Difference between revisions of "2004 AMC 8 Problems/Problem 24"
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Latest revision as of 11:35, 20 January 2024
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 
?
![[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); pair D=(0,0), C=(10,0), B=(10,8), A=(0,8); pair E=(4,8), F=(10,3), G=(6,0), H=(0,5); draw(A--B--C--D--cycle); draw(E--F--G--H--cycle); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$E$",E,N); label("$F$",(10.8,3)); label("$G$",G,S); label("$H$",H,W); label("$4$",A--E,N); label("$6$",B--E,N); label("$5$",(10.8,5.5)); label("$3$",(10.8,1.5)); label("$4$",G--C,S); label("$6$",G--D,S); label("$5$",D--H,W); label("$3$",A--H,W); draw((3,7.25)--(7.56,1.17)); label("$d$",(3,7.25)--(7.56,1.17), NE); [/asy]](http://latex.artofproblemsolving.com/d/2/3/d23e9bb30960feaeb516f8d23d48c117676bfbd0.png)
Excellent Solution
The area of the parallelogram can be found in two ways. The first is by taking rectangle and subtracting the areas of the triangles cut out to create parallelogram . This is
![\[(4+6)(5+3) - 2 \cdot \frac12 \cdot 6 \cdot 5 - 2 \cdot \frac12 \cdot 3 \cdot 4 = 80 - 30 - 12 = 38\]](http://latex.artofproblemsolving.com/4/b/5/4b5dda11f7c6e3b72d3a36f4ddd6595f9721b68d.png)
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.
![\[5d = 38 \rightarrow d = \boxed{\textbf{(C)}\ 7.6}\]](http://latex.artofproblemsolving.com/5/1/f/51f918965efba4986427b9d116eb2a04b263958f.png)
~Ak
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
