Difference between revisions of "1950 AHSME Problems/Problem 47"
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==Solution 2== | ==Solution 2== | ||
You can also use area to solve this problem. Draw the triangle, and note that the small triangle formed by taking away the rectangle and the two small portions left is similar to the big triangle. Therefore the smaller triangle (similar to the big one) has an area of <math>\dfrac{2x(h-x)}{2}=x(h-x)</math>. Now note that because the rectangle is right, and the two other pieces are complementary triangles, we can add them together to create one triangle. Therefore the area of these two triangles is <math>\dfrac{x(b-2x)}{2}</math>. This is due to the fact that we subtract <math>2x</math> from the base of the triangle. Lastly the area of the rectangle is <math>2x^2</math>. These areas together sum to the area of the big rectangle, which is <math>\dfrac{bh}{2}</math>. Solving we get that <math>x = \dfrac{bh}{2h + b}</math>. The answer is <math>\boxed{\textbf{(C)}}</math>. | You can also use area to solve this problem. Draw the triangle, and note that the small triangle formed by taking away the rectangle and the two small portions left is similar to the big triangle. Therefore the smaller triangle (similar to the big one) has an area of <math>\dfrac{2x(h-x)}{2}=x(h-x)</math>. Now note that because the rectangle is right, and the two other pieces are complementary triangles, we can add them together to create one triangle. Therefore the area of these two triangles is <math>\dfrac{x(b-2x)}{2}</math>. This is due to the fact that we subtract <math>2x</math> from the base of the triangle. Lastly the area of the rectangle is <math>2x^2</math>. These areas together sum to the area of the big rectangle, which is <math>\dfrac{bh}{2}</math>. Solving we get that <math>x = \dfrac{bh}{2h + b}</math>. The answer is <math>\boxed{\textbf{(C)}}</math>. | ||
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| + | ==Video Solution== | ||
| + | https://www.youtube.com/watch?v=l4lAvs2P_YA&t=169s | ||
| + | |||
| + | ~MathProblemSolvingSkills.com | ||
| + | |||
==See Also== | ==See Also== | ||
Latest revision as of 16:50, 15 July 2023
Problem
A rectangle inscribed in a triangle has its base coinciding with the base 
of the triangle. If the altitude of the triangle is 
, and the altitude 
of the rectangle is half the base of the rectangle, then:
Solution 1
Draw the triangle, and note that the small triangle formed by taking away the rectangle and the two small portions left is similar to the big triangle, so the proportions of the heights is equal to the proportions of the sides. In particular, we get 
. The answer is 
.
Solution 2
You can also use area to solve this problem. Draw the triangle, and note that the small triangle formed by taking away the rectangle and the two small portions left is similar to the big triangle. Therefore the smaller triangle (similar to the big one) has an area of 
. Now note that because the rectangle is right, and the two other pieces are complementary triangles, we can add them together to create one triangle. Therefore the area of these two triangles is 
. This is due to the fact that we subtract 
from the base of the triangle. Lastly the area of the rectangle is 
. These areas together sum to the area of the big rectangle, which is 
. Solving we get that 
. The answer is .
Video Solution
https://www.youtube.com/watch?v=l4lAvs2P_YA&t=169s
~MathProblemSolvingSkills.com
See Also
| 1950 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 46 |
Followed by Problem 48 | |
| 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 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
