1990 AHSME Problems/Problem 7

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Problem

A triangle with integral sides has perimeter $8$. The area of the triangle is

$\text{(A) } 2\sqrt{2}\quad \text{(B) } \frac{16}{9}\sqrt{3}\quad \text{(C) }2\sqrt{3} \quad \text{(D) } 4\quad \text{(E) } 4\sqrt{2}$

Solution

The shortest side must be $1$ or $2$. However none of $(1,1,6),(1,2,5),(1,3,4)$ form triangles, so the shortest side must be $2$. Then $(2,2,4)$ is degenerate, so the sides must be $(2,3,3)$.

This can be cut in half and reassembled into a rectangle with one side $1$ and diagonal $3$. By Pythagoras its area is then $2\sqrt2$ which means $\fbox{A}$

See also

1990 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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