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

(Solution)
(Solution 2)
 
Line 66: Line 66:
 
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 into the base, we have  
 
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 into the base, we have  
 
<cmath>\text{area}  = \frac{1}{2}(7)(2\sqrt{3})\Rightarrow \boxed{7\sqrt{3}}</cmath>
 
<cmath>\text{area}  = \frac{1}{2}(7)(2\sqrt{3})\Rightarrow \boxed{7\sqrt{3}}</cmath>
 +
 +
== Video Solution ==
 +
https://youtu.be/4_x1sgcQCp4?t=1195
 +
 +
~ pi_is_3.14
  
 
== See Also ==
 
== See Also ==

Latest revision as of 20:34, 24 January 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 into the base, we have \[\text{area}  = \frac{1}{2}(7)(2\sqrt{3})\Rightarrow \boxed{7\sqrt{3}}\]

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS