Difference between revisions of "2021 Fall AMC 10B Problems/Problem 13"
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==Solution 1== | ==Solution 1== | ||
− | Let's split the triangle down the middle: | + | Let's split the triangle down the middle and label it: |
<asy> | <asy> | ||
Line 40: | Line 40: | ||
draw(E--H); | draw(E--H); | ||
draw(K--(0.75,0)); | draw(K--(0.75,0)); | ||
+ | label("$A$",A,W); | ||
+ | label("$B$",B,S); | ||
+ | label("$C$",C,S); | ||
+ | label("$D$",D,E); | ||
+ | label("$E$",E,W); | ||
+ | label("$F$",F,S); | ||
+ | label("$G$",G,S); | ||
+ | label("$H$",H,E); | ||
+ | label("$I$",I,W); | ||
+ | label("$J$",J,E); | ||
+ | label("$K$",K,N); | ||
+ | |||
Revision as of 15:38, 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:
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.