Difference between revisions of "2004 AMC 8 Problems/Problem 24"
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<math>\textbf{(A)}\ 6.8\qquad \textbf{(B)}\ 7.1\qquad \textbf{(C)}\ 7.6\qquad \textbf{(D)}\ 7.8\qquad \textbf{(E)}\ 8.1</math> | <math>\textbf{(A)}\ 6.8\qquad \textbf{(B)}\ 7.1\qquad \textbf{(C)}\ 7.6\qquad \textbf{(D)}\ 7.8\qquad \textbf{(E)}\ 8.1</math> | ||
− | Solution == | + | == Solution == |
The area of the parallelogram can be found in two ways. The first is by taking rectangle <math>ABCD</math> and subtracting the areas of the triangles cut out to create parallelogram <math>EFGH</math>. This is | The area of the parallelogram can be found in two ways. The first is by taking rectangle <math>ABCD</math> and subtracting the areas of the triangles cut out to create parallelogram <math>EFGH</math>. This is | ||
<cmath>(4+6)(5+3) - 2 \cdot \frac12 \cdot 6 \cdot 5 - 2 \cdot \frac12 \cdot 3 \cdot 4 = 80 - 30 - 12 = 38</cmath> | <cmath>(4+6)(5+3) - 2 \cdot \frac12 \cdot 6 \cdot 5 - 2 \cdot \frac12 \cdot 3 \cdot 4 = 80 - 30 - 12 = 38</cmath> |
Revision as of 01:45, 24 May 2025
Problem
In the figure, is a rectangle and
is a parallelogram. Using the measurements given in the figure, what is the length
of the segment that is perpendicular to
and
?
Solution
The area of the parallelogram can be found in two ways. The first is by taking rectangleand subtracting the areas of the triangles cut out to create parallelogram
. This is
The second way is by multiplying the base of the parallelogram such as
with its altitude
, which is perpendicular to both bases.
is a
triangle so
. Set these two representations of the area together.
:)
~Ak ~Smiley face by Shreyansh Medatati
Video Solution by Sohil Rathi (Omega Learn)
https://youtu.be/abSgjn4Qs34?t=4
~ pi is 3.14
See Also
2004 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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.