2015 AMC 8 Problem 19

Attempt Problem 19 of the 2015 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 2015 AMC 8 solutions, or check the answer key.

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

19.

A triangle with vertices at A=(1,3),A=(1,3), B=(5,1),B=(5,1), and C=(4,4)C=(4,4) is plotted on a 6×56\times5 grid. What fraction of the grid is covered by the triangle?
16 \dfrac{1}{6}
15 \dfrac{1}{5}
14 \dfrac{1}{4}
13 \dfrac{1}{3}
12 \dfrac{1}{2}
Answer: A
Concepts:coordinate geometrytriangle area
Small hint:
Put the triangle inside a small rectangle aligned to the grid.
Big hint:
Subtract the three corner right triangles from the rectangle.
Video solution:
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Written solution:
The total area of the grid is 6⋅5=30.6\cdot5=30. To find the area of the triangle, place it inside the 44-by-33 rectangle shown below.
The rectangle has area 4⋅3=12.4\cdot3=12. The three corner triangles have areas 12(4)(2)=4,\dfrac12(4)(2)=4, 12(1)(3)=32,\dfrac12(1)(3)=\dfrac32, and 12(3)(1)=32.\dfrac12(3)(1)=\dfrac32. Therefore, [△ABC]=12−4−32−32=5.\begin{aligned} [\triangle ABC] &= 12-4-\dfrac32-\dfrac32\\ &=5. \end{aligned}
Therefore, the fraction of the area is 530=16.\dfrac{5}{30} = \dfrac{1}{6} .
Thus, the correct answer is A .
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