Difference between revisions of "2009 AMC 10A Problems/Problem 10"
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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 = \boxed{7\sqrt 3}</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 = \boxed{7\sqrt 3}</math>. | ||
| − | Proof: Consider the [[Pythagorean theorem]] for each of the triangles <math>ABC</math>, <math>ABD</math>, and <math>CBD</math>. We get: | + | ''Proof'': Consider the [[Pythagorean theorem]] for each of the triangles <math>ABC</math>, <math>ABD</math>, and <math>CBD</math>. We get: |
# <math>AB^2 + BC^2 = AC^2 = (AD+DC)^2 = AD^2 + DC^2 + 2 \cdot AD \cdot DC</math>. | # <math>AB^2 + BC^2 = AC^2 = (AD+DC)^2 = AD^2 + DC^2 + 2 \cdot AD \cdot DC</math>. | ||
# <math>AB^2 = AD^2 + BD^2</math> | # <math>AB^2 = AD^2 + BD^2</math> | ||
# <math>BC^2 = BD^2 + CD^2</math> | # <math>BC^2 = BD^2 + CD^2</math> | ||
| − | Substituting equations 2 and 3 into the left hand side of equation 1, we get <math>BD^2 = AD \cdot DC \blacksquare</math> | + | Substituting equations 2 and 3 into the left hand side of equation 1, we get <math>BD^2 = AD \cdot DC</math>. <math>\blacksquare</math> |
== 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}} | ||
Revision as of 17:17, 19 February 2009
Problem
Triangle 
has a right angle at 
. Point 
is the foot of the altitude from , 
, and 
. What is the area of 
?
![[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]](http://latex.artofproblemsolving.com/4/b/e/4bea2faf81f4ba6dca40fdb9ce720c9ed82b168c.png)
Solution
It is a well-known fact that in any right triangle with the right angle at and the foot of the altitude from onto 
we have 
. (See below for a proof.) Then 
, and the area of the triangle is 
.
Proof: Consider the Pythagorean theorem for each of the triangles , 
, and 
. We get:
.
.
Substituting equations 2 and 3 into the left hand side of equation 1, we get 
. 
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 | ||
