Difference between revisions of "1998 AHSME Problems/Problem 8"
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== Solution 3 == | == Solution 3 == | ||
− | Divide the | + | Divide the pentagon into 2 small congruent trapezoids by extending the common shorter base of the 2 larger trapezoids. |
Since each of the smaller trapezoids has its area half each of the larger trapezoids, and each of them has a base $\frac{1}{2}, we have | Since each of the smaller trapezoids has its area half each of the larger trapezoids, and each of them has a base $\frac{1}{2}, we have | ||
<cmath>b_{large}+\frac{1}{2}=2(b_{small}+ \frac{1}{2}</cmath> | <cmath>b_{large}+\frac{1}{2}=2(b_{small}+ \frac{1}{2}</cmath> | ||
<cmath>x+\frac{1}{2}=2((1-x)+0\frac{1}{2}</cmath> | <cmath>x+\frac{1}{2}=2((1-x)+0\frac{1}{2}</cmath> | ||
− | <cmath>x=\frac{5}{6}</cmath> | + | <cmath>x=\frac{5}{6}\boxed{D}</cmath> |
+ | |||
+ | ~ Nafer | ||
== See also == | == See also == |
Revision as of 10:25, 2 December 2019
Problem
A square with sides of length is divided into two congruent trapezoids and a pentagon, which have equal areas, by joining the center of the square with points on three of the sides, as shown. Find , the length of the longer parallel side of each trapezoid.
Solution
Solution 1
Then . Let the shorter side of be and the base of be such that ; then implies that , and since it follows that and .
Solution 2
The area of the trapezoid is , and the shorter base and height are both . Therefore,
Solution 3
Divide the pentagon into 2 small congruent trapezoids by extending the common shorter base of the 2 larger trapezoids.
Since each of the smaller trapezoids has its area half each of the larger trapezoids, and each of them has a base $\frac{1}{2}, we have
~ Nafer
See also
1998 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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 |
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