Difference between revisions of "2011 AMC 8 Problems/Problem 20"
m (Added Solution and See Also headers) |
|||
| Line 28: | Line 28: | ||
==Solution== | ==Solution== | ||
| + | |||
| + | <asy> | ||
| + | unitsize(1.5mm); | ||
| + | defaultpen(linewidth(.9pt)+fontsize(10pt)); | ||
| + | dotfactor=3; | ||
| + | |||
| + | pair A,B,C,D,X,Y; | ||
| + | A=(9,12); B=(59,12); C=(75,0); D=(0,0); X=(9,0); Y=(59,0); | ||
| + | draw(A--B--C--D--cycle); | ||
| + | draw(A--X); draw(B--Y); | ||
| + | |||
| + | pair[] ps={A,B,C,D,X,Y}; | ||
| + | dot(ps); | ||
| + | |||
| + | label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); | ||
| + | label("$X$",X,SE); label("$Y$",Y,S); | ||
| + | label("$a$",D--X,S); label("$b$",Y--C,S); | ||
| + | label("$15$",D--A,NW); label("$50$",B--A,N); label("$20$",B--C,NE); label("$12$",X--A,E); label("$12$",Y--B,W); | ||
| + | </asy> | ||
| + | |||
| + | If you draw altitudes from <math>A</math> and <math>B</math> to <math>CD,</math> the trapezoid will be divided into two right triangles and a rectangle. You can find the values of <math>a</math> and <math>b</math> with the [[Pythagorean theorem]]. | ||
| + | |||
| + | <cmath>a=\sqrt{15^2-12^2}=\sqrt{81}=9</cmath> | ||
| + | |||
| + | <cmath>b=\sqrt{20^2-12^2}=\sqrt{256}=16</cmath> | ||
| + | |||
| + | <math>ABYX</math> is a rectangle so <math>XY=AB=50.</math> | ||
| + | |||
| + | <cmath>CD=a+XY+b=9+50+16=75</cmath> | ||
| + | |||
| + | The area of the trapezoid is | ||
| + | |||
| + | <cmath>12\cdot \frac{(50+75)}{2} = 6(125) = \boxed{\textbf{(D)}\ 750}</cmath> | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2011|num-b=19|num-a=21}} | {{AMC8 box|year=2011|num-b=19|num-a=21}} | ||
Revision as of 21:14, 25 November 2011
Quadrilateral 
is a trapezoid, 
, 
, 
, and the altitude is 
. What is the area of the trapeziod?
![[asy] pair A,B,C,D; A=(3,20); B=(35,20); C=(47,0); D=(0,0); draw(A--B--C--D--cycle); dot((0,0)); dot((3,20)); dot((35,20)); dot((47,0)); label("A",A,N); label("B",B,N); label("C",C,S); label("D",D,S); draw((19,20)--(19,0)); dot((19,20)); dot((19,0)); draw((19,3)--(22,3)--(22,0)); label("12",(21,10),E); label("50",(19,22),N); label("15",(1,10),W); label("20",(41,12),E);[/asy]](http://latex.artofproblemsolving.com/a/5/b/a5bd7719b521c33191726670908546d54161aeda.png)
Solution
![[asy] unitsize(1.5mm); defaultpen(linewidth(.9pt)+fontsize(10pt)); dotfactor=3; pair A,B,C,D,X,Y; A=(9,12); B=(59,12); C=(75,0); D=(0,0); X=(9,0); Y=(59,0); draw(A--B--C--D--cycle); draw(A--X); draw(B--Y); pair[] ps={A,B,C,D,X,Y}; dot(ps); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$X$",X,SE); label("$Y$",Y,S); label("$a$",D--X,S); label("$b$",Y--C,S); label("$15$",D--A,NW); label("$50$",B--A,N); label("$20$",B--C,NE); label("$12$",X--A,E); label("$12$",Y--B,W); [/asy]](http://latex.artofproblemsolving.com/c/d/e/cde7b3b44e3f5ba857b5151d425928901478ec30.png)
If you draw altitudes from 
and 
to 
the trapezoid will be divided into two right triangles and a rectangle. You can find the values of 
and 
with the Pythagorean theorem.
![\[a=\sqrt{15^2-12^2}=\sqrt{81}=9\]](http://latex.artofproblemsolving.com/2/7/5/2755d6e1e591261df5a6462b4497b28773090c56.png)
![\[b=\sqrt{20^2-12^2}=\sqrt{256}=16\]](http://latex.artofproblemsolving.com/c/e/6/ce6556edf7ce29e6cdbce56c0514b89382528d87.png)
is a rectangle so 
![\[CD=a+XY+b=9+50+16=75\]](http://latex.artofproblemsolving.com/9/8/4/9843a0050be4143ca4e871f1cef98b398ac84e66.png)
The area of the trapezoid is
![\[12\cdot \frac{(50+75)}{2} = 6(125) = \boxed{\textbf{(D)}\ 750}\]](http://latex.artofproblemsolving.com/8/3/5/83558d77c0d5b33f316ba373de8af2f83e88e8dd.png)
See Also
| 2011 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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 | ||
