Difference between revisions of "1989 AJHSME Problems/Problem 15"

Latest revision as of 23:03, 4 July 2013

Problem

The area of the shaded region $\text{BEDC}$ in parallelogram $\text{ABCD}$ is

$[asy] unitsize(10); pair A,B,C,D,E; A=origin; B=(4,8); C=(14,8); D=(10,0); E=(4,0); draw(A--B--C--D--cycle); fill(B--E--D--C--cycle,gray); label("A",A,SW); label("B",B,NW); label("C",C,NE); label("D",D,SE); label("E",E,S); label("$10$",(9,8),N); label("$6$",(7,0),S); label("$8$",(4,4),W); draw((3,0)--(3,1)--(4,1)); [/asy]$

$\text{(A)}\ 24 \qquad \text{(B)}\ 48 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 80$

Solution 1

Let $[ABC]$ denote the area of figure $ABC$ .

Clearly, $[BEDC]=[ABCD]-[ABE]$ . Using basic area formulas,

$[ABCD]=(BC)(BE)=80$ $[ABE]=(BE)(AE)/2 = 4(AE)$

Since $AE+ED=BC=10$ and $ED=6$ , $AE=4$ and the area of $\triangle ABE$ is $4(4)=16$ .

Finally, we have $[BEDC]=80-16=64\rightarrow \boxed{\text{D}}$

Solution 2

Notice that $BEDC$ is a trapezoid. Therefore its area is $8\left(\frac{6+10}{2}\right)=8\left(\frac{16}{2}\right)=8(8)=64\Rightarrow \mathrm{(D)}$

1989 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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.

@@ Line 38: / Line 38: @@
 {{AJHSME box|year=1989|num-b=14|num-a=16}}
 [[Category:Introductory Geometry Problems]]
+{{MAA Notice}}

Page

Toolbox

Search

Difference between revisions of "1989 AJHSME Problems/Problem 15"

Latest revision as of 23:03, 4 July 2013

Contents

Problem

Solution 1

Solution 2

See Also