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
Small hint:
First find BDBD using the small right triangle.
Big hint:
Then recognize the larger triangle as a 55-1212-1313 right triangle and subtract the small triangle’s area.
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)=12⋅12⋅5−12⋅4⋅3=30−6=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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