Difference between revisions of "1987 AIME Problems/Problem 6"
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[[Image:AIME_1987_Problem_6.png]] | [[Image:AIME_1987_Problem_6.png]] | ||
== Solution == | == Solution == | ||
− | Since <math>XY = WZ</math> and <math>PQ = PQ</math> and the [[area]]s of the [[trapezoid]]s <math>\displaystyle PQZW</math> and <math>\displaystyle PQYX</math> are the same, the heights of the trapezoids are the same. Thus both trapezoids have | + | Since <math>XY = WZ</math> and <math>PQ = PQ</math> and the [[area]]s of the [[trapezoid]]s <math>\displaystyle PQZW</math> and <math>\displaystyle PQYX</math> are the same, the heights of the trapezoids are the same. Thus both trapezoids have area <math>\frac{1}{2} \cdot \frac{19}{2}(XY + PQ) = \frac{19}{4}(XY + 87)</math>. This number is also equal to one quarter the area of the entire rectangle, which is <math>\frac{19\cdot AB}{4}</math>, so we have <math>AB = XY + 87</math>. |
In addition, we see that the [[perimeter]] of the rectangle is <math>2AB + 38 = XA + AD + DW + WZ + ZC + CB + BY + YX = 4XY</math>, so <math>AB + 19 = 2XY</math>. | In addition, we see that the [[perimeter]] of the rectangle is <math>2AB + 38 = XA + AD + DW + WZ + ZC + CB + BY + YX = 4XY</math>, so <math>AB + 19 = 2XY</math>. |
Revision as of 21:52, 6 March 2007
Problem
Rectangle is divided into four parts of equal area by five segments as shown in the figure, where , and is parallel to . Find the length of (in cm) if cm and cm.
Solution
Since and and the areas of the trapezoids and are the same, the heights of the trapezoids are the same. Thus both trapezoids have area . This number is also equal to one quarter the area of the entire rectangle, which is , so we have .
In addition, we see that the perimeter of the rectangle is , so .
Solving these two equations gives .
See also
1987 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |