Difference between revisions of "1988 AHSME Problems/Problem 3"
(Created page with "==Problem== <asy> draw((0,0)--(1,0)--(1,4)--(0,4)--(0,0)--(0,1)--(-1,1)--(-1,2)); draw((-1,2)--(0,2)--(0,4)--(-1,4)--(-1,5)--(1,5)--(1,6)--(0,6)); draw((0,6)--(0,5)--(3,5)--(3,6...") |
Quantummech (talk | contribs) (→Solution) |
||
Line 20: | Line 20: | ||
==Solution== | ==Solution== | ||
− | + | We first notice that the paper strips cover up part of the others. Since the width of the overlap is <math>1</math> and the length of the overlap is <math>1</math>, the area of | |
− | + | each of the strips with the overlap is <math>(10\cdot 1)-1=9</math>. Since there are 4 strips, <math>4\cdot 9=36 \implies \boxed{\text{A}}</math>. | |
== See also == | == See also == |
Latest revision as of 05:26, 31 August 2015
Problem
Four rectangular paper strips of length and width are put flat on a table and overlap perpendicularly as shown. How much area of the table is covered?
Solution
We first notice that the paper strips cover up part of the others. Since the width of the overlap is and the length of the overlap is , the area of each of the strips with the overlap is . Since there are 4 strips, .
See also
1988 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.