Problem:
A square with side length 3 is inscribed in an isosceles triangle with one side of the square along the base of the triangle. A square with side length 2 has two vertices on the other square and the other two on sides of the triangle, as shown. What is the area of the triangle?
Answer Choices:
A. 1941β
B. 2041β
C. 2143β
D. 2221β
E. 2343β
Solution:
Letβs split the triangle down the middle and label it:
We see that β³ADGβΌβ³BEGβΌβ³CFG by AA similarity. BE=23β because AG cuts the side length of the square in half; similarly, CF=1. Let CG=h, then by side ratios,
hh+2β=23ββ2(h+2)=3hβ2h+4=3hβh=4
Now the height of the triangle is AG=4+2+3=9. By side ratios,
49β=1ADββAD=49β
The area of the triangle is AGβ
AD=9β
49β=481β=(B) 2041ββ.
The problems on this page are the property of the MAA's American Mathematics Competitions