Difference between revisions of "2011 AMC 8 Problems/Problem 20"
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− | Quadrilateral <math>ABCD</math> is a trapezoid, <math>AD = 15</math>, <math>AB = 50</math>, <math>BC = 20</math>, and the altitude is <math>12</math>. What is the area of the | + | ==Problem== |
+ | Quadrilateral <math>ABCD</math> is a trapezoid, <math>AD = 15</math>, <math>AB = 50</math>, <math>BC = 20</math>, and the altitude is <math>12</math>. What is the area of the trapezoid? | ||
<asy> | <asy> | ||
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<math> \textbf{(A) }600\qquad\textbf{(B) }650\qquad\textbf{(C) }700\qquad\textbf{(D) }750\qquad\textbf{(E) }800 </math> | <math> \textbf{(A) }600\qquad\textbf{(B) }650\qquad\textbf{(C) }700\qquad\textbf{(D) }750\qquad\textbf{(E) }800 </math> | ||
+ | |||
+ | ==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> | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/51K3uCzntWs?t=2521 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/MjxiQ9MZiHk | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC8 box|year=2011|num-b=19|num-a=21}} | ||
+ | {{MAA Notice}} |
Latest revision as of 11:31, 15 July 2024
Problem
Quadrilateral is a trapezoid,
,
,
, and the altitude is
. What is the area of the trapezoid?
Solution
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.
is a rectangle so
The area of the trapezoid is
Video Solution by OmegaLearn
https://youtu.be/51K3uCzntWs?t=2521
~ pi_is_3.14
Video Solution by WhyMath
~savannahsolver
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.