Difference between revisions of "1994 AJHSME Problems/Problem 19"
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==Solution== | ==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 | + | 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 side length. The area of <math>ABCD</math> is <math>(4+2+2)^2 = \boxed{\text{(E)}\ 64}</math>. |
==See Also== | ==See Also== | ||
{{AJHSME box|year=1994|num-b=18|num-a=20}} | {{AJHSME box|year=1994|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:12, 21 February 2021
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
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 side length. 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.