Difference between revisions of "2009 AMC 10A Problems/Problem 10"

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\mathrm{(E)}\ 42
 
\mathrm{(E)}\ 42
 
</math>
 
</math>
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[[Category: Introductory Geometry Problems]]
  
== Solution ==
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== Solution 1==
  
 
It is a well-known fact that in any right triangle <math>ABC</math> with the right angle at <math>B</math> and <math>D</math> the foot of the altitude from <math>B</math> onto <math>AC</math> we have <math>BD^2 = AD\cdot CD</math>. (See below for a proof.) Then <math>BD = \sqrt{ 3\cdot 4 } = 2\sqrt 3</math>, and the area of the triangle <math>ABC</math> is <math>\frac{AC\cdot BD}2 = 7\sqrt3\Rightarrow\boxed{\text{(B)}}</math>.
 
It is a well-known fact that in any right triangle <math>ABC</math> with the right angle at <math>B</math> and <math>D</math> the foot of the altitude from <math>B</math> onto <math>AC</math> we have <math>BD^2 = AD\cdot CD</math>. (See below for a proof.) Then <math>BD = \sqrt{ 3\cdot 4 } = 2\sqrt 3</math>, and the area of the triangle <math>ABC</math> is <math>\frac{AC\cdot BD}2 = 7\sqrt3\Rightarrow\boxed{\text{(B)}}</math>.
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Substituting equations 2 and 3 into the left hand side of equation 1, we get <math>BD^2 =  AD \cdot DC</math>.  
 
Substituting equations 2 and 3 into the left hand side of equation 1, we get <math>BD^2 =  AD \cdot DC</math>.  
  
Alternatively, note that <math>\triangle ABD \sim \triangle BCD \Longrightarrow \frac{AD}{BD} = \frac{BD}{CD}</math>. <math>\blacksquare</math>
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Alternatively, note that <math>\triangle ABD \sim \triangle BCD \Longrightarrow \frac{AD}{BD} = \frac{BD}{CD}</math>.
 +
 
 +
== Solution 2 ==
 +
 
 +
For those looking for a dumber solution, we can use Pythagoras and manipulation of area formulas as well to solve the problem.
 +
 
 +
Assume the length of <math>BD</math> is equal to <math>h</math>. Then, by Pythagoras, we have,
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 +
<cmath>AB^2 = h^2 + 9 \Rightarrow AB = \sqrt{h^2 + 9}</cmath>
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<cmath>BC^2 = h^2 + 16 \Rightarrow BC = \sqrt{h^2 + 16}</cmath>
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 +
Then, by area formulas, we know that
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<cmath>\frac{1}{2}\sqrt{(h^2+9)(h^2+16)} = \frac{1}{2}(7)(h)</cmath>
 +
 
 +
Squaring and solving the subsequent equation yields our solution that <math>h^2 = 12 \Rightarrow h = 2\sqrt{3}.</math> Since the area of the triangle is half of this quantity multiplied by the base, we have
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<cmath>\text{area}  = \frac{1}{2}(7)(2\sqrt{3})\Rightarrow \boxed{7\sqrt{3}}</cmath>
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 +
== Solution 3 (Power of a point)==
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Draw the circumcircle <math>\omega</math> of the <math>\Delta ABC</math>. Because <math>\Delta ABC</math> is a right angle triangle, AC is the diameter of the circumcircle. By applying [[Power of a Point Theorem]], we can have <math>BD=DE</math> and <math>AD\cdot CD=BD^2</math> <math>\Rightarrow BD=\sqrt{3\times 4}=2\sqrt{3}</math>. Then we have <math>S_{[ABC]}=\frac{1}{2}(7)(2\sqrt{3})=\boxed{7\sqrt{3}}</math>
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<asy>
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unitsize(6mm);
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defaultpen(linewidth(.8pt)+fontsize(8pt));
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dotfactor=4;
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pair B=(0,0), C=(sqrt(28),0), A=(0,sqrt(21)), E=(6*sqrt(28)/7,8*sqrt(21)/7);
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pair D=foot(B,A,C);
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pair[] ps={B,C,A,D};
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filldraw(Circle((sqrt(28)/2,sqrt(21)/2),sqrt(49)/2),white,black);
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draw(A--B--C--cycle);
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draw(B--D);
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draw(E--D);
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draw(rightanglemark(B,D,C));
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dot(ps);
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label("$A$",A,NW);
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label("$B$",B,SW);
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label("$E$",E,NE);
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label("$C$",C,SE);
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label("$D$",D,NE);
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label("$3$",midpoint(A--D),NE);
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label("$4$",midpoint(D--C),NE);
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</asy>
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~Bran_Qin
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== Video Solution ==
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https://youtu.be/4_x1sgcQCp4?t=1195
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 +
~ pi_is_3.14
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC10 box|year=2009|ab=A|num-b=9|num-a=11}}
 
{{AMC10 box|year=2009|ab=A|num-b=9|num-a=11}}
 +
{{MAA Notice}}

Latest revision as of 10:49, 29 July 2021

Problem

Triangle $ABC$ has a right angle at $B$. Point $D$ is the foot of the altitude from $B$, $AD=3$, and $DC=4$. What is the area of $\triangle ABC$?

[asy] unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=4;  pair B=(0,0), C=(sqrt(28),0), A=(0,sqrt(21)); pair D=foot(B,A,C); pair[] ps={B,C,A,D};  draw(A--B--C--cycle); draw(B--D); draw(rightanglemark(B,D,C));  dot(ps); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,NE); label("$3$",midpoint(A--D),NE); label("$4$",midpoint(D--C),NE); [/asy]

$\mathrm{(A)}\ 4\sqrt3 \qquad \mathrm{(B)}\ 7\sqrt3 \qquad \mathrm{(C)}\ 21 \qquad \mathrm{(D)}\ 14\sqrt3  \qquad \mathrm{(E)}\ 42$

Solution 1

It is a well-known fact that in any right triangle $ABC$ with the right angle at $B$ and $D$ the foot of the altitude from $B$ onto $AC$ we have $BD^2 = AD\cdot CD$. (See below for a proof.) Then $BD = \sqrt{ 3\cdot 4 } = 2\sqrt 3$, and the area of the triangle $ABC$ is $\frac{AC\cdot BD}2 = 7\sqrt3\Rightarrow\boxed{\text{(B)}}$.

Proof: Consider the Pythagorean theorem for each of the triangles $ABC$, $ABD$, and $CBD$. We get:

  1. $AB^2 + BC^2 = AC^2 = (AD+DC)^2 = AD^2 + DC^2 + 2 \cdot AD \cdot DC$.
  2. $AB^2 = AD^2 + BD^2$
  3. $BC^2 = BD^2 + CD^2$

Substituting equations 2 and 3 into the left hand side of equation 1, we get $BD^2 =  AD \cdot DC$.

Alternatively, note that $\triangle ABD \sim \triangle BCD \Longrightarrow \frac{AD}{BD} = \frac{BD}{CD}$.

Solution 2

For those looking for a dumber solution, we can use Pythagoras and manipulation of area formulas as well to solve the problem.

Assume the length of $BD$ is equal to $h$. Then, by Pythagoras, we have,

\[AB^2 = h^2 + 9 \Rightarrow AB = \sqrt{h^2 + 9}\] \[BC^2 = h^2 + 16 \Rightarrow BC = \sqrt{h^2 + 16}\]

Then, by area formulas, we know that

\[\frac{1}{2}\sqrt{(h^2+9)(h^2+16)} = \frac{1}{2}(7)(h)\]

Squaring and solving the subsequent equation yields our solution that $h^2 = 12 \Rightarrow h = 2\sqrt{3}.$ Since the area of the triangle is half of this quantity multiplied by the base, we have \[\text{area}  = \frac{1}{2}(7)(2\sqrt{3})\Rightarrow \boxed{7\sqrt{3}}\]

Solution 3 (Power of a point)

Draw the circumcircle $\omega$ of the $\Delta ABC$. Because $\Delta ABC$ is a right angle triangle, AC is the diameter of the circumcircle. By applying Power of a Point Theorem, we can have $BD=DE$ and $AD\cdot CD=BD^2$ $\Rightarrow BD=\sqrt{3\times 4}=2\sqrt{3}$. Then we have $S_{[ABC]}=\frac{1}{2}(7)(2\sqrt{3})=\boxed{7\sqrt{3}}$


[asy] unitsize(6mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=4;  pair B=(0,0), C=(sqrt(28),0), A=(0,sqrt(21)), E=(6*sqrt(28)/7,8*sqrt(21)/7); pair D=foot(B,A,C); pair[] ps={B,C,A,D};   filldraw(Circle((sqrt(28)/2,sqrt(21)/2),sqrt(49)/2),white,black); draw(A--B--C--cycle); draw(B--D); draw(E--D); draw(rightanglemark(B,D,C));  dot(ps); label("$A$",A,NW); label("$B$",B,SW); label("$E$",E,NE); label("$C$",C,SE); label("$D$",D,NE); label("$3$",midpoint(A--D),NE); label("$4$",midpoint(D--C),NE); [/asy] ~Bran_Qin

Video Solution

https://youtu.be/4_x1sgcQCp4?t=1195

~ pi_is_3.14

See Also

2009 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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 AMC 10 Problems and Solutions

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