2021 AMC Fall 10B Problem 2

The shaded shape is an arrowhead pointing up. Its corners are $(1,0)$, $(3,5)$, $(5,0)$, and the notch point $(3,2)$. 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 $(1,0)$, $(5,0)$, $(3,5)$. Its base runs along the $x$-axis from $1$ to $5$, so the base is $4$. Its top is at height $5$. Area:
$$\tfrac12 \times 4 \times 5 = 10.$$
2. The dent. It is the triangle with corners $(1,0)$, $(5,0)$, $(3,2)$. Same base $4$, height $2$. Area:
$$\tfrac12 \times 4 \times 2 = 4.$$
3. Shaded part. Big triangle minus dent:
$$10 - 4 = 6.$$

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

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: $\frac12 \times 4 \times (5 - 2) = 6$. The shaded region has the same area as a triangle with base $4$ and height $3$.

Cut the arrowhead along the vertical line $x = 3$ into a left half and a right half. The left half has corners $(1,0)$, $(3,5)$, $(3,2)$. Its side on the line $x = 3$ goes from height $2$ to height $5$, so it has length $3$, and the corner $(1,0)$ is $2$ units away from that line. Its area is $\frac12 \times 3 \times 2 = 3$. The right half is a mirror copy, also $3$. Total: $6$.

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 $\boxed{\textbf{(B)}\ 6}$.