Difference between revisions of "2009 AMC 10B Problems/Problem 18"
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== Solution 1 (Coordinate Geo)== | == Solution 1 (Coordinate Geo)== | ||
− | Set A to (0,0). Since M is the midpoint of the diagonal, it would be (4,-3). The | + | Set A to <math>(0,0)</math>. Since M is the midpoint of the diagonal, it would be <math>(4,-3)</math>. The diagonal AC would be the line <math>y = -\frac{3x}{4}</math>. Since ME is perpendicular to AC, its line would be in the form <math>y = \frac{4x}{3} + b</math>. Plugging in <math>4</math> and <math>-3</math> for <math>x</math> and <math>y</math> would give <math>b = \frac{25}{3}</math>. To find the x-intercept of <math>y = \frac{4x}{3} + \frac{25}{3}</math> we plug in <math>0</math> for <math>y</math> and get <math>x = \frac{25}{4}</math>. Then, using the Shoelace Formula for <math>(0,0)</math> , <math>(4,-3)</math>, and <math>(\frac{25}{4}, 0)</math>, we find the area is <math>\frac{75}{8}</math>. |
== Solution 2 == | == Solution 2 == |
Revision as of 23:28, 18 December 2019
Contents
Problem
Rectangle has
and
. Point
is the midpoint of diagonal
, and
is on
with
. What is the area of
?
Solution 1 (Coordinate Geo)
Set A to . Since M is the midpoint of the diagonal, it would be
. The diagonal AC would be the line
. Since ME is perpendicular to AC, its line would be in the form
. Plugging in
and
for
and
would give
. To find the x-intercept of
we plug in
for
and get
. Then, using the Shoelace Formula for
,
, and
, we find the area is
.
Solution 2
By the Pythagorean theorem we have , hence
.
The triangles and
have the same angle at
and a right angle, thus all their angles are equal, and therefore these two triangles are similar.
The ratio of their sides is , hence the ratio of their areas is
.
And as the area of triangle is
, the area of triangle
is
.
Solution 3 (Only Pythagorean Theorem)
Draw as shown from the diagram. Since
is of length
, we have that
is of length
, because of the midpoint
. Through the Pythagorean theorem, we know that
, which means
. Define
to be
for the sake of clarity. We know that
. From here, we know that
. From here, we can write the expression
. Now, remember
.
, since we set
in the start of the solution. Now to find the area
See Also
2009 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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.