Difference between revisions of "2015 AMC 8 Problems/Problem 25"
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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>. | ||
+ | |||
+ | ===Solution 3=== | ||
+ | <asy> | ||
+ | draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); | ||
+ | filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); | ||
+ | filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); | ||
+ | filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); | ||
+ | filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); | ||
+ | path arc = arc((2.5,4),1.5,0,90); | ||
+ | pair P = intersectionpoint(arc,(0,5)--(5,5)); | ||
+ | pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp; | ||
+ | draw(P--Pp--Ppp--Pppp--cycle); | ||
+ | </asy> | ||
+ | |||
+ | Let's find the area of the triangles and the unit squares: on each side, there are 2 triangles. They both have 1 leg of length 1, and let's label the other legs x for one of thr triangles and y for the other. Note that x+y=3. | ||
+ | The area of each of the triangles is <math>\frac{x}{2}</math> and <math>\frac{y}{2}</math>, and there are 4 of each. So now we need to find <math>(4)\frac{x}{2} + (4)\frac{y}{2}</math>. | ||
+ | <math>(4)\frac{x}{2} + (4)\frac{y}{2}</math> | ||
+ | <math>\Rightarrow~~4\left(\frac{x}{2}+ \frac{y}{2}\right)</math> | ||
+ | <math>\Rightarrow~~4\left(\frac{x+y}{2}\right)</math> | ||
+ | Remember that x+y=3, so substituting this in we find that the area of all of the triangles is <math>4\left(\frac{3}{2}\right) = 6</math>. | ||
+ | The area of the 4 unit squares is 4, so the area of the square we need is <math>25- (4+6) = 15 \Rightarrow \boxed{(C)}</math> | ||
==See Also== | ==See Also== |
Revision as of 16:58, 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?
Contents
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 .
Solution 3
Let's find the area of the triangles and the unit squares: on each side, there are 2 triangles. They both have 1 leg of length 1, and let's label the other legs x for one of thr triangles and y for the other. Note that x+y=3. The area of each of the triangles is and , and there are 4 of each. So now we need to find . Remember that x+y=3, so substituting this in we find that the area of all of the triangles is . The area of the 4 unit squares is 4, so the area of the square we need 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.