Difference between revisions of "1991 AJHSME Problems/Problem 10"
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==Solution== | ==Solution== | ||
| − | The base is <math>\overline{BC}=4</math>. The height has a length of the difference of the y-coordinates of A and B, which is 2. Therefore the area is <math>4\cdot 2=8\Rightarrow \boxed{\mathrm{B}</math>. | + | The base is <math>\overline{BC}=4</math>. The height has a length of the difference of the y-coordinates of A and B, which is 2. Therefore the area is <math>4\cdot 2=8\Rightarrow \boxed{\mathrm{B}}</math>. |
==See Also== | ==See Also== | ||
Revision as of 21:18, 7 August 2009
Problem 10
The area in square units of the region enclosed by parallelogram 
is
![[asy] unitsize(24); pair A,B,C,D; A=(-1,0); B=(0,2); C=(4,2); D=(3,0); draw(A--B--C--D); draw((0,-1)--(0,3)); draw((-2,0)--(6,0)); draw((-.25,2.75)--(0,3)--(.25,2.75)); draw((5.75,.25)--(6,0)--(5.75,-.25)); dot(origin); dot(A); dot(B); dot(C); dot(D); label("$y$",(0,3),N); label("$x$",(6,0),E); label("$(0,0)$",origin,SE); label("$D (3,0)$",D,SE); label("$C (4,2)$",C,NE); label("$A$",A,SW); label("$B$",B,NW); [/asy]](http://latex.artofproblemsolving.com/0/0/6/006625b37683a704cc02bcf460b1e73c2e0ecc24.png)
Solution
The base is 
. The height has a length of the difference of the y-coordinates of A and B, which is 2. Therefore the area is 
.
See Also
| 1991 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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 | ||
