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

(Solution)
m
Line 18: Line 18:
 
==Solution==
 
==Solution==
  
Let <math>!ABC!</math> denote the area of figure <math>ABC</math>.
+
Let <math>[ABC]</math> denote the area of figure <math>ABC</math>.
  
Clearly, <math>!BEDC!=!ABCD!-!ABE!</math>.  Using basic area formulas,
+
Clearly, <math>[BEDC]=[ABCD]-[ABE]</math>.  Using basic area formulas,
<cmath>!ABCD!=(BC)(BE)=80</cmath>
+
 
<cmath>!ABE!=(BE)(AE)/2 = 4(AE)</cmath>
+
<center><math>[ABCD]=(BC)(BE)=80</math></center>
 +
 
 +
<center><math>[ABE]=(BE)(AE)/2 = 4(AE)</math></center>
  
 
Since <math>AE+ED=BC=10</math> and <math>ED=6</math>, <math>AE=4</math> and the area of <math>\triangle ABE</math> is <math>4(4)=16</math>.
 
Since <math>AE+ED=BC=10</math> and <math>ED=6</math>, <math>AE=4</math> and the area of <math>\triangle ABE</math> is <math>4(4)=16</math>.
  
Finally, we have <math>!BEDC!=80-16=64\rightarrow \boxed{\text{D}}</math>
+
Finally, we have <math>[BEDC]=80-16=64\rightarrow \boxed{\text{D}}</math>
  
 
==See Also==
 
==See Also==

Revision as of 22:23, 25 April 2010

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

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

See Also

1989 AJHSME (ProblemsAnswer KeyResources)
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