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Difficulty: HardPythagorean Theorem and Special Right Triangles

In right triangle ABCABC, the measure of ∠B\angle B is 90∘90^\circ, the measure of ∠A\angle A is 60∘60^\circ, and the hypotenuse ACAC has a length of 2020 centimeters. Point DD lies on leg BCBC such that the length of segment BDBD is 232\sqrt{3} centimeters. A line segment DEDE is drawn perpendicular to ACAC such that EE lies on ACAC. What is the length, in centimeters, of segment AEAE?

Answer: 8 cm

Answer

8
The correct answer is 8. By solving for the angles and side lengths of the two nested 30∘-60∘-90∘30^\circ\text{-}60^\circ\text{-}90^\circ right triangles, we find that the segment BCBC is 10310\sqrt{3} cm, making DC=83DC = 8\sqrt{3} cm. Using the ratio of sides for the smaller right triangle DECDEC, we find EC=12EC = 12 cm, which leaves AE=20−12=8AE = 20 - 12 = 8 cm.

Step-by-Step Solution

1
Determine the third angle of right triangle ABCABC.
∠C=30∘\angle C = 30^\circ
The sum of angles in a triangle is 180∘180^\circ. Since ∠B=90∘\angle B = 90^\circ and ∠A=60∘\angle A = 60^\circ, we have ∠C=180∘−90∘−60∘=30∘\angle C = 180^\circ - 90^\circ - 60^\circ = 30^\circ.
2
Calculate the length of the side BCBC.
BC=103BC = 10\sqrt{3} cm
In the 30∘-60∘-90∘30^\circ\text{-}60^\circ\text{-}90^\circ triangle ABCABC, the side BCBC is opposite the 60∘60^\circ angle, so its length is the hypotenuse ACAC multiplied by sin⁡(60∘)\sin(60^\circ) or 32\frac{\sqrt{3}}{2}. Thus, BC=20×32=103BC = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}.
3
Find the length of segment DCDC.
DC=83DC = 8\sqrt{3} cm
Since point DD lies on segment BCBC, the length of DCDC is the total length of BCBC minus the length of BDBD. Since BD=23BD = 2\sqrt{3}, we have DC=103−23=83DC = 10\sqrt{3} - 2\sqrt{3} = 8\sqrt{3}.
4
Determine the properties of the right triangle DECDEC.
△DEC\triangle DEC is a 30∘-60∘-90∘30^\circ\text{-}60^\circ\text{-}90^\circ triangle with hypotenuse DC=83DC = 8\sqrt{3} cm.
Since segment DEDE is perpendicular to ACAC, ∠DEC=90∘\angle DEC = 90^\circ. Triangle DECDEC shares the angle ∠C=30∘\angle C = 30^\circ with triangle ABCABC, which makes it a 30∘-60∘-90∘30^\circ\text{-}60^\circ\text{-}90^\circ triangle where DCDC is the hypotenuse.
5
Calculate the length of segment ECEC.
EC=12EC = 12 cm
In the 30∘-60∘-90∘30^\circ\text{-}60^\circ\text{-}90^\circ triangle DECDEC, the leg ECEC is adjacent to the 30∘30^\circ angle, so its length is the hypotenuse DCDC multiplied by cos⁡(30∘)\cos(30^\circ) or 32\frac{\sqrt{3}}{2}. Thus, EC=83×32=12EC = 8\sqrt{3} \times \frac{\sqrt{3}}{2} = 12.
6
Calculate the length of segment AEAE.
AE=8AE = 8 cm
Since point EE lies on segment ACAC, we can find AEAE by subtracting ECEC from ACAC. Thus, AE=AC−EC=20−12=8AE = AC - EC = 20 - 12 = 8.

Key Concept

Using the properties of 30∘-60∘-90∘30^\circ\text{-}60^\circ\text{-}90^\circ special right triangles to find missing lengths in composite geometric configurations.
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Topics

Pythagorean Theorem and Special Right Triangles

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