Difference between revisions of "1994 AJHSME Problems/Problem 19"
Mrdavid445 (talk | contribs) (Created page with "==Problem== Around the outside of a <math>4</math> by <math>4</math> square, construct four semicircles (as shown in the figure) with the four sides of the square as their diame...") |
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<math>\text{(A)}\ 16 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 48 \qquad \text{(E)}\ 64</math> | <math>\text{(A)}\ 16 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 48 \qquad \text{(E)}\ 64</math> | ||
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| + | ==Solution== | ||
| + | The radius of each semicircle is <math>2</math>, half the sidelength of the square. The line straight down the middle of square <math>ABCD</math> is the sum of two radii and the length of the smaller square, which is equivalent to its sidelength. The area of <math>ABCD</math> is <math>(4+2+2)^2 = \boxed{\text{(E)}\ 64}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AJHSME box|year=1994|num-b=18|num-a=20}} | ||
Revision as of 00:18, 23 December 2012
Problem
Around the outside of a 
by square, construct four semicircles (as shown in the figure) with the four sides of the square as their diameters. Another square, 
, has its sides parallel to the corresponding sides of the original square, and each side of is tangent to one of the semicircles. The area of the square is
![[asy] pair A,B,C,D; A = origin; B = (4,0); C = (4,4); D = (0,4); draw(A--B--C--D--cycle); draw(arc((2,1),(1,1),(3,1),CCW)--arc((3,2),(3,1),(3,3),CCW)--arc((2,3),(3,3),(1,3),CCW)--arc((1,2),(1,3),(1,1),CCW)); draw((1,1)--(3,1)--(3,3)--(1,3)--cycle); dot(A); dot(B); dot(C); dot(D); dot((1,1)); dot((3,1)); dot((1,3)); dot((3,3)); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); [/asy]](http://latex.artofproblemsolving.com/1/6/4/1647e4063573eec7bb48f68f685ea124d226c3bb.png)
Solution
The radius of each semicircle is 
, half the sidelength of the square. The line straight down the middle of square is the sum of two radii and the length of the smaller square, which is equivalent to its sidelength. The area of is 
.
See Also
| 1994 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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 | ||
