AMC 10 A AMC 12 A Problem 13 AOC BOC 30 Solution Answer B Let r

# Amc 10 a amc 12 a problem 13 aoc boc 30 solution

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2008 AMC 10 A, Problem #162008 AMC 12 A, Problem #13AOC=BOC= 30.”SolutionAnswer (B):LetrandRbe the radii of the smaller and larger circles,respectively. LetEbe the center of the smaller circle, letOCbe the radiusof the larger circle that containsE, and letDbe the point of tangencyof the smaller circle toOA. ThenOE=R-r, and because4EDOisa306090triangle,OE= 2DE= 2r. Thus2r=R-r, sorR=13.The ratio of the areas is(13)2=19.ABCDEODifficulty:HardNCTM Standard:Geometry Standard: analyze characteristics and properties of two- and three-dimensional geometric shapes and develop mathematical arguments about geometric relationships.Mathworld.com Classification:Geometry>Plane Geometry>Triangles>Special Triangles>Other Triangles>30-60-90 Triangle
An equilateral triangle has side length 6. What is thearea of the region containing all points that are outsidethe triangle and not more than 3 units from a pointof the triangle?93 + 1·2π2008 AMC 10 A, Problem #17“The region consists of three rectangles together withthree120sectors of circles.”SolutionAnswer (B):The region consists of three rectangles with length 6 andwidth 3 together with three120sectors of circles with radius 3.63The combined area of the three120sectors is the same as the area of acircle with radius 3, so the area of the region is3·6·3 +π·32= 54 + 9π.Difficulty:HardNCTM Standard:Geometry Standard: analyze characteristics and properties of two- and three-dimensional geometric shapes and develop mathematical arguments about geometric relationships.Mathworld.com Classification:Geometry>Plane Geometry>Triangles>Special Triangles>Other Triangles>Triangle
A right triangle has perimeter 32 and area 20. Whatis the length of its hypotenuse?2008 AMC 10 A, Problem #18“Set up two equations from given information withthree sides, and with the Pythagorean Theorem, wecan generate another equation.”SolutionAnswer (B):Letxbe the length of the hypotenuse, and letyandzbethe lengths of the legs. The given conditions imply thaty2+z2=x2,y+z= 32-x,andyz= 40.Thus(32-x)2= (y+z)2=y2+z2+ 2yz=x2+ 80,from which1024-64x= 80, andx=594.Note: Solving the system of equations yields leg lengths of18(69 +2201)and18(69-2201),so a triangle satisfying the given conditions does in fact exist.Difficulty:HardNCTM Standard:Geometry Standard: analyze properties and determine attributes of two- andthree-dimensional objects.Mathworld.com Classification:Geometry>Plane Geometry>Triangles>Special Triangles>Other Triangles>TriangleGeometry>Plane Geometry>Triangles>Triangle Properties>Pythagorean Theorem
RectanglePQRSlies in a plane withPQ=RS= 2andQR=SP= 6.The rectangle is rotated90clockwise aboutR, then rotated90clockwise aboutthe point thatSmoved to after the first rotation.What is the length of the path traveled by pointP?2008 AMC 10 A, Problem #19“Sketch the rotations.”SolutionAnswer (C):LetP0andS0denote the positions ofPandS, respectively,after the rotation aboutR, and letP00denote the final position ofP. Inthe rotation that movesPto positionP0, the pointP

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