1968 AHSME Problems/Problem 31
Problem
In this diagram, not drawn to scale, Figures and are equilateral triangular regions with respective areas of and square inches. Figure is a square region with area square inches. Let the length of segment be decreased by % of itself, while the lengths of and remain unchanged. The percent decrease in the area of the square is:
Solution
Given an equilateral triangle with side length , the area is given by . Setting this equation equal to the area of triangle , , we find that . Because triangle is also equilateral, it is similar to trinagle , and, because it has a quarter of the area of , it has of the side length. Thus, its sides have a length of . Square initially has an area of , so it's side length starts at . The initial length , therefore, is . Because decreases by %, it becomes of its inital value, which is . Because the sides of the triangles remain unchanged, this decrease of must come from the side length of the square. Thus, the square's final side is , which gives an area of square inches. to is a decrease of %. Therefore, our answer is .
See also
1968 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 30 |
Followed by Problem 32 | |
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 • 31 • 32 • 33 • 34 • 35 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.