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Difference between revisions of "1988 AJHSME Problems/Problem 17"

(New page: ==Problem== The shaded region formed by the two intersecting perpendicular rectangles, in square units, is <asy> fill((0,0)--(6,0)--(6,-3.5)--(9,-3.5)--(9,0)--(10,0)--(10,2)--(9,2)--(9,4...)
 
 
(One intermediate revision by one other user not shown)
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==Problem==
 
==Problem==
  
The shaded region formed by the two intersecting perpendicular rectangles, in square units, is
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The shaded region formed by the two intersecting [[perpendicular]] [[rectangle|rectangles]], in square units, is
  
 
<asy>
 
<asy>
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==Solution==
 
==Solution==
  
Looking at the diagram, the shaded region is the union of two rectangles, with a small rectangle as overlap.  If we just add the areas of the two rectangles, then we'll overcount the small rectangle, so we must subtract that area to get the desired area.
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Looking at the diagram, the shaded region is the union of two rectangles, with a small rectangle as overlap.  If we just add the areas of the two rectangles, then we'll overcount the small rectangle, so we must [[subtraction|subtract]] that [[area]] to get the desired area.
  
 
The areas of the two larger rectangles are <math>2\cdot 10=20</math> and <math>3\cdot 8=24</math>, and the area of the small rectangle is <math>2\cdot 3=6</math>.  The desired area is thus <math>20+24-6=38 \rightarrow \boxed{\text{B}}</math>.
 
The areas of the two larger rectangles are <math>2\cdot 10=20</math> and <math>3\cdot 8=24</math>, and the area of the small rectangle is <math>2\cdot 3=6</math>.  The desired area is thus <math>20+24-6=38 \rightarrow \boxed{\text{B}}</math>.
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==See Also==
 
==See Also==
  
[[1988 AJHSME Problems]]
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{{AJHSME box|year=1988|num-b=16|num-a=18}}
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 22:56, 4 July 2013

Problem

The shaded region formed by the two intersecting perpendicular rectangles, in square units, is

[asy] fill((0,0)--(6,0)--(6,-3.5)--(9,-3.5)--(9,0)--(10,0)--(10,2)--(9,2)--(9,4.5)--(6,4.5)--(6,2)--(0,2)--cycle,black); label("2",(0,.9),W); label("3",(7.3,4.5),N); draw((0,-3.3)--(0,-5.3),linewidth(1)); draw((0,-4.3)--(3.7,-4.3),linewidth(1)); label("10",(4.7,-3.7),S); draw((5.7,-4.3)--(10,-4.3),linewidth(1)); draw((10,-3.3)--(10,-5.3),linewidth(1)); draw((11,4.5)--(13,4.5),linewidth(1)); draw((12,4.5)--(12,2),linewidth(1)); label("8",(11.3,1),E); draw((12,0)--(12,-3.5),linewidth(1)); draw((11,-3.5)--(13,-3.5),linewidth(1)); [/asy]

$\text{(A)}\ 23 \qquad \text{(B)}\ 38 \qquad \text{(C)}\ 44 \qquad \text{(D)}\ 46 \qquad \text{(E)}\ \text{unable to be determined from the information given}$

Solution

Looking at the diagram, the shaded region is the union of two rectangles, with a small rectangle as overlap. If we just add the areas of the two rectangles, then we'll overcount the small rectangle, so we must subtract that area to get the desired area.

The areas of the two larger rectangles are $2\cdot 10=20$ and $3\cdot 8=24$, and the area of the small rectangle is $2\cdot 3=6$. The desired area is thus $20+24-6=38 \rightarrow \boxed{\text{B}}$.

See Also

1988 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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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