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Difficulty: HardPythagorean Theorem and Special Right Triangles

In the right trapezoid ABCDABCD below, ABAB is parallel to CDCD, and the measures of ∠A\angle A and ∠D\angle D are both 90∘90^\circ. The length of CDCD is 77, the length of BCBC is 88, and the measure of ∠B\angle B is 60∘60^\circ. What is the length of the diagonal BDBD?

Answer: 13

Answer

13
By drawing altitude CECE perpendicular to ABAB, we form rectangle AECDAECD and right triangle △CEB\triangle CEB. Since ∠B=60∘\angle B = 60^\circ, △CEB\triangle CEB is a 30∘30^\circ-60∘60^\circ-90∘90^\circ triangle with hypotenuse BC=8BC = 8. The side opposite 30∘30^\circ is BE=8/2=4BE = 8/2 = 4, and the side opposite 60∘60^\circ is CE=43CE = 4\sqrt{3}. Since opposite sides of rectangle AECDAECD are equal, we find AD=CE=43AD = CE = 4\sqrt{3} and AE=CD=7AE = CD = 7. Thus, AB=AE+BE=7+4=11AB = AE + BE = 7 + 4 = 11. Finally, we apply the Pythagorean Theorem to right triangle △DAB\triangle DAB: BD2=AD2+AB2=(43)2+112=48+121=169BD^2 = AD^2 + AB^2 = (4\sqrt{3})^2 + 11^2 = 48 + 121 = 169, yielding BD=13BD = 13.

Step-by-Step Solution

1
Decompose the trapezoid by drawing an altitude from CC perpendicular to ABAB, meeting it at EE.
This forms a rectangle AECDAECD and a right triangle CEBCEB.
Decomposing the figure allows us to use right triangle trigonometry and parallel line relationships to determine missing side lengths.
2
Calculate the lengths of the legs of right triangle CEBCEB.
BE=4BE = 4 and CE=43CE = 4\sqrt{3}.
Since ∠B=60∘\angle B = 60^\circ, △CEB\triangle CEB is a 30∘30^\circ-60∘60^\circ-90∘90^\circ triangle. The shorter leg BEBE is half the hypotenuse BCBC, and the longer leg CECE is the shorter leg times 3\sqrt{3}.
3
Determine the lengths of ADAD and ABAB.
AD=43AD = 4\sqrt{3} and AB=11AB = 11.
In the rectangle AECDAECD, opposite sides are equal, so AD=CE=43AD = CE = 4\sqrt{3} and AE=CD=7AE = CD = 7. Thus, the base AB=AE+BE=7+4=11AB = AE + BE = 7 + 4 = 11.
4
Use the Pythagorean Theorem in right triangle DABDAB to solve for BDBD.
BD=13BD = 13.
In right triangle DABDAB, the legs are AD=43AD = 4\sqrt{3} and AB=11AB = 11. The hypotenuse BDBD satisfies BD2=AD2+AB2=(43)2+112=48+121=169BD^2 = AD^2 + AB^2 = (4\sqrt{3})^2 + 11^2 = 48 + 121 = 169, so BD=169=13BD = \sqrt{169} = 13.

Key Concept

Solving multi-step geometry problems by decomposing shapes into rectangles and special right triangles (30∘30^\circ-60∘60^\circ-90∘90^\circ), then applying the Pythagorean Theorem.
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Topics

Pythagorean Theorem and Special Right Triangles

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