Difference between revisions of "2004 AMC 8 Problems/Problem 24"
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<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> | ||
The second way is by multiplying the base of the parallelogram such as <math>\overline{FG}</math> with its altitude <math>d</math>, which is perpendicular to both bases. <math>\triangle FGC</math> is a <math>3-4-5</math> triangle so <math>\overline{FG} = 5</math>. Set these two representations of the area together. | The second way is by multiplying the base of the parallelogram such as <math>\overline{FG}</math> with its altitude <math>d</math>, which is perpendicular to both bases. <math>\triangle FGC</math> is a <math>3-4-5</math> triangle so <math>\overline{FG} = 5</math>. Set these two representations of the area together. | ||
| − | <cmath>5d = 38 \rightarrow d = \boxed{\textbf{( | + | <cmath>5d = 38 \rightarrow d = \boxed{\textbf{(C)}\ 7.6}</cmath> |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2004|num-b=23|num-a=25}} | {{AMC8 box|year=2004|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 14:39, 22 December 2021
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(altitude of parallelogram ) 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)
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)
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. 
