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2002 AMC 8 Problems/Problem 20: Difference between revisions

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==Solution 4 (Dummed down)==
The area of triangle <math>XYZ</math> is <math>8</math> square inches. You can turn the triangle into a rectangle by drawing a triangle on the left side with the hypotenuse of it being <math>YAX</math> and another triangle on the right side, with its hypotenuse being <math>XBZ</math>. After drawing the square, you can cut it into <math>8</math> squares. The area of the rectangle is <math>16</math> square inches because the triangle on the left is half of <math>XYCZ</math> and there's another triangle on the other side, equaling them to <math>8</math> square inches. We will be focusing on the left side of the rectangle of <math>XC</math> because it includes the shaded region. It's split into <math>4</math> equal squares. The area of this triangle is <math>8</math> square inches because the total area of the rectangle is <math>16</math> square inches and <math>16</math> divided by <math>2</math> is <math>8</math>. There are <math>4</math> sections, so you would do <math>8</math> divided by <math>4</math> in order to find the area of one square. That means that the area of the top right square is <math>2</math> inches and because it's not needed, we will subtract <math>2</math> from <math>8</math> to get rid of it. If you turn around the paper, you notice that the square with half of the shaded region, and the square above it is the shaded region, except flipped and turned. Therefore, the remaining of the area of it which is <math>6</math> should then be divided by <math>2</math> in order to find the shaded region since the shaded region is equal to the other square and half a square.
<math>6</math> divided by <math>2</math> is <math>3</math>, so the answer is <math>3</math>.
==Video Solution==
==Video Solution==
https://www.youtube.com/watch?v=zwy5U5IQi88  ~David
https://www.youtube.com/watch?v=zwy5U5IQi88  ~David

Revision as of 09:00, 14 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 3

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

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