Difference between revisions of "1988 AJHSME Problems/Problem 17"
5849206328x (talk | contribs) (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...) |
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==Problem== | ==Problem== | ||
− | The shaded region formed by the two intersecting perpendicular rectangles, in square units, is | + | 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. | + | 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== | ||
− | + | {{AJHSME box|year=1988|num-b=16|num-a=18}} | |
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] |
Revision as of 21:23, 3 June 2009
Problem
The shaded region formed by the two intersecting perpendicular rectangles, in square units, is
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 and , and the area of the small rectangle is . The desired area is thus .
See Also
1988 AJHSME (Problems • Answer Key • Resources) | ||
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 |