Difference between revisions of "2017 AMC 10B Problems/Problem 22"

Revision as of 16:16, 20 February 2017

Problem

The diameter $\overline{AB}$ of a circle of radius $2$ is extended to a point $D$ outside the circle so that $BD=3$ . Point $E$ is chosen so that $ED=5$ and line $ED$ is perpendicular to line $AD$ . Segment $\overline{AE}$ intersects the circle at a point $C$ between $A$ and $E$ . What is the area of $\triangle ABC$ ?

$\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}$

Solution

Solution 1

Notice that $ADE$ and $ABC$ are right triangles. Then $AE = \sqrt{7^2+5^2} = \sqrt{74}$ . $\sin{DAE} = \frac{5}{\sqrt{74}} = \sin{BAE} = \sin{BAC} = \frac{BC}{4}$ , so $BC = \frac{20}{\sqrt{74}}$ . We also find that $AC = \frac{28}{\sqrt{74}}$ , and thus the area of $ABC$ is $\frac{\frac{20}{\sqrt{74}}\cdot\frac{28}{\sqrt{74}}}{2} = \frac{\frac{560}{74}}{2} = \boxed{\textbf{(D) } \frac{140}{37}}$ .

Solution 2

We note that $\triangle ACB ~ \triangle ADE$ by $AA$ similarity. Also, since the area of $\triangle ADE = \frac{7 \cdot 5}2 = \frac{35}2$ and $AE = \sqrt{74}$ , $\frac{[ABC]}{[ADE]} = \frac{[ABC]}{\frac{35}2} = \left(\frac{4}{\sqrt{74}}\right)^2$ , so the area of $\triangle ABC = \boxed{\textbf{(D) } \frac{140}{37}}$ .

Solution 3

As stated before, note that $\triangle ACB ~ \triangle ADE$ . By similarity, we note that $\frac{\overline{AC}}{\overline{BC}}$ is equivalent to $\frac{7}{5}$ . We set $\overline{AC}$ to $7x$ and $\overline{BC}$ to $5x$ . By the Pythagorean Theorem, $(7x)^2+(5x)^2$ = 4^2. Combining, $49x^2+25x^2=16$ . We can add and divide to get $x^2=\frac{8}{37}$ . We square root and rearrange to get $x=\frac{2\sqrt{74}}{37}$ . We know that the legs of the triangle are $7x$ and $5x$ . Mulitplying $x$ by 7 and 5 eventually gives us $\frac{14\sqrt{74}}{37}$ x $\frac{10\sqrt{74}}{37}$ . We divide this by 2, since $\frac{1}{2}bh$ is the formula for a triangle. This gives us $\boxed{\textbf{(D) } \frac{140}{37}}$ .

@@ Line 11: / Line 11: @@
 ===Solution 2===
 We note that <math>\triangle ACB ~ \triangle ADE</math> by <math>AA</math> similarity. Also, since the area of <math>\triangle ADE = \frac{7 \cdot 5}2 = \frac{35}2</math> and <math>AE = \sqrt{74}</math>, <math>\frac{[ABC]}{[ADE]} = \frac{[ABC]}{\frac{35}2} = \left(\frac{4}{\sqrt{74}}\right)^2</math>, so the area of <math>\triangle ABC = \boxed{\textbf{(D) } \frac{140}{37}}</math>.
+===Solution 3===
+As stated before, note that <math>\triangle ACB ~ \triangle ADE</math>. By similarity, we note that <math>\frac{\overline{AC}}{\overline{BC}}</math> is equivalent to <math>\frac{7}{5}</math>. We set <math>\overline{AC}</math> to <math>7x</math> and <math>\overline{BC}</math> to <math>5x</math>. By the Pythagorean Theorem, <math>(7x)^2+(5x)^2</math> = 4^2. Combining, <math>49x^2+25x^2=16</math>. We can add and divide to get <math>x^2=\frac{8}{37}</math>. We square root and rearrange to get <math>x=\frac{2\sqrt{74}}{37}</math>. We know that the legs of the triangle are <math>7x</math> and <math>5x</math>. Mulitplying <math>x</math> by 7 and 5 eventually gives us <math>\frac{14\sqrt{74}}{37}</math>x<math>\frac{10\sqrt{74}}{37}</math>. We divide this by 2, since <math>\frac{1}{2}bh</math> is the formula for a triangle. This gives us <math>\boxed{\textbf{(D) } \frac{140}{37}}</math>.
 ==See Also==
 {{AMC10 box|year=2017|ab=B|num-b=21|num-a=23}}
 {{MAA Notice}}

2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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
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