Difference between revisions of "2005 AMC 10A Problems/Problem 25"
Erics son07 (talk | contribs) (→Solution 4) |
|||
(5 intermediate revisions by 2 users not shown) | |||
Line 7: | Line 7: | ||
We have | We have | ||
<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*} | ||
Line 45: | Line 47: | ||
− | 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> | ||
Line 163: | Line 166: | ||
==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