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...") |
|||
| Line 10: | Line 10: | ||
== Solution == | == Solution == | ||
| − | <math>\fbox{A}</math> | + | 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>. |
| + | |||
| + | 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 04:49, 4 February 2016
Problem
A triangle with integral sides has perimeter 
. The area of the triangle is
Solution
The shortest side must be 
or 
. However none of 
form triangles, so the shortest side must be . Then 
is degenerate, so the sides must be 
.
This can be cut in half and reassembled into a rectangle with one side and diagonal 
. By Pythagoras its area is then 
which means 
See also
| 1990 AHSME (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
