Difference between revisions of "2021 Fall AMC 10B Problems/Problem 13"
(→Solution 1) |
(→Solution 1) |
||
Line 52: | Line 52: | ||
</asy> | </asy> | ||
− | We see that <math>\bigtriangleup ADG \sim \bigtriangleup BEG \sim \bigtriangleup CFG</math> by AA similarity. | + | We see that <math>\bigtriangleup ADG \sim \bigtriangleup BEG \sim \bigtriangleup CFG</math> by AA similarity. |
+ | <math>BE = \frac{3}{2}</math> because <math>AK</math> cuts the side length of the square in half; similarly, <math>CF = 1</math>. Let <math>CG = h</math>: then by side ratios, | ||
+ | |||
+ | <cmath>\frac{h+2}{h} = \frac{\frac{3}{2}}{1} \implies 2(h+2) = 3h \implies h = 4</cmath>. | ||
+ | |||
+ | Now the height of the triangle is <math>AG = 4+2+3 = 9</math>. By side ratios, | ||
+ | <math></math>\frac{9}{4} = \frac{AD}{1} \implies AD = \frac{9}{4}<math>. | ||
+ | |||
+ | The area of the triangle is </math>AG\cdot AD = 9 \cdot \frac{9}{4} = \frac{81}{4} = \boxed{B}$ | ||
+ | |||
+ | ~KingRavi | ||
==Solution 2== | ==Solution 2== |
Revision as of 17:02, 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, $$ (Error compiling LaTeX. Unknown error_msg)\frac{9}{4} = \frac{AD}{1} \implies AD = \frac{9}{4}$.
The area of the triangle is$ (Error compiling LaTeX. Unknown error_msg)AG\cdot AD = 9 \cdot \frac{9}{4} = \frac{81}{4} = \boxed{B}$
~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.