Difference between revisions of "2002 AMC 8 Problems/Problem 20"
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==Problem== | ==Problem== | ||
− | Points <math> | + | The area of triangle <math>XYZ</math> is 8 square inches. Points <math>A</math> and <math>B</math> are midpoints of congruent segments <math>\overline{XY}</math> and <math>\overline{XZ}</math>. Altitude <math>\overline{XC}</math> bisects <math>\overline{YZ}</math>. The area (in square inches) of the shaded region is |
<asy> | <asy> | ||
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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 3== | ||
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- sarah07 | - sarah07 | ||
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+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=zwy5U5IQi88 ~David | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2002|num-b=19|num-a=21}} | {{AMC8 box|year=2002|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 10:00, 14 June 2024
Contents
Problem
The area of triangle is 8 square inches. Points
and
are midpoints of congruent segments
and
. Altitude
bisects
. The area (in square inches) of the shaded region is
Solution 3
We know the area of triangle is
square inches. The area of a triangle can also be represented as
or in this problem
. By solving, we have
With SAS congruence, triangles and
are congruent. Hence, triangle
. (Let's say point
is the intersection between line segments
and
.) We can find the area of the trapezoid
by subtracting the area of triangle
from
.
We find the area of triangle by the
formula-
.
is
of
from solution 1. The area of
is
.
Therefore, the area of the shaded area- trapezoid has area
.
- 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.