Difference between revisions of "2020 AMC 8 Problems/Problem 18"
Sweetmango77 (talk | contribs) (→Solution 3) |
m (→Solution 1) |
||
Line 10: | Line 10: | ||
<asy> draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot("$A$",(8,0), 1.25*S); dot("$B$",(8,15), 1.25*N); dot("$C$",(-8,15), 1.25*N); dot("$D$",(-8,0), 1.25*S); dot("$E$",(17,0), 1.25*S); dot("$F$",(-17,0), 1.25*S); label("$16$",(0,0),N); label("$9$",(12.5,0),N); label("$9$",(-12.5,0),N); dot("$O$", (0,0), 1.25*S); draw((0,0)--(-8,15));</asy> | <asy> draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot("$A$",(8,0), 1.25*S); dot("$B$",(8,15), 1.25*N); dot("$C$",(-8,15), 1.25*N); dot("$D$",(-8,0), 1.25*S); dot("$E$",(17,0), 1.25*S); dot("$F$",(-17,0), 1.25*S); label("$16$",(0,0),N); label("$9$",(12.5,0),N); label("$9$",(-12.5,0),N); dot("$O$", (0,0), 1.25*S); draw((0,0)--(-8,15));</asy> | ||
− | Let <math>O</math> be the center of the semicircle. The diameter of the semicircle is <math>9+16+9=34</math>, so <math>OC = 17</math>. By symmetry, <math>O</math> is in fact the midpoint of <math>DA</math>, so <math>OD=OA=\frac{16}{2}= </math>. By the Pythagorean theorem in right-angled triangle <math>ODC< | + | Let <math>O</math> be the center of the semicircle. The diameter of the semicircle is <math>9+16+9=34</math>, so <math>OC = 17</math>. By symmetry, <math>O</math> is in fact the midpoint of <math>DA</math>, so <math>OD=OA=\frac{16}{2}= </math>8<math>. By the Pythagorean theorem in right-angled triangle </math>ODC<math> (or </math>OBA<math>), we have that </math>CD<math> (or </math>AB<math>) is </math>\sqrt{17^2-8^2}=15<math>. Accordingly, the area of </math>ABCD<math> is </math>16\cdot 15=\boxed{\textbf{(A) }240}$. |
==Solution 2 (coordinate geometry)== | ==Solution 2 (coordinate geometry)== |
Revision as of 21:47, 24 November 2020
Problem
Rectangle is inscribed in a semicircle with diameter
as shown in the figure. Let
and let
What is the area of
Solution 1
Let be the center of the semicircle. The diameter of the semicircle is
, so
. By symmetry,
is in fact the midpoint of
, so
8
ODC
OBA
CD
AB
\sqrt{17^2-8^2}=15
ABCD
16\cdot 15=\boxed{\textbf{(A) }240}$.
Solution 2 (coordinate geometry)
Let the midpoint of segment be the origin. Evidently, point
and
. Since points
and
share
-coordinates with
and
respectively, it suffices to find the
-coordinate of
(which will be the height of the rectangle) and multiply this by
(which we know is
). The radius of the semicircle is
, so the whole circle has equation
; as already stated,
has the same
-coordinate as
, i.e.
, so substituting this into the equation shows that
. Since
at
, the y-coordinate of
is
. Therefore, the answer is
.
(Note that the synthetic solution (Solution 1) is definitely faster and more elegant. However, this is the solution that you should use if you can't see any other easier strategy.)
Solution 3
We can use a result from the Art of Problem Solving Introduction to Algebra book: for a semicircle with diameter , such that the
part is on one side and the
part is on the other side, the height from the end of the
side (or the start of the
side) is
. To use this, we scale the figure down by
; then the height is
. Now, scaling back up by
, the height
is
. The answer is
.
-SweetMango77; edited by Sevenoptimus, who made it easier to read
Video Solution
2020 AMC 8 (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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.