1996 AMC 8 Problem 22

Attempt Problem 22 of the 1996 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1996 AMC 8 solutions, or check the answer key.

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

22.

The horizontal and vertical distances between adjacent points equal 11 unit. The area of triangle ABCABC is
14\dfrac{1}{4}
12\dfrac{1}{2}
34\dfrac{3}{4}
11
54\dfrac{5}{4}
Answer: B
Concepts:Pick’s Theoremtriangle area
Small hint:
Assign coordinates to A,A, B,B, and CC and use the determinant formula for area
Big hint:
Alternatively use Pick’s theorem, area =I+B2−1,= I + \tfrac{B}{2} - 1, with I=0I = 0 interior points and B=3B = 3 boundary points
Solution:
Taking A=(0,0),A = (0, 0), B=(3,2),B = (3, 2), and C=(4,3),C = (4, 3), the determinant formula gives 12∣3⋅3−2⋅4∣=12.\frac12\left\lvert 3\cdot3-2\cdot4\right\rvert=\frac12.
Equivalently, by Pick’s theorem with no interior lattice points and 33 boundary points, the area is 0+32−1=12.0 + \tfrac32 - 1 = \tfrac12.
Thus, the correct answer is B .
LIVE by Po-Shen Loh
Programs

LIVE Classes

Self-Paced Classes

Stars

Scholarship

About Us

Who Is Po-Shen Loh?

Email Contact

Newsletter

Jobs

Become a Star

For a more Thought Full world.

© 2026 Expii, Inc.