2017 AMC 8 Problem 18

Attempt Problem 18 of the 2017 AMC 8 below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2017 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

18.

In the non-convex quadrilateral ABCDABCD shown below, BCD\angle BCD is a right angle, AB=12,AB=12, BC=4,BC=4, CD=3,CD=3, and AD=13.AD=13. What is the area of quadrilateral ABCD?ABCD?

12 12

24 24

26 26

30 30

36 36

Answer: B
Concepts:Pythagorean TheoremPythagorean Triplearea
Difficulty rating: 1430
Video solution:
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Written solution:

Since BCD\angle BCD is a right angle, we can apply the Pythagorean theorem to BCD\triangle BCD to get that BD=5.\overline{BD} = 5. We also get that DBA\angle DBA is right since the sides of BDA\triangle BDA form a Pythagorean triple.

Then the area of ABCDABCD is equal to area(BDA)area(BCD)=121251243=306=24.\begin{align*} \text{area}(\triangle &BDA) - \text{area}(\triangle BCD) \\ &= \dfrac{1}{2} \cdot 12 \cdot 5 - \dfrac{1}{2} \cdot 4 \cdot 3 \\ &= 30 - 6 \\ &= 24. \end{align*}

Thus, B is the correct answer.

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