AMC 10 Step by Step

2021 AMC Fall 10B Problem 2

P1-10~2 minPrint

What is the area of the shaded figure shown below? figure

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Key insight

Fill in the notch to get a big triangle, then take the notch back out; both triangles sit on the same base of length 4.

Solution

The shaded shape is an arrowhead pointing up. Its corners are (1,0)(1,0), (3,5)(3,5), (5,0)(5,0), and the notch point (3,2)(3,2). A shape with a dent is awkward, so fill the dent in and then take it back out.

1. Big triangle (dent filled in). Corners (1,0)(1,0), (5,0)(5,0), (3,5)(3,5). Its base runs along the xx-axis from 11 to 55, so the base is 44. Its top is at height 55. Area:
12×4×5=10.\tfrac12 \times 4 \times 5 = 10.
2. The dent. It is the triangle with corners (1,0)(1,0), (5,0)(5,0), (3,2)(3,2). Same base 44, height 22. Area:
12×4×2=4.\tfrac12 \times 4 \times 2 = 4.
3. Shaded part. Big triangle minus dent:
10−4=6.10 - 4 = 6.

The answer is (B) 6\boxed{\textbf{(B)}\ 6}.

Why this works

A shape with a bite taken out is usually easiest as "whole shape minus bite." Here both triangles share the same base, so you can even do it in one step: 12×4×(5−2)=6\frac12 \times 4 \times (5 - 2) = 6. The shaded region has the same area as a triangle with base 44 and height 33.

Alternative approach

Cut the arrowhead along the vertical line x=3x = 3 into a left half and a right half. The left half has corners (1,0)(1,0), (3,5)(3,5), (3,2)(3,2). Its side on the line x=3x = 3 goes from height 22 to height 55, so it has length 33, and the corner (1,0)(1,0) is 22 units away from that line. Its area is 12×3×2=3\frac12 \times 3 \times 2 = 3. The right half is a mirror copy, also 33. Total: 66.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The shaded region has vertices (1,0)(1,0), (3,5)(3,5), (5,0)(5,0) and (3,2)(3,2); it is an arrowhead pointing up, with a notch at (3,2)(3,2).

Fill in the notch: the large triangle with vertices (1,0)(1,0), (5,0)(5,0), (3,5)(3,5) has base 44 (along the xx-axis) and height 55, so its area is 12⋅4⋅5=10\frac12 \cdot 4 \cdot 5 = 10.

The notch itself is the triangle with vertices (1,0)(1,0), (5,0)(5,0), (3,2)(3,2): base 44, height 22, area 12⋅4⋅2=4\frac12 \cdot 4 \cdot 2 = 4.

Shaded area =10−4=6= 10 - 4 = 6.

The answer is (B) 6\boxed{\textbf{(B)}\ 6}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.