Difference between revisions of "2002 AMC 8 Problems/Problem 16"
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| − | == Problem | + | == Problem == |
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label(scale(0.65)*"4", (2.3,-0.4)); | label(scale(0.65)*"4", (2.3,-0.4)); | ||
label(scale(0.65)*"3", (4.3,1.5));</asy> | label(scale(0.65)*"3", (4.3,1.5));</asy> | ||
| − | |||
<math> \textbf{(A)}\ X+Z=W+Y\qquad\textbf{(B)}\ W+X=Z\qquad\textbf{(C)}\ 3X+4Y=5Z\qquad</math> | <math> \textbf{(A)}\ X+Z=W+Y\qquad\textbf{(B)}\ W+X=Z\qquad\textbf{(C)}\ 3X+4Y=5Z\qquad</math> | ||
<math>\textbf{(D)}\ X+W=\frac{1}{2}(Y+Z)\qquad\textbf{(E)}\ X+Y=Z </math> | <math>\textbf{(D)}\ X+W=\frac{1}{2}(Y+Z)\qquad\textbf{(E)}\ X+Y=Z </math> | ||
| + | |||
| + | ==Solution== | ||
| + | The area of a right triangle can be found by using the legs of triangle as the base and height. In the three isosceles triangles, the length of their second leg is the same as the side that is connected to the <math>3-4-5</math> triangle. | ||
| + | |||
| + | <cmath>\begin{align*} | ||
| + | W&=(3)(4)/2 = 6\\ | ||
| + | X&=(3)(3)/2=4.5\\ | ||
| + | Y&=(4)(4)/2 = 8\\ | ||
| + | Z&=(5)(5)/2 = 12.5 | ||
| + | \end{align*}</cmath> | ||
| + | |||
| + | Plugging into the answer choices, the only that works is <math>\boxed{\textbf{(E)}\ X+Y=Z}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2002|num-b=15|num-a=17}} | ||
Revision as of 18:56, 23 December 2012
Problem
Right isosceles triangles are constructed on the sides of a 3-4-5 right triangle, as shown. A capital letter represents the area of each triangle. Which one of the following is true?
![[asy] /* AMC8 2002 #16 Problem */ draw((0,0)--(4,0)--(4,3)--cycle); draw((4,3)--(-4,4)--(0,0)); draw((-0.15,0.1)--(0,0.25)--(.15,0.1)); draw((0,0)--(4,-4)--(4,0)); draw((4,0.2)--(3.8,0.2)--(3.8,-0.2)--(4,-0.2)); draw((4,0)--(7,3)--(4,3)); draw((4,2.8)--(4.2,2.8)--(4.2,3)); label(scale(0.8)*"$Z$", (0, 3), S); label(scale(0.8)*"$Y$", (3,-2)); label(scale(0.8)*"$X$", (5.5, 2.5)); label(scale(0.8)*"$W$", (2.6,1)); label(scale(0.65)*"5", (2,2)); label(scale(0.65)*"4", (2.3,-0.4)); label(scale(0.65)*"3", (4.3,1.5));[/asy]](http://latex.artofproblemsolving.com/a/f/b/afbe72ee3e1ba66d7e02b6fd47981276584473cd.png)
Solution
The area of a right triangle can be found by using the legs of triangle as the base and height. In the three isosceles triangles, the length of their second leg is the same as the side that is connected to the 
triangle.
Plugging into the answer choices, the only that works is 
.
See Also
| 2002 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
