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Difference between revisions of "2002 AMC 8 Problems/Problem 20"

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<math> \textbf{(A)}\ 1\frac{1}2\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 2\frac{1}2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 3\frac{1}2 </math>
 
<math> \textbf{(A)}\ 1\frac{1}2\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 2\frac{1}2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 3\frac{1}2 </math>
  
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==Solution 3==
+
==Solution 1==
 
We know the area of triangle <math>XYZ</math> is <math>8</math> square inches. The area of a triangle can also be represented as <math>\frac{bh}{2}</math> or in this problem <math>\frac{XC\cdot YZ}{2}</math>. By solving, we have <cmath>\frac{XC\cdot YZ}{2} = 8,</cmath> <cmath>XC\cdot YZ = 16.</cmath>  
 
We know the area of triangle <math>XYZ</math> is <math>8</math> square inches. The area of a triangle can also be represented as <math>\frac{bh}{2}</math> or in this problem <math>\frac{XC\cdot YZ}{2}</math>. By solving, we have <cmath>\frac{XC\cdot YZ}{2} = 8,</cmath> <cmath>XC\cdot YZ = 16.</cmath>  
  

Revision as of 19:36, 26 June 2024

Problem

The area of triangle $XYZ$ is 8 square inches. Points $A$ and $B$ are midpoints of congruent segments $\overline{XY}$ and $\overline{XZ}$. Altitude $\overline{XC}$ bisects $\overline{YZ}$. The area (in square inches) of the shaded region is

[asy] /* AMC8 2002 #20 Problem */ draw((0,0)--(10,0)--(5,4)--cycle); draw((2.5,2)--(7.5,2)); draw((5,4)--(5,0)); fill((0,0)--(2.5,2)--(5,2)--(5,0)--cycle, mediumgrey); label(scale(0.8)*"$X$", (5,4), N); label(scale(0.8)*"$Y$", (0,0), W); label(scale(0.8)*"$Z$", (10,0), E); label(scale(0.8)*"$A$", (2.5,2.2), W); label(scale(0.8)*"$B$", (7.5,2.2), E); label(scale(0.8)*"$C$", (5,0), S); fill((0,-.8)--(1,-.8)--(1,-.95)--cycle, white);[/asy]

$\textbf{(A)}\ 1\frac{1}2\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 2\frac{1}2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 3\frac{1}2$

Solution 1

We know the area of triangle $XYZ$ is $8$ square inches. The area of a triangle can also be represented as $\frac{bh}{2}$ or in this problem $\frac{XC\cdot YZ}{2}$. By solving, we have \[\frac{XC\cdot YZ}{2} = 8,\] \[XC\cdot YZ = 16.\]

With SAS congruence, triangles $XCY$ and $XCZ$ are congruent. Hence, triangle $XCY = \frac{8}{2} = 4$. (Let's say point $D$ is the intersection between line segments $XC$ and $AB$.) We can find the area of the trapezoid $ADCY$ by subtracting the area of triangle $XAD$ from $4$.

We find the area of triangle $XAD$ by the $\frac{bh}{2}$ formula- $\frac{XD\cdot AD}{2} = \frac{\frac{XC}{2}\cdot AD}{2}$. $AD$ is $\frac{1}{4}$ of $YZ$ from solution 1. The area of $XAD$ is \[\frac{\frac{XC}{2}\cdot \frac{YZ}{4}}{2} = \frac{16}{16} = 1\].

Therefore, the area of the shaded area- trapezoid $ADCY$ has area $4-1 = \boxed{\text{(D)}\ 3}$.


- sarah07

Video Solution

https://www.youtube.com/watch?v=zwy5U5IQi88 ~David

See Also

2002 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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

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