Difference between revisions of "2001 AMC 12 Problems/Problem 22"
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As <math>E</math> is the midpoint of <math>CD</math>, the height from <math>E</math> onto <math>AC</math> is <math>1/2</math> of the height from <math>D</math> onto <math>AC</math>. Therefore we have <math>[EHJ] = \frac{6}{35} \cdot \frac 12 \cdot [ACD] = \frac 3{35} \cdot 35 = \boxed{3}</math>. | As <math>E</math> is the midpoint of <math>CD</math>, the height from <math>E</math> onto <math>AC</math> is <math>1/2</math> of the height from <math>D</math> onto <math>AC</math>. Therefore we have <math>[EHJ] = \frac{6}{35} \cdot \frac 12 \cdot [ACD] = \frac 3{35} \cdot 35 = \boxed{3}</math>. | ||
+ | |||
+ | === Solution 3 === | ||
+ | Because we see that there are only lines and there is a rectangle, we can coordbash (place this figure on coordinates). We can find <math>H</math> and <math>J</math> by intersecting lines, and then we calculate the area of <math>EHJ</math> using shoelace formula. This yields <math>\boxed{3}</math>. | ||
== See Also == | == See Also == |
Revision as of 14:09, 12 August 2020
Problem
In rectangle , points
and
lie on
so that
and
is the midpoint of
. Also,
intersects
at
and
at
. The area of the rectangle
is
. Find the area of triangle
.
Solution
Solution 1
Note that the triangles and
are similar, as they have the same angles. Hence
.
Also, triangles and
are similar, hence
.
We can now compute as
. We have:
.
is
of
, as these two triangles have the same base
, and
is
of
, therefore also the height from
onto
is
of the height from
. Hence
.
is
of
, as the base
is
of the base
, and the height from
is
of the height from
. Hence
.
is
of
for similar reasons, hence
.
Therefore .
Solution 2
As in the previous solution, we note the similar triangles and prove that is in
and
in
of
.
We can then compute that .
As is the midpoint of
, the height from
onto
is
of the height from
onto
. Therefore we have
.
Solution 3
Because we see that there are only lines and there is a rectangle, we can coordbash (place this figure on coordinates). We can find and
by intersecting lines, and then we calculate the area of
using shoelace formula. This yields
.
See Also
2001 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.