Difference between revisions of "2020 AMC 8 Problems/Problem 18"
Vincentwant (talk | contribs) |
Vincentwant (talk | contribs) |
||
Line 26: | Line 26: | ||
draw(arc((0,0),17,360,0)); 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*SE); dot("$B$",(8,15), 1.25*N); dot("$C$",(-8,15), 1.25*N); dot("$D$",(-8,0), 1.25*SW); dot("$E$",(17,0), 1.25*E); dot("$F$",(-17,0), 1.25*W); label("$16$",(0,0),N); label("$9$",(12.5,0),N); label("$9$",(-12.5,0),N); draw((-8,-15)--(-8,0)--(8,0)--(8,-15)--cycle); dot("$B'$",(8,-15), 1.25*S); dot("$C'$",(-8,-15), 1.25*S); | draw(arc((0,0),17,360,0)); 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*SE); dot("$B$",(8,15), 1.25*N); dot("$C$",(-8,15), 1.25*N); dot("$D$",(-8,0), 1.25*SW); dot("$E$",(17,0), 1.25*E); dot("$F$",(-17,0), 1.25*W); label("$16$",(0,0),N); label("$9$",(12.5,0),N); label("$9$",(-12.5,0),N); draw((-8,-15)--(-8,0)--(8,0)--(8,-15)--cycle); dot("$B'$",(8,-15), 1.25*S); dot("$C'$",(-8,-15), 1.25*S); | ||
</asy> | </asy> | ||
− | By Power of a Point, <math>FD\cdot DE = CD\cdot C'D</math>. By symmetry, <math>CD = C'D</math>. <math>FD = 9</math> and <math>DE = 25</math>. Substituting in these values, <math>9\cdot 25 = CD^2</math>, giving <math>CD^2 = 225</math> and <math>CD = 15</math>. | + | By Power of a Point, <math>FD\cdot DE = CD\cdot C'D</math>. By symmetry, <math>CD = C'D</math>. We see that <math>FD = 9</math> and <math>DE = 25</math>. Substituting in these values, <math>9\cdot 25 = CD^2</math>, giving <math>CD^2 = 225</math> and <math>CD = 15</math>. |
The area of the rectangle is therefore <math>15\cdot 16 = \boxed{\textbf{(A) }240}</math>. | The area of the rectangle is therefore <math>15\cdot 16 = \boxed{\textbf{(A) }240}</math>. | ||
==Video Solution== | ==Video Solution== |
Revision as of 09:50, 8 December 2020
Contents
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 . By the Pythagorean theorem in right-angled triangle (or ), we have that (or ) is . Accordingly, the area of is .
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 Sidenote: 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 formula, we scale the figure down by ; this will give the height a length of . Now, scaling back up by , the height is . The answer is then . -SweetMango77
Solution 4 (Power Of A Point)
Draw the other half of the circle as follows: By Power of a Point, . By symmetry, . We see that and . Substituting in these values, , giving and . The area of the rectangle is therefore .
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.