2017 AMC 10B Problem 15

Attempt Problem 15 of the 2017 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2017 AMC 10B solutions, or check the answer key.

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15.

Rectangle ABCDABCD has AB=3AB=3 and BC=4.BC=4. Point EE is the foot of the perpendicular from BB to diagonal AC‾.\overline{AC}. What is the area of △ADE?\triangle ADE?

1 1

4225 \dfrac{42}{25}

2815 \dfrac{28}{15}

2 2

5425 \dfrac{54}{25}

Answer: E
Concepts:similarityarea ratioright triangle
Difficulty rating: 1600
Small Hint:

Use similarity between △ABE\triangle ABE and △ACB\triangle ACB

Big Hint:

Then compare △EAD\triangle EAD to △CDA\triangle CDA using their bases on ACAC

Solution:

The area of △CDA\triangle CDA is 3⋅42=6\dfrac{3\cdot4}{2}=6. Since EE lies on ACAC, triangles EADEAD and CDACDA share the same altitude from DD, so [EAD]=[CDA]⋅AEAC[EAD]=[CDA]\cdot \dfrac{AE}{AC}.

By the Pythagorean Theorem, AC=5AC=5. Also △ABE∼△ACB\triangle ABE\sim \triangle ACB, so AEAB=ABAC=35\dfrac{AE}{AB}=\dfrac{AB}{AC}=\dfrac35, giving AE=95AE=\dfrac95. Thus AEAC=925\dfrac{AE}{AC}=\dfrac{9}{25}.

Therefore [EAD]=6⋅925=5425[EAD]=6\cdot \dfrac{9}{25}=\dfrac{54}{25}. Thus, E is the correct answer.

Problem 14#14
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