Difference between revisions of "2015 AMC 8 Problems/Problem 25"
(→Solution 1) |
Math101010 (talk | contribs) |
||
Line 52: | Line 52: | ||
We wish to find the area of the larger triangle. The area of the larger square is composed of the smaller square and the four triangles. The triangles have base <math>3</math> and height <math>1</math>, so the combined area of the four triangles is <math>4 \cdot \frac 32=6</math>. The area of the smaller square is <math>9</math>. We add these to see that the area of the large square is <math>9+6=\boxed{\mathrm{(C) \ } 15}</math>. | We wish to find the area of the larger triangle. The area of the larger square is composed of the smaller square and the four triangles. The triangles have base <math>3</math> and height <math>1</math>, so the combined area of the four triangles is <math>4 \cdot \frac 32=6</math>. The area of the smaller square is <math>9</math>. We add these to see that the area of the large square is <math>9+6=\boxed{\mathrm{(C) \ } 15}</math>. | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AMC8 box|year=2015|num-b=24|after=Last Problem}} | ||
+ | {{MAA Notice}} |
Revision as of 16:53, 25 November 2015
One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space?
Solution 1
We draw a diagram as shown.
Let us focus on the big triangles taking up the rest of the space. The triangles on top of the unit square between the inscribed square, are similiar to the 4 big triangles by AA. Let the height of a big triangle be
then
.
Which means
This means the area of each triangle is
This the area of the square is
Solution 2
We draw a diagram as shown:
We wish to find the area of the larger triangle. The area of the larger square is composed of the smaller square and the four triangles. The triangles have base and height
, so the combined area of the four triangles is
. The area of the smaller square is
. We add these to see that the area of the large square is
.
See Also
2015 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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.