Difference between revisions of "2005 AMC 10A Problems/Problem 25"
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<cmath>\frac{[ADE]}{[ABC]} = \frac{AD}{AB} \cdot \frac{AE}{AC} = \frac{19}{25} \cdot \frac{14}{42} = \frac{19}{75}.</cmath> | <cmath>\frac{[ADE]}{[ABC]} = \frac{AD}{AB} \cdot \frac{AE}{AC} = \frac{19}{25} \cdot \frac{14}{42} = \frac{19}{75}.</cmath> | ||
+ | |||
+ | (Area of a triangle is base times height, so the area ratio of triangles, that have a common vertex (height) and bases on a common line, is the base length ratio. This is applied twice, using different pairs of bases, and corresponding altitudes for height.). | ||
<asy> | <asy> | ||
Line 33: | Line 35: | ||
</asy> | </asy> | ||
− | + | <math>[BCED] = [ABC] - [ADE]</math>, so | |
<cmath> | <cmath> | ||
\begin{align*} | \begin{align*} | ||
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− | Note: If it is hard to understand why <cmath>\frac{[ADE]}{[ABC]} = \frac{AD}{AB} \cdot \frac{AE}{AC}</cmath>, you can use the fact that the area of a triangle equals <math>\frac{1}{2} \cdot ab \cdot \sin(C)</math>. If angle <math>DAE = Z</math>, we have that <cmath>\frac{[ADE]}{[ABC]} = \frac{\frac{1}{2} \cdot 19 \cdot 14 \cdot \sin(Z)}{\frac{1}{2} \cdot 25 \cdot 42 \cdot \sin(Z)} = \frac{19 \cdot 14}{25 \cdot 42} = \frac{ab}{cd}</cmath>. | + | Note: If it is hard to understand why <cmath>\frac{[ADE]}{[ABC]} = \frac{AD}{AB} \cdot \frac{AE}{AC}</cmath>, you can use the fact that the area of a triangle equals <math>\frac{1}{2} \cdot ab \cdot \sin(C)</math>. If angle <math>DAE = Z</math>, we have that <cmath>\frac{[ADE]}{[ABC]} = \frac{\frac{1}{2} \cdot 19 \cdot 14 \cdot \sin(Z)}{\frac{1}{2} \cdot 25 \cdot 42 \cdot \sin(Z)} = \frac{19 \cdot 14}{25 \cdot 42} = \frac{ab}{cd}</cmath>. |
==Video Solution== | ==Video Solution== | ||
− | |||
− | ==Solution 2 | + | https://youtu.be/VXyOJWcpi00 |
+ | |||
+ | ==Solution 2 == | ||
<asy> | <asy> | ||
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~ LaTeX changes by tkfun | ~ LaTeX changes by tkfun | ||
+ | |||
+ | |||
+ | == Solution 5 == | ||
+ | Let the area of triangle ABC be denoted by [ABC] and the area of quadrilateral ABCD be denoted by [ABCD]. | ||
+ | |||
+ | <cmath>\text{Diagram:}</cmath> | ||
+ | <asy> | ||
+ | unitsize(0.15 cm); | ||
+ | |||
+ | pair A, B, C, D, E; | ||
+ | |||
+ | A = (191/39,28*sqrt(1166)/39); | ||
+ | B = (0,0); | ||
+ | C = (39,0); | ||
+ | D = (6*A + 19*B)/25; | ||
+ | E = (28*A + 14*C)/42; | ||
+ | |||
+ | draw(A--B--C--cycle); | ||
+ | draw(D--E); | ||
+ | |||
+ | label("$A$", A, N); | ||
+ | label("$B$", B, SW); | ||
+ | label("$C$", C, SE); | ||
+ | label("$D$", D, W); | ||
+ | label("$E$", E, NE); | ||
+ | label("$19$", (A + D)/2, W); | ||
+ | label("$6$", (B + D)/2, W); | ||
+ | label("$14$", (A + E)/2, NE); | ||
+ | label("$28$", (C + E)/2, NE); | ||
+ | </asy> | ||
+ | |||
+ | Let the area of <math>\triangle ABC</math> be <math>x</math>. <math>\triangle ABE</math> and <math>\triangle BEC</math> share a height, and the ratio of their bases are <math>1:2</math>, so the area of <math>\triangle ABE</math> is <math>\frac{x}{3}</math>. | ||
+ | |||
+ | Similarly, <math>\triangle AED</math> and <math>\triangle DEB</math> share a height, and the ratio of their bases is <math>19:6</math>, so the ratio of <math>\frac{[AED]}{[AEB]}=\frac{19}{25}</math>. Therefore, | ||
+ | <cmath>[AED]=\frac{19}{25}\cdot\left[AEB\right]=\frac{19}{25}\cdot\frac{1}{3}\cdot\left[ABC\right]=\frac{19}{25}\cdot\frac{1}{3}\cdot x=\frac{19}{75}x</cmath> | ||
+ | <cmath>[DECB]=[ABC]-[AED]=x-\frac{19}{75}x=\frac{56}{75}x</cmath> | ||
+ | The ratio <math>\frac{[AED]}{[DECB]}=\frac{\frac{19}{75}}{\frac{56}{75}}=\frac{19}{56}</math> which is answer choice <math>\boxed{\textbf{(D) } \frac{19}{56}}</math>. | ||
+ | |||
+ | |||
+ | ~JH. L | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2005|ab=A|num-b=24|after=Last Problem}} | {{AMC10 box|year=2005|ab=A|num-b=24|after=Last Problem}} | ||
− | + | [[Category:Triangle Area Ratio Problems]] | |
{{MAA Notice}} | {{MAA Notice}} | ||
https://ivyleaguecenter.files.wordpress.com/2017/11/amc-10-picture.jpg | https://ivyleaguecenter.files.wordpress.com/2017/11/amc-10-picture.jpg |
Latest revision as of 13:31, 13 October 2024
Contents
Problem
In we have , , and . Points and are on and respectively, with and . What is the ratio of the area of triangle to the area of the quadrilateral ?
Solution 1
We have
(Area of a triangle is base times height, so the area ratio of triangles, that have a common vertex (height) and bases on a common line, is the base length ratio. This is applied twice, using different pairs of bases, and corresponding altitudes for height.).
, so
Note: If it is hard to understand why , you can use the fact that the area of a triangle equals . If angle , we have that .
Video Solution
Solution 2
We can let .
Since , .
So, .
This means that .
Thus,
-Conantwiz2023
Solution 3 (trig)
Using this formula:
Since the area of is equal to the area of minus the area of ,
.
Therefore, the desired ratio is
Note: was not used in this problem.
Solution 4
Let be on such that then we have Since we have Thus and Finally, after some calculations, .
~ Nafer
~ LaTeX changes by tkfun
Solution 5
Let the area of triangle ABC be denoted by [ABC] and the area of quadrilateral ABCD be denoted by [ABCD].
Let the area of be . and share a height, and the ratio of their bases are , so the area of is .
Similarly, and share a height, and the ratio of their bases is , so the ratio of . Therefore, The ratio which is answer choice .
~JH. L
See also
2005 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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.
https://ivyleaguecenter.files.wordpress.com/2017/11/amc-10-picture.jpg