Difference between revisions of "2017 AMC 10B Problems/Problem 15"
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==Problem== | ==Problem== | ||
− | + | Rectangle <math>ABCD</math> has <math>AB=3</math> and <math>BC=4</math>. Point <math>E</math> is the foot of the perpendicular from <math>B</math> to diagonal <math>\overline{AC}</math>. What is the area of <math>\triangle AED</math>? | |
+ | |||
+ | <math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{42}{25}\qquad\textbf{(C)}\ \frac{28}{15}\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{54}{25}</math> | ||
+ | |||
+ | ==Solution 1== | ||
+ | |||
+ | <asy> | ||
+ | pair A,B,C,D,E; | ||
+ | A=(0,4); | ||
+ | B=(3,4); | ||
+ | C=(3,0); | ||
+ | D=(0,0); | ||
+ | draw(A--B--C--D--cycle); | ||
+ | label("$A$",A,N); | ||
+ | label("$B$",B,N); | ||
+ | label("$C$",C,S); | ||
+ | label("$D$",D,S); | ||
+ | E=foot(B,A,C); | ||
+ | draw(E--B); | ||
+ | draw(A--C); | ||
+ | draw(rightanglemark(B,E,C)); | ||
+ | label("$E$",E,N); | ||
+ | draw(D--E); | ||
+ | label("$3$",A--B,N); | ||
+ | label("$4$",B--C,E); | ||
+ | </asy> | ||
+ | |||
+ | First, note that <math>AC=5</math> because <math>ABC</math> is a right triangle. In addition, we have <math>AB\cdot BC=2[ABC]=AC\cdot BE</math>, so <math>BE=\frac{12}{5}</math>. Using similar triangles within <math>ABC</math>, we get that <math>AE=\frac{9}{5}</math> and <math>CE=\frac{16}{5}</math>. | ||
+ | |||
+ | Let <math>F</math> be the foot of the perpendicular from <math>E</math> to <math>AB</math>. Since <math>EF</math> and <math>BC</math> are parallel, <math>\Delta AFE</math> is similar to <math>\Delta ABC</math>. Therefore, we have <math>\frac{AF}{AB}=\frac{AE}{AC}=\frac{9}{25}</math>. Since <math>AB=3</math>, <math>AF=\frac{27}{25}</math>. Note that <math>AF</math> is an altitude of <math>\Delta AED</math> from <math>AD</math>, which has length <math>4</math>. Therefore, the area of <math>\Delta AED</math> is <math>\frac{1}{2}\cdot\frac{27}{25}\cdot4=\boxed{\textbf{(E)}\frac{54}{25}}.</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | From similar triangles, we know that <math>AE=\frac{9}{5}</math> (see Solution 1). Furthermore, we also know that <math>AD=4</math> from the rectangle. Using the sine formula for area, we have | ||
+ | <cmath>\frac{1}{2}(AE)(AD)(\sin(m\angle EAD)).</cmath> | ||
+ | But, note that <math>\sin(m\angle EAD)=\sin(m\angle CAD)=\frac{O}{H}=\frac{3}{5}</math>. Thus, we see that | ||
+ | <cmath>[AED]=\frac{1}{2}\cdot \frac{9}{5}\cdot 4\cdot\frac{3}{5}=\boxed{\textbf{(E)}\frac{54}{25}}.</cmath> | ||
+ | ~coolwiz | ||
+ | |||
+ | ==Solution 3== | ||
+ | Alternatively, we can use coordinates. Denote <math>D</math> as the origin. We find the equation for <math>AC</math> as <math>y=-\frac{4}{3}x+4</math>, and <math>BE</math> as <math>y=\frac{3}{4}x+\frac{7}{4}</math>. Solving for <math>x</math> yields <math>\frac{27}{25}</math>. Our final answer then becomes <math>\frac{1}{2}\cdot\frac{27}{25}\cdot4=\boxed{\textbf{(E)}\frac{54}{25}}.</math> | ||
+ | |||
+ | ==Solution 4== | ||
+ | We note that the area of <math>ABE</math> must equal the area of <math>AED</math> because they share the base <math>AE</math> and the height of both is the altitude of congruent triangles. Therefore, we find the area of <math>ABE</math> to be <math>\frac{1}{2}*\frac{9}{5}*\frac{12}{5}=\boxed{\textbf{(E)}\frac{54}{25}}.</math> | ||
+ | |||
+ | ==Solution 5== | ||
+ | We know all right triangles are 5-4-3, so the areas are proportional to the square of corresponding sides. Area of <math>ABE</math> is <math> (\dfrac{3}{5})^2</math> of <math>ABC = \frac{54}{25}</math>. Using similar logic in Solution 4, Area of <math>AED</math> is the same as <math>ABE</math>. | ||
+ | |||
+ | ==Solution 6== | ||
+ | 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{\textbf{(E)}\frac{54}{25}}</math>. | ||
+ | |||
+ | - Fasolinka | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/GrCtzL0S-Uo?t=369 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution by Math4All999== | ||
+ | https://www.youtube.com/watch?v=hSrgGKxduho | ||
+ | |||
+ | ~Math4All999 | ||
− | |||
− | |||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=B|num-b=14|num-a=16}} | {{AMC10 box|year=2017|ab=B|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 01:20, 4 December 2023
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
Video Solution by OmegaLearn
https://youtu.be/GrCtzL0S-Uo?t=369
~ pi_is_3.14
Video Solution by Math4All999
https://www.youtube.com/watch?v=hSrgGKxduho
~Math4All999
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.