Difference between revisions of "2001 AMC 12 Problems/Problem 22"
(→Solution 3) |
m (→Solution 3) |
||
Line 66: | Line 66: | ||
=== Solution 3 === | === Solution 3 === | ||
− | Because we see that there are only lines and there is a rectangle, we can coordbash (place this figure on coordinates). Because this is a general figure, we can assume the sides are <math>7</math> and <math>10</math>. 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>. | + | Because we see that there are only lines and there is a rectangle, we can coordbash (place this figure on coordinates). Because this is a general figure, we can assume the sides are <math>7</math> and <math>10</math> (or any other two positive numbers that multiply to 70). 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 20:58, 15 August 2021
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). Because this is a general figure, we can assume the sides are and
(or any other two positive numbers that multiply to 70). 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.