Difference between revisions of "2017 AMC 10B Problems/Problem 15"
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− | Drop an altitude from <math>E</math> to <math>BC</math> and call its foot <math>X</math>. We have that <math>EX \cdot BC=BE \cdot EC</math> since both are equal to two times the area of <math>BEC</math>. Since <math>BC=4</math>, <math>BE=\frac{12}{5}</math>, and <math>EC=\frac{16}{5}</math>, we can calculate that <math>EX=\frac{48}{25}</math>. If <math>EX</math> is extended to meet <math>AD</math> at point <math>Y</math>, <math>EY=3-\frac{48}{25}=\frac{27}{25}</math>. Therefore, <math>[AED]=\frac{EY \cdot AD}{2}=\frac{\frac{27}{25} \cdot 4}{2}=\frac{54}{25}</math> | + | Drop an altitude from <math>E</math> to <math>BC</math> and call its foot <math>X</math>. We have that <math>EX \cdot BC=BE \cdot EC</math> since both are equal to two times the area of <math>BEC</math>. Since <math>BC=4</math>, <math>BE=\frac{12}{5}</math>, and <math>EC=\frac{16}{5}</math>, we can calculate that <math>EX=\frac{48}{25}</math>. If <math>EX</math> is extended to meet <math>AD</math> at point <math>Y</math>, <math>EY=3-\frac{48}{25}=\frac{27}{25}</math>. Therefore, <math>[AED]=\frac{EY \cdot AD}{2}=\frac{\frac{27}{25} \cdot 4}{2}=\boxed{(E)\frac{54}{25}}</math>. |
Fasolinka | Fasolinka |
Revision as of 17:49, 29 October 2021
Contents
Problem
Rectangle has
and
. Point
is the foot of the perpendicular from
to diagonal
. What is the area of
?
Solution 1
First, note that because
is a right triangle. In addition, we have
, so
. Using similar triangles within
, we get that
and
.
Let be the foot of the perpendicular from
to
. Since
and
are parallel,
is similar to
. Therefore, we have
. Since
,
. Note that
is an altitude of
from
, which has length
. Therefore, the area of
is
Solution 2
From similar triangles, we know that (see Solution 1). Furthermore, we also know that
from the rectangle. Using the sine formula for area, we have
But, note that
. Thus, we see that
~coolwiz
Solution 3
Alternatively, we can use coordinates. Denote as the origin. We find the equation for
as
, and
as
. Solving for
yields
. Our final answer then becomes
Solution 4
We note that the area of must equal the area of
because they share the base
and the height of both is the altitude of congruent triangles. Therefore, we find the area of
to be
Solution 5
We know all right triangles are 5-4-3, so the areas are proportional to the square of corresponding sides. Area of is
of
. Using similar logic in Solution 4, Area of
is the same as
.
Solution 6
Drop an altitude from to
and call its foot
. We have that
since both are equal to two times the area of
. Since
,
, and
, we can calculate that
. If
is extended to meet
at point
,
. Therefore,
.
Fasolinka
See Also
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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.