Difference between revisions of "1987 AIME Problems/Problem 6"
Mrdavid445 (talk | contribs) |
(→Solution) |
||
Line 12: | Line 12: | ||
---- | ---- | ||
− | + | ==Solution 2== | |
− | |||
Let <math>YB=a</math>, <math>CZ=b</math>, <math>AX=c</math>, and <math>WD=d</math>. First we drop a perpendicular from <math>Q</math> to a point <math>R</math> on <math>BC</math> so <math>QR=H</math>. Since <math>XY = WZ</math> and <math>PQ = PQ</math> and the [[area]]s of the [[trapezoid]]s <math>PQZW</math> and <math>PQYX</math> are the same, the heights of the trapezoids are both <math>\frac{19}{2}</math>.From here, we have that <math>[BYQZC]=\frac{a+h}{2}*19/2+\frac{b+h}{2}*19/2=19/2* \frac{a+b+2h}{2}</math>. We are told that this area is equal to <math>[PXYQ]=\frac{19}{2}* \frac{XY+87}{2}=\frac{19}{2}* \frac{a+b+106}{2}</math>. Setting these equal to each other and solving gives <math>H=53</math>. In the same way, we find that the perpendicular from <math>P</math> to <math>AD</math> is <math>53</math>. So <math>AB=53*2+87=193</math> | Let <math>YB=a</math>, <math>CZ=b</math>, <math>AX=c</math>, and <math>WD=d</math>. First we drop a perpendicular from <math>Q</math> to a point <math>R</math> on <math>BC</math> so <math>QR=H</math>. Since <math>XY = WZ</math> and <math>PQ = PQ</math> and the [[area]]s of the [[trapezoid]]s <math>PQZW</math> and <math>PQYX</math> are the same, the heights of the trapezoids are both <math>\frac{19}{2}</math>.From here, we have that <math>[BYQZC]=\frac{a+h}{2}*19/2+\frac{b+h}{2}*19/2=19/2* \frac{a+b+2h}{2}</math>. We are told that this area is equal to <math>[PXYQ]=\frac{19}{2}* \frac{XY+87}{2}=\frac{19}{2}* \frac{a+b+106}{2}</math>. Setting these equal to each other and solving gives <math>H=53</math>. In the same way, we find that the perpendicular from <math>P</math> to <math>AD</math> is <math>53</math>. So <math>AB=53*2+87=193</math> | ||
Revision as of 21:49, 9 March 2012
Contents
[hide]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 .
Solution 2
Let , , , and . First we drop a perpendicular from to a point on so . Since and and the areas of the trapezoids and are the same, the heights of the trapezoids are both .From here, we have that . We are told that this area is equal to . Setting these equal to each other and solving gives . In the same way, we find that the perpendicular from to is . So
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 |