Difference between revisions of "1990 AHSME Problems/Problem 7"

(Created page with "== Problem == A triangle with integral sides has perimeter <math>8</math>. The area of the triangle is <math>\text{(A) } 2\sqrt{2}\quad \text{(B) } \frac{16}{9}\sqrt{3}\quad \t...")
 
 
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== Solution ==
 
== Solution ==
<math>\fbox{A}</math>
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The shortest side must be <math>1</math> or <math>2</math>. However none of <math>(1,1,6),(1,2,5),(1,3,4)</math> form triangles, so the shortest side must be <math>2</math>. Then <math>(2,2,4)</math> is degenerate, so the sides must be <math>(2,3,3)</math>.
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This can be cut in half and reassembled into a rectangle with one side <math>1</math> and diagonal <math>3</math>. By Pythagoras its area is then <math>2\sqrt2</math> which means <math>\fbox{A}</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 03:49, 4 February 2016

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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions

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