1966 AHSME Problems/Problem 14
Contents
Problem
The length of rectangle 
is 5 inches and its width is 3 inches. Diagonal 
is divided into three equal segments by points 
and 
. The area of triangle 
, expressed in square inches, is:
Solution
Draw the rectangle with 
= 
and 
= 
. We created our diagonal, and use the Pythagorean Theorem to find the length of , which is 
. Since 
breaks the diagonal into equal parts, the lenght of is 
. The only other thing we need is the height of . Realize that the height of is also the height of right triangle 
using as the base. The area of is 
(using the side lengths of the rectangle). The height of to base is divided by 
(remember, we multiply by 
because we are finding the height from the area of a triangle which is 
). That simplfies to 
which equal to 
. Now doing all the arithmetic, 
= 
= 
.
Solution 2
Draw the rectangle with = and = . Next draw a height from vertex 
to diagonal intersecting at 
and similarly from 
to intersecting at 
. Let's call this height 
and we can notice that 
= 
= . Since the points and are trisecting diagonal the bases and heights of triangles 
, , 
, 
, 
, and 
have the same areas. Hence the area of one of these triangles such as is 
= . So our answer is 
.
See also
| 1966 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
