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

Revision as of 17:57, 30 April 2024

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:

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

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

Solution 5

Let $\overline{BD} = x$ , and $\angle ABD = \theta$ . Since $m\angle ABC = 90, DBC = 90 - \theta$ . $\tan{\theta} = \frac{3}{x}$ , and $\tan{90-\theta} = \frac{4}{x}$ . Using trig identities, we can simplify the second one to $\cot{\theta} = \frac{4}{x}$ . Let $y = \tan{\theta}$ . We get $xy = 3$ , and using the fact that $\cot{a} = \frac{1}{\tan{a}}$ , $\frac{x}{y} = 4$ . Solving for x and y(using substitution), we get $y^2 = \frac{3}{4}$ . We could substitute back in and solve, but there is an easier way. Squaring $xy = 3$ , we get $x^2y^2 = 9$ . Since we know $y^2 = \frac{3}{4}$ , $x = \sqrt{12}$ . Using the Pythagorean Theorem in the original triangle, we get the legs of the triangle to be $\sqrt{9+x^2}$ and $\sqrt{16+x^2}$ . Putting in $x$ , we get the equation for the area, which is $\frac{\sqrt{28} \cdot \sqrt{21}}{2} = \frac{14\sqrt{2}}{2} = \boxed{\textbf{(B) } 7\sqrt{3}}$

~idk12345678

Video Solution by OmegaLearn

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

~ pi_is_3.14

@@ Line 106: / Line 106: @@
 ~Arcticturn
+===Solution 5===
+Let <math>\overline{BD} = x</math>, and <math>\angle ABD = \theta</math>. Since <math>m\angle ABC = 90, DBC = 90 - \theta</math>. <math>\tan{\theta} = \frac{3}{x}</math>, and <math>\tan{90-\theta} = \frac{4}{x}</math>. Using trig identities, we can simplify the second one to <math>\cot{\theta} = \frac{4}{x}</math>. Let <math>y = \tan{\theta}</math>. We get <math>xy = 3</math>, and using the fact that <math>\cot{a} = \frac{1}{\tan{a}}</math>, <math>\frac{x}{y} = 4</math>. Solving for x and y(using substitution), we get <math>y^2 = \frac{3}{4}</math>. We could substitute back in and solve, but there is an easier way. Squaring <math>xy = 3</math>, we get <math>x^2y^2 = 9</math>. Since we know <math>y^2 = \frac{3}{4}</math>, <math>x = \sqrt{12}</math>. Using the Pythagorean Theorem in the original triangle, we get the legs of the triangle to be <math>\sqrt{9+x^2}</math> and <math>\sqrt{16+x^2}</math>. Putting in <math>x</math>, we get the equation for the area, which is <math>\frac{\sqrt{28} \cdot \sqrt{21}}{2} = \frac{14\sqrt{2}}{2} = \boxed{\textbf{(B) } 7\sqrt{3}}</math>
+~idk12345678
 == Video Solution by OmegaLearn==

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

Page

Toolbox

Search