2021 AMC Fall 10B Problem 2

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

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

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

Shaded area $= 10 - 4 = 6$.

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

A concave polygon is often easiest to handle as a convex shape minus a bite. Both triangles here share the same base, so the answer is simply $\frac12 \cdot 4 \cdot (5 - 2)$: the shaded region has the same area as a triangle with base $4$ and height $3$.

Shoelace formula on $(1,0), (3,5), (5,0), (3,2)$ in order: the cross terms give $(1\cdot5 - 3\cdot0) + (3\cdot0 - 5\cdot5) + (5\cdot2 - 3\cdot0) + (3\cdot0 - 1\cdot2) = 5 - 25 + 10 - 2 = -12$, and half the absolute value is $6$.

• Reading the notch vertex (3,2) as (3,1) or (3,3) off the grid, or adding the two triangles instead of subtracting.
• Treating the figure as two triangles split by the vertical line $x=3$ but using the wrong heights for them.