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Difference between revisions of "1952 AHSME Problems/Problem 24"

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== Problem==
  
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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:
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<asy>
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unitsize(7);
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defaultpen(linewidth(.8pt)+fontsize(10pt));
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pair A,B,C,D,E;
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A=(0,0); B=(20,0); C=(36/5,48/5); D=(10,0); E=(10,75/10);
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draw(A--B--C--cycle); draw(D--E);
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label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,S); label("$E$",E,NE);
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draw(rightanglemark(B,D,E,30));
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</asy>
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<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>
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==Solution==
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<math> [ADEC]=[BCA]-[BDE] </math>
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Note that, as right triangles sharing <math> \angle B </math>, <math> \triangle BDE \sim \triangle BCA </math>.
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<math> \overline{BD}=\frac{20}{2}=10 </math>
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Because the sides of <math> \triangle BCA </math> are in the ratio <math> 3:4:5 </math>, <math> \overline{BC}=16 </math>.
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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>.
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<math> [BCA]=\frac{12 \cdot 16}{2}=96</math>, and <math> [BDE]=\frac{96 \cdot 25}{64}=\frac{75}{2} </math>.
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<math> [ADEC]=\frac{192-75}{2}=\boxed{\textbf{(B)}\ 58\frac{1}{2}} </math>
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==See also==
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{{AHSME 50p box|year=1952|num-b=23|num-a=25}}
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{{MAA Notice}}

Latest revision as of 00:06, 25 January 2014

Problem

In the figure, it is given that angle $C = 90^{\circ}$, $\overline{AD} = \overline{DB}$, $DE \perp AB$, $\overline{AB} = 20$, and $\overline{AC} = 12$. The area of quadrilateral $ADEC$ 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]

$\textbf{(A)}\ 75\qquad\textbf{(B)}\ 58\frac{1}{2}\qquad\textbf{(C)}\ 48\qquad\textbf{(D)}\ 37\frac{1}{2}\qquad\textbf{(E)}\ \text{none of these}$

Solution

$[ADEC]=[BCA]-[BDE]$

Note that, as right triangles sharing $\angle B$, $\triangle BDE \sim \triangle BCA$.

$\overline{BD}=\frac{20}{2}=10$

Because the sides of $\triangle BCA$ are in the ratio $3:4:5$, $\overline{BC}=16$.

The sides of our triangles are in the ratio $\frac{10}{16}=\frac{5}{8}$, and the ratio of their areas is $\left(\frac{5}{8}\right)^2=\frac{25}{64}$.

$[BCA]=\frac{12 \cdot 16}{2}=96$, and $[BDE]=\frac{96 \cdot 25}{64}=\frac{75}{2}$.

$[ADEC]=\frac{192-75}{2}=\boxed{\textbf{(B)}\ 58\frac{1}{2}}$

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

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