feet long and the bottom of the ladder is moved $5$ feet closer to the wall, what is the new height the ladder reaches on the wall?

$10\sqrt{3} + 5\sqrt{3}$ feet
$5\sqrt{15}$ feet (correct answer)
$15\sqrt{3}$ feet
$5\sqrt{11}$ feet

Explanation: Initially, the ladder forms a 30°-60°-90° triangle with the wall. The distance from wall is 20 × cos(60°) = 10 feet, so the new distance is 5 feet. Using the Pythagorean theorem: new height = √(20² - 5²) = √(400 - 25) = √375 = √(25 × 15) = 5√15 feet. Choice A incorrectly adds the initial height to some offset. Choice C assumes the new angle is still 60°. Choice D has an arithmetic error in the radical simplification.

Question 3

In a $30°-60°-90°$ triangle, the side opposite the $30°$ angle has length $x$ . A second $45°-45°-90°$ triangle has the same perimeter as the first triangle. What is the length of each leg of the $45°-45°-90°$ triangle in terms of $x$ ?

$\frac{x(3 + \sqrt{3})}{2 + \sqrt{2}}$ (correct answer)
$\frac{x(3 + \sqrt{3})}{2(1 + \sqrt{2})}$
$\frac{x(2 + \sqrt{3})}{1 + \sqrt{2}}$
$\frac{x(3 + \sqrt{3})}{1 + \sqrt{2}}$

Explanation: In the 30°-60°-90° triangle with shortest side x, the sides are x, x√3, and 2x. Perimeter = x + x√3 + 2x = x(3 + √3). In the 45°-45°-90° triangle with legs of length y, the sides are y, y, and y√2. Perimeter = 2y + y√2 = y(2 + √2). Setting equal: x(3 + √3) = y(2 + √2), so y = x(3 + √3)/(2 + √2). Choice B has an extra factor of 2 in denominator. Choice C uses wrong coefficients. Choice D omits the factor of 2 in the denominator.

Question 4

In the coordinate plane, point A is at the origin and point B is at $(6, 6\sqrt{3})$ . Point C is chosen so that triangle ABC is a 30-60-90 triangle with the right angle at A. If C is in the first quadrant, what are the coordinates of point C?

$(6\sqrt{3}, -6)$
$(6\sqrt{3}, 6)$ (correct answer)
$(-6\sqrt{3}, 6)$
$(3\sqrt{3}, -3)$

Explanation: Vector

\overrightarrow{AB} = (6, 6\sqrt{3})

has length

\sqrt{6^2 + (6\sqrt{3})^2} = \sqrt{36 + 108} = \sqrt{144} = 12

. For a right angle at A, vector

\overrightarrow{AC}

must be perpendicular to

\overrightarrow{AB}

. A vector perpendicular to

(6, 6\sqrt{3})

(-6\sqrt{3}, 6)

(6\sqrt{3}, -6)

. Since C must be in the first quadrant, we need both coordinates positive. We can scale the perpendicular vector: if

\overrightarrow{AC} = (6\sqrt{3}, 6)

, then C =

(6\sqrt{3}, 6)

. We can verify:

\overrightarrow{AB} \cdot \overrightarrow{AC} = 6(6\sqrt{3}) + 6\sqrt{3}(6) = 36\sqrt{3} + 36\sqrt{3} = 0

, confirming perpendicularity.

Question 5

In a 30-60-90 triangle, the side opposite the 30° angle has length $x$ . If this triangle is used as the base of a right circular cone with the hypotenuse as the slant height, what is the lateral surface area of the cone?

$4\pi x^2$
$\pi x^2 \sqrt{3}$
$2\pi x^2 \sqrt{3}$ (correct answer)
$2\pi x^2$

Explanation: This problem combines 30-60-90 triangle properties with cone surface area formulas. When you see special right triangles in 3D geometry problems, first identify all side lengths, then determine how the triangle relates to the solid figure. In a 30-60-90 triangle, the sides are in the ratio

1 : \sqrt{3} : 2

. If the side opposite the 30° angle has length

x

, then the side opposite the 60° angle has length

x\sqrt{3}

, and the hypotenuse has length

2x

. When this triangle forms the base of a cone with the hypotenuse as slant height, the cone's radius equals the triangle's base (

x\sqrt{3}

) and the slant height is

2x

. The lateral surface area formula for a cone is

\pi r \ell

, where

r

is the radius and

\ell

is the slant height. Substituting: Lateral surface area =

\pi \cdot x\sqrt{3} \cdot 2x = 2\pi x^2\sqrt{3}

Looking at the wrong answers: Choice A (

4\pi x^2

) likely results from using

2x

as both radius and slant height. Choice B (

\pi x^2\sqrt{3}

) correctly identifies the component parts but forgets to multiply by 2 in the slant height. Choice D (

2\pi x^2

) uses the correct slant height but mistakes the radius as

x

instead of

x\sqrt{3}

. For special right triangle problems involving 3D figures, always write out the complete side ratio first, then carefully identify which sides correspond to which parts of the solid. Double-check that you're using the correct geometric measurements in your formulas.

Question 6

An equilateral triangle has side length $12$ . A circle is inscribed in the triangle, and then a square is inscribed in the circle. What is the side length of the square?

$2\sqrt{3}$
$4\sqrt{3}$
$2\sqrt{6}$ (correct answer)
$4\sqrt{6}$

Explanation: The equilateral triangle with side 12 has height 12 × (√3/2) = 6√3. The inscribed circle has radius r = (1/3) × height = 2√3. When a square is inscribed in a circle of radius r, the diagonal of the square equals the diameter 2r = 4√3. If the square has side s, then s√2 = 4√3, so s = 4√3/√2 = 4√3/√2 × √2/√2 = 4√6/2 = 2√6. Choice A uses radius instead of diameter. Choice B forgets to divide by √2. Choice D uses diameter directly without the √2 factor.

Question 7

In triangle ABC, angle A = 30°, angle B = 60°, and the altitude from C to side AB has length $6\sqrt{3}$ . What is the length of side BC?

18
$6\sqrt{3}$
$12\sqrt{3}$
12 (correct answer)

Explanation: When you encounter a triangle problem with given angles and an altitude, think about how the altitude creates right triangles that you can solve using basic trigonometry. Since angles A = 30° and B = 60°, angle C must be 90° (since angles sum to 180°). This means triangle ABC is a 30-60-90 right triangle, which has special properties you should memorize. Let's call the foot of the altitude from C to AB point D. The altitude CD =

6\sqrt{3}

creates two right triangles. In right triangle BCD, angle B = 60° and the side opposite to B (which is CD) equals

6\sqrt{3}

. Using trigonometry:

\sin(60°) = \frac{CD}{BC} = \frac{6\sqrt{3}}{BC}

Since

\sin(60°) = \frac{\sqrt{3}}{2}

, we have:

\frac{\sqrt{3}}{2} = \frac{6\sqrt{3}}{BC}

Solving for BC:

BC = \frac{6\sqrt{3} \times 2}{\sqrt{3}} = 12

Looking at the wrong answers: Choice A (18) results from incorrectly using

\sin(30°)

instead of

\sin(60°)

. Choice B (

6\sqrt{3}

) is simply the given altitude length—a common trap where students confuse which measurement they're solving for. Choice C (

12\sqrt{3}

) comes from forgetting to simplify the radical expression properly. Study tip: For 30-60-90 triangles, memorize that sides are in the ratio

1 : \sqrt{3} : 2

. Also, when using trigonometry with altitudes, always identify which angle you're working with in the resulting right triangle.

Question 8

A rhombus has side length 8 and one of its angles measures 60°. What is the length of the shorter diagonal?

$4\sqrt{3}$
8 (correct answer)
$8\sqrt{3}$
$4\sqrt{2}$

Explanation: In a rhombus with side length 8 and one angle of 60°, the opposite angle is also 60°, and the other two angles are 120°. The diagonals of a rhombus bisect each other at right angles. Consider the triangle formed by two adjacent sides and the shorter diagonal. This is an isosceles triangle with two sides of length 8 and vertex angle 60°. Since this is isosceles with a 60° angle, it's actually equilateral, so the shorter diagonal also has length 8.

Question 9

A regular hexagon is inscribed in a circle of radius 8. If the hexagon is divided into 6 congruent triangles by drawing lines from the center to each vertex, what is the area of the entire hexagon?

$96\sqrt{3}$ (correct answer)
$48\sqrt{3}$
$64\sqrt{3}$
$32\sqrt{3}$

Explanation: When a regular hexagon is inscribed in a circle of radius 8, connecting the center to each vertex creates 6 congruent triangles. Each triangle has two sides of length 8 (the radii) and a central angle of 60°. This makes each triangle equilateral with all sides of length 8. The area of each equilateral triangle is

\frac{\sqrt{3}}{4} \cdot 8^2 = \frac{\sqrt{3}}{4} \cdot 64 = 16\sqrt{3}

. The total area of the hexagon is

6 \cdot 16\sqrt{3} = 96\sqrt{3}