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Difference between revisions of "2004 AMC 8 Problems/Problem 24"

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==Problem==
 
==Problem==
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In the figure, <math>ABCD</math> is a rectangle and <math>EFGH</math> is a parallelogram. Using the measurements given in the figure, what is the length <math>d</math> of the segment that is perpendicular(altitude of parallelogram <math>EFGH</math>) to <math>\overline{HE}</math> and <math>\overline{FG}</math>?
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In the figure, <math>ABCD</math> is a rectangle and <math>EFGH</math> is a parallelogram. Using the measurements given in the figure, what is the length <math>d</math> of the segment that is perpendicular to <math>\overline{HE}</math> and <math>\overline{FG}</math>?
  
 
<asy>
 
<asy>
Line 38: Line 38:
 
<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>
  
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==Solution==
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==Excellent 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>
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The second step 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.
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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{(C)}\ 7.6}</cmath>
 
<cmath>5d = 38 \rightarrow d = \boxed{\textbf{(C)}\ 7.6}</cmath>
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 +
~Ak
 +
 +
==Video Solution by Sohil Rathi (Omega Learn)==
 +
https://youtu.be/abSgjn4Qs34?t=4
 +
 +
~ pi_is_3.14
  
 
==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}}

Latest revision as of 11:35, 20 January 2024

Problem

In the figure, $ABCD$ is a rectangle and $EFGH$ is a parallelogram. Using the measurements given in the figure, what is the length $d$ of the segment that is perpendicular to $\overline{HE}$ and $\overline{FG}$?

[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]

$\textbf{(A)}\ 6.8\qquad \textbf{(B)}\ 7.1\qquad \textbf{(C)}\ 7.6\qquad \textbf{(D)}\ 7.8\qquad \textbf{(E)}\ 8.1$

Excellent Solution

The area of the parallelogram can be found in two ways. The first is by taking rectangle $ABCD$ and subtracting the areas of the triangles cut out to create parallelogram $EFGH$. This is \[(4+6)(5+3) - 2 \cdot \frac12 \cdot 6 \cdot 5 - 2 \cdot \frac12 \cdot 3 \cdot 4 = 80 - 30 - 12 = 38\] The second way is by multiplying the base of the parallelogram such as $\overline{FG}$ with its altitude $d$, which is perpendicular to both bases. $\triangle FGC$ is a $3-4-5$ triangle so $\overline{FG} = 5$. Set these two representations of the area together. \[5d = 38 \rightarrow d = \boxed{\textbf{(C)}\ 7.6}\]

~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

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