Difference between revisions of "2021 Fall AMC 10B Problems/Problem 13"
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Now the height of the triangle is <math>AG = 4+2+3 = 9</math>. By side ratios, | Now the height of the triangle is <math>AG = 4+2+3 = 9</math>. By side ratios, | ||
− | < | + | <cmath>\frac{9}{4} = \frac{AD}{1} \implies AD = \frac{9}{4}</cmath>. |
− | The area of the triangle is < | + | The area of the triangle is <math>AG\cdot AD = 9 \cdot \frac{9}{4} = \frac{81}{4} = \boxed{B}</math> |
~KingRavi | ~KingRavi |
Revision as of 17:03, 24 November 2021
Contents
[hide]Problem
A square with side length is inscribed in an isosceles triangle with one side of the square along the base of the triangle. A square with side length 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?
Solution 1
Let's split the triangle down the middle and label it:
We see that by AA similarity. because cuts the side length of the square in half; similarly, . Let : then by side ratios,
.
Now the height of the triangle is . By side ratios, .
The area of the triangle is
~KingRavi
Solution 2
By similarity, the height is and the base is . Thus the area is , or .
~Hefei417, or 陆畅 Sunny from China
See Also
2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.