Difference between revisions of "1952 AHSME Problems/Problem 24"
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| + | == Problem== | ||
| + | In the figure, it is given that angle <math> C = 90^{\circ} </math>, <math> \overline{AD} = \overline{DB} </math>, <math> DE \perp AB </math>, <math> \overline{AB} = 20 </math>, and <math> \overline{AC} = 12 </math>. The area of quadrilateral <math> ADEC </math> is: | ||
| + | <asy> | ||
| + | unitsize(7); | ||
| + | defaultpen(linewidth(.8pt)+fontsize(10pt)); | ||
| + | pair A,B,C,D,E; | ||
| + | A=(0,0); B=(20,0); C=(36/5,48/5); D=(10,0); E=(10,75/10); | ||
| + | draw(A--B--C--cycle); draw(D--E); | ||
| + | label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,S); label("$E$",E,NE); | ||
| + | draw(rightanglemark(B,D,E,30)); | ||
| + | </asy> | ||
| + | |||
| + | <math> \textbf{(A)}\ 75\qquad\textbf{(B)}\ 58\frac{1}{2}\qquad\textbf{(C)}\ 48\qquad\textbf{(D)}\ 37\frac{1}{2}\qquad\textbf{(E)}\ \text{none of these} </math> | ||
| + | |||
| + | ==Solution== | ||
| + | |||
| + | <math> [ADEC]=[BCA]-[BDE] </math> | ||
| + | |||
| + | Note that, as right triangles sharing <math> \angle B </math>, <math> \triangle BDE \sim \triangle BCA </math>. | ||
| + | |||
| + | <math> \overline{BD}=\frac{20}{2}=10 </math> | ||
| + | |||
| + | Because the sides of <math> \triangle BCA </math> are in the ratio <math> 3:4:5 </math>, <math> \overline{BC}=16 </math>. | ||
| + | |||
| + | The sides of our triangles are in the ratio <math> \frac{10}{16}=\frac{5}{8} </math>, and the ratio of their areas is <math> \left(\frac{5}{8}\right)^2=\frac{25}{64} </math>. | ||
| + | |||
| + | <math> [BCA]=\frac{12 \cdot 16}{2}=96</math>, and <math> [BDE]=\frac{96 \cdot 25}{64}=\frac{75}{2} </math>. | ||
| + | |||
| + | <math> [ADEC]=\frac{192-75}{2}=\boxed{\textbf{(B)}\ 58\frac{1}{2}} </math> | ||
| + | |||
| + | ==See also== | ||
| + | {{AHSME 50p box|year=1952|num-b=23|num-a=25}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 01:06, 25 January 2014
Problem
In the figure, it is given that angle 
, 
, 
, 
, and 
. The area of quadrilateral 
is:
![[asy] unitsize(7); defaultpen(linewidth(.8pt)+fontsize(10pt)); pair A,B,C,D,E; A=(0,0); B=(20,0); C=(36/5,48/5); D=(10,0); E=(10,75/10); draw(A--B--C--cycle); draw(D--E); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,S); label("$E$",E,NE); draw(rightanglemark(B,D,E,30)); [/asy]](http://latex.artofproblemsolving.com/4/4/3/443fde7549f310f3da05d18725232c0f573b5f8a.png)
Solution
![$[ADEC]=[BCA]-[BDE]$](http://latex.artofproblemsolving.com/5/e/a/5ea314f60b9f1201ff49f3392b635dbc8c906ec3.png)
Note that, as right triangles sharing 
, 
.
Because the sides of 
are in the ratio 
, 
.
The sides of our triangles are in the ratio 
, and the ratio of their areas is 
.
![$[BCA]=\frac{12 \cdot 16}{2}=96$](http://latex.artofproblemsolving.com/4/2/c/42c3ccd38d4c44d008a442ac2ecb6233c7c93c50.png)
, and ![$[BDE]=\frac{96 \cdot 25}{64}=\frac{75}{2}$](http://latex.artofproblemsolving.com/b/2/0/b202984851a2c3c9e24c5462a8f2150218ccddf4.png)
.
![$[ADEC]=\frac{192-75}{2}=\boxed{\textbf{(B)}\ 58\frac{1}{2}}$](http://latex.artofproblemsolving.com/5/2/9/52957d6d5a338637d53e97774fa87394e66ac9ee.png)
See also
| 1952 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
