Difference between revisions of "2017 AMC 10B Problems/Problem 22"
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===Solution 3=== | ===Solution 3=== | ||
− | As stated before, note that <math>\triangle ACB | + | As stated before, note that <math>\triangle ACB</math> is similar to <math>\triangle ADE</math>. By similarity, we note that <math>\frac{\overline{AC}}{\overline{BC}}</math> is equivalent to <math>\frac{7}{5}</math>. We set <math>\overline{AC}</math> to <math>7x</math> and <math>\overline{BC}</math> to <math>5x</math>. By the Pythagorean Theorem, <math>(7x)^2+(5x)^2 = 4^2</math>. Combining, <math>49x^2+25x^2=16</math>. We can add and divide to get <math>x^2=\frac{8}{37}</math>. We square root and rearrange to get <math>x=\frac{2\sqrt{74}}{37}</math>. We know that the legs of the triangle are <math>7x</math> and <math>5x</math>. Multiplying <math>x</math> by <math>7</math> and <math>5</math> eventually gives us <math>\frac {14\sqrt{74}}{37}</math> <math>\frac {10\sqrt{74}}{37}</math>. We divide this by <math>2</math>, since <math>\frac{1}{2}bh</math> is the formula for a triangle. This gives us <math>\boxed{\textbf{(D) } \frac{140}{37}}</math>. |
===Solution 4=== | ===Solution 4=== |
Revision as of 19:35, 30 December 2020
Contents
[hide]Problem
The diameter of a circle of radius
is extended to a point
outside the circle so that
. Point
is chosen so that
and line
is perpendicular to line
. Segment
intersects the circle at a point
between
and
. What is the area of
?
Solutions
![[asy] size(10cm); pair A, B, C, D, E, O; A = (-2,0); B = (2,0); C = (2*cos(1.24),2*sin(1.24)); D = (5,0); E = (5,5); O = (A+B)/2; dot(A); dot(B); dot(C); dot(D); dot(E); dot(O); draw(Circle((A+B)/2,2)); draw(A--D--E--C--A); draw(C--B); draw(rightanglemark(A,C,B,5)); draw(rightanglemark(A,D,E,5)); label("$A$",A,W); label("$B$",B,SE); label("$D$",D,SE); label("$E$",E,NE); label("$C$",C,N); label("$2$",(O+B)/2,S); label("$3$",(B+D)/2,S); label("$5$",(D+E)/2,NE); [/asy]](http://latex.artofproblemsolving.com/a/b/8/ab8a0aedc923c0f0d7ffd36e507de0b9ef4606a2.png)
Solution 1
Notice that and
are right triangles. Then
.
, so
. We also find that
, and thus the area of
is
.
Solution 2
We note that by
similarity. Also, since the area of
and
,
, so the area of
.
Solution 3
As stated before, note that is similar to
. By similarity, we note that
is equivalent to
. We set
to
and
to
. By the Pythagorean Theorem,
. Combining,
. We can add and divide to get
. We square root and rearrange to get
. We know that the legs of the triangle are
and
. Multiplying
by
and
eventually gives us
. We divide this by
, since
is the formula for a triangle. This gives us
.
Solution 4
Let's call the center of the circle that segment is the diameter of,
. Note that
is an isosceles right triangle. Solving for side
, using the Pythagorean theorem, we find it to be
. Calling the point where segment
intersects circle
, the point
, segment
would be
. Also, noting that
is a right triangle, we solve for side
, using the Pythagorean Theorem, and get
. Using Power of Point on point
, we can solve for
. We can subtract
from
to find
and then solve for
using Pythagorean theorem once more.
= (Diameter of circle
+
)
=
=
=
-
=
Now to solve for :
-
=
+
=
=
Note that is a right triangle because the hypotenuse is the diameter of the circle. Solving for area using the bases
and
, we get the area of triangle
to be
.
Coordinate Geo
Drawing the picture, we realize that the equation for the line from A to E is , and the equation for the circle is
plugging in
for y we get
so
, that means
the height is and the base is
, so the area is
-harsha12345
See Also
2017 AMC 10B (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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.