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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 ==
+
== 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>
+
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 to solve the problem.
 +
 
 +
Assume the length of <math>BD</math> is equal to <math>h</math>. Then, by Pythagoras, we have,
 +
 
 +
<cmath>AB^2 = h^2 + 9 \Rightarrow AB = \sqrt{h^2 + 9}</cmath>
 +
<cmath>BC^2 = h^2 + 16 \Rightarrow BC = \sqrt{h^2 + 16}</cmath>
 +
 
 +
Then, by area formulas, we know:
 +
 
 +
<cmath>\frac{1}{2}\sqrt{(h^2+9)(h^2+16)} = \frac{1}{2}(7)(h)</cmath>
 +
 
 +
Squaring and solving the above 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
 +
<cmath>\text{area}  = \frac{1}{2}(7)(2\sqrt{3})\Rightarrow \boxed{7\sqrt{3}}</cmath>
 +
 
 +
== Solution 3 (Power of a point)==
 +
 
 +
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>
 +
 
 +
 
 +
<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
 +
 
 +
===Solution 4 (Fakesolve)/Answer Choices===
 +
The area of the triangle is <math>\frac{7x}{2}</math>. We want to fine what <math>x</math> is. Now, we try each answer choice.
 +
 
 +
<math>A): 4 \sqrt{3}</math>. I am too lazy to go over this, but we immediately see that this is very improbably due to the area being <math>\frac{7x}{2}</math>. This does not work.
 +
<math>B): 7 \sqrt{3}</math>. This is promising. This means that <math>x = 2\sqrt{3}</math>. Now, applying Pythagorean Theorem, we have the vertical sides is <math>\sqrt {21}</math> and the horizontal side is <math>\sqrt {28}</math>. Multiplying these and dividing by <math>2</math> indeed gives us <math>7 \sqrt {3}</math> as desired. Therefore, the answer is <math>\boxed{7 \sqrt{3}}</math>
 +
 
 +
~Arcticturn
 +
 
 +
== Video Solution by OmegaLearn==
 +
https://youtu.be/4_x1sgcQCp4?t=1195
 +
 
 +
~ 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 03:17, 23 January 2023

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 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:

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

Squaring and solving the above 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

Solution 4 (Fakesolve)/Answer Choices

The area of the triangle is $\frac{7x}{2}$. We want to fine what $x$ is. Now, we try each answer choice.

$A): 4 \sqrt{3}$. I am too lazy to go over this, but we immediately see that this is very improbably due to the area being $\frac{7x}{2}$. This does not work. $B): 7 \sqrt{3}$. This is promising. This means that $x = 2\sqrt{3}$. Now, applying Pythagorean Theorem, we have the vertical sides is $\sqrt {21}$ and the horizontal side is $\sqrt {28}$. Multiplying these and dividing by $2$ indeed gives us $7 \sqrt {3}$ as desired. Therefore, the answer is $\boxed{7 \sqrt{3}}$

~Arcticturn

Video Solution by OmegaLearn

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

~ pi_is_3.14

See Also

2009 AMC 10A (Problems • Answer Key • Resources)
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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