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Difficulty: MediumPythagorean Theorem and Special Right Triangles

In right trapezoid ABCDABCD, side ABAB is parallel to side DCDC, and ∠DAB=90∘\angle DAB = 90^\circ. The lengths of the sides are AD=12AD = 12, AB=9AB = 9, and BC=13BC = 13, with DC>ABDC > AB. Point EE lies on the line containing segment DCDC such that segment BEBE is perpendicular to segment BCBC. What is the length of segment CECE?

  1. A
    16.916.9
  2. B
    25.025.0
  3. C
    28.828.8
  4. 33.833.8Answer
  5. E
    36.036.0

Answer

The length of segment CECE is 33.833.8.
Dropping altitude BPBP from BB to side DCDC creates a rectangle ABPDABPD with height BP=12BP = 12 and a right triangle △BPC\triangle BPC with legs BP=12BP = 12 and hypotenuse BC=13BC = 13. Applying the Pythagorean theorem yields PC=5PC = 5. Since segment BEBE is perpendicular to BCBC, triangle △BCE\triangle BCE is a right triangle with right angle at BB and altitude BPBP to hypotenuse CECE. Using the leg rule BC2=PC×CEBC^2 = PC \times CE, we find 132=5×CE13^2 = 5 \times CE, which simplifies to CE=169/5=33.8CE = 169/5 = 33.8.

Step-by-Step Solution

1
Find the length of segment PCPC by dropping a perpendicular from BB to DCDC.
PC=5PC = 5
Let PP be the foot of the perpendicular from BB to line DCDC. Since ABCDABCD is a right trapezoid with AD⊥DCAD \perp DC, ABPDABPD forms a rectangle. Thus BP=AD=12BP = AD = 12 and DP=AB=9DP = AB = 9. In right triangle △BPC\triangle BPC, BP2+PC2=BC2  ⟹  122+PC2=132  ⟹  PC=169−144=5BP^2 + PC^2 = BC^2 \implies 12^2 + PC^2 = 13^2 \implies PC = \sqrt{169 - 144} = 5.
2
Set up a similar triangle relationship for right triangle △BCE\triangle BCE.
△BPC∼△EBC\triangle BPC \sim \triangle EBC
In △BCE\triangle BCE, ∠CBE=90∘\angle CBE = 90^\circ and BP⊥CEBP \perp CE. Thus, right triangle △BPC\triangle BPC is similar to right triangle △EBC\triangle EBC because they share ∠C\angle C.
3
Solve for the length of CECE.
CE=33.8CE = 33.8
By the geometric mean theorem (or ratio of corresponding sides in similar triangles), BCCE=PCBC  ⟹  BC2=PC×CE\frac{BC}{CE} = \frac{PC}{BC} \implies BC^2 = PC \times CE. Substituting BC=13BC = 13 and PC=5PC = 5 yields 132=5×CE  ⟹  169=5×CE  ⟹  CE=1695=33.813^2 = 5 \times CE \implies 169 = 5 \times CE \implies CE = \frac{169}{5} = 33.8.

Key Concept

Pythagorean Theorem and Geometric Mean Theorem (Similar Right Triangles)
Estimated Time:1m 40s
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Topics

Pythagorean Theorem and Special Right Triangles

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