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MAΘ Theta CPAV Solutions 2018

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MAΘ Theta CPAV Solutions 2018 MAΘ Theta CPAV Solutions 2018 Since the perimeter of the square is 40, one side of the square is 10 & its area is 100. Since the 1. B diameter of the circle is 8 its radius is 4. One quarter of the circle is inside the square; its area is 1 1 퐴 = 휋(42) = 휋(16) = 4휋. The area inside the square but outside the circle is 100 − 4휋. 4 4 Since the diameter of the sphere is 20 feet, its radius is 10 feet. 4 4 4000휋 1200휋 푉 = 휋푟3 = 휋103 = ; 퐴 = 4휋푟2 = 4휋102 = 400휋 = ; 3 3 3 3 2. C 1200휋 4000휋 5200휋 퐴 + 푉 = + = 3 3 3 3. B Since the diameter is 18 퐶 = 휋푑 = 18휋 4. A 3 3 288 2 푉 = 푎 ⁄ = 12 ⁄ = 1728⁄ = 288⁄ = √ ⁄ = 144√2 6√2 6√2 6√2 √2 2 20 Use the given circumference to find the radius of the circle. 퐶 = 2휋푟, 40 = 2휋푟, 20 = 휋푟, ⁄휋 = 푟 . If the inscribed angle is 45∘ its arc is 90∘ and has length 10, one quarter of the circumference. The chord of the 5. D segment is the hypotenuse of an isosceles right triangle whose legs are radii of the circle. 20 20√2 ℎ = 푙푒√2 = ⁄휋 √2 = ⁄휋 20√2 Perimeter of the segment is arc length plus chord: 10 + ⁄휋 Since the diameter of the sphere is the space diagonal of the inscribed cube, we must first find the diameter of the sphere. 6. B 4 125휋 3 4 125휋(3√3) 125 27 5 3 푉 = 휋푟3; √ = 휋푟3; = 푟3; √ = 푟3; √ = 푟 and diameter is 5√3 3 2 3 8휋 8 2 Space diagonal of a cube is 푎√3 = 5√3; 푎 = 5 and surface area is 6푎2 = 6 ∙ 52 = 150 Radius found using the circular cross-section area 퐴 = 휋푟2; 9휋 = 휋푟2; 푟 = 3; ℎ = 3푟 == 9 Volume is equal to volume of cylinder minus volume of hemisphere. 푉 = 휋푟2ℎ − 2⁄ 휋푟3 = 81휋 − 2⁄ ∙ 27휋 = 81휋 − 18휋 = 63휋 7. D 3 3 Surface area equal to cylinder lateral area plus one base area & surface area of hemisphere. 퐴 = 2휋푟ℎ + 휋푟2 + 2휋푟2 = 2휋푟ℎ + 3휋푟2 = 54휋 + 27휋 = 81휋 Sum of volume and surface area: 푆푢푚 = 63휋 + 81휋 = 144휋 8. D Regular hexahedron is a cube with squares as faces. Since perimeter of a face is 16 an edge is 4. Volume is lwh or (4)(4)(4) = 64. MAΘ Theta CPAV Solutions 2018 Perimeter of segment equals diameter plus semicircle length. 2 6 12 6휋 9. C 퐴 = 휋푟 = 36; 푟 = ⁄ 푎푛푑 푑 = ⁄ ; 푎푟푐푙푒푛푡ℎ푠푒푚 = 휋푟 = ⁄ √휋 √휋 √휋 12 6휋 (12 + 6휋) (12 + 6휋)√휋 6√휋(2 + 휋) Perimeter is ⁄ + ⁄ = ⁄ = ⁄휋 = ⁄휋 √휋 √휋 √휋 Space Diagonal of cube: 퐷 = 푎√3; √6 = 푎√3; 푎 = √6⁄ = √2 10. A √3 2 Surface Area of cube: 퐴 = 6푎2 = 6 (√2 ) = 6 ∙ 2 = 12 Surface area of sphere: 4휋푟2 = 256휋; 푟2 = 64; 푟 = 8. 4 4 11. B Volume of sphere: 푉 = 휋푟3 = 휋(512) = 2048휋⁄ 3 3 3 The greatest distance between two vertices of a regular hexagon is a diameter or 2a. 2a =10; a=5. 12. B Area of regular hexagon is equal to area of six equilateral triangles with side ‘a’. 푎2√3 25√3 150√3 75√3 퐴 = 6 ∙ = 6 ∙ = = 4 4 4 2 If diagonal of square is 6, a side of square is 6⁄ = 3√2. Four of the sides of the octagon have one √2 13. B third the length of a side of the square, that is √2. Each of the other four sides is the hypotenuse of an isosceles right triangle with legs equal to √2. One of those sides is √2 ∙ √2 = 2. Perimeter of octagon: 8 + 4√2 4 3 푉 = 휋푟3 = 288휋; 푟3 = 288 ∙ ; 푟3 = 216; 푟 = 6 3 4 14. B Surface area of the solid equals surface area of hemisphere plus circular cross-section area 2 2 2 2 푆퐴 = 2휋푟 + 휋푟 = 3휋푟 = 3휋(6 ) = 108휋 푎2 3 3푎2 3 푉 = 퐵ℎ; 108√3 = 퐵 ∙ 2; 퐵 = 54√3 ; 퐵 = 6 ∙ √ ; 54√3 = √ ; 푎2 = 36; 푎 = 6; 4 2 15. D 푝 = 6푎 = 36 푆퐴 = 2퐵 + 푝ℎ; 푆퐴 = 2(54√3) + 2(36) = 108√3 + 72 MAΘ Theta CPAV Solutions 2018 16. A 4 3 4 3 4 1372휋 퐶 = 2휋푟 = 14휋; 푟 = 7 푉 = ⁄3 휋푟 = ⁄3 휋(7 ) = ⁄3 휋(343) = ⁄3 The cross-section forms a 1:3 similar smaller cone. 17. D 2 2 4 Use area of the larger base to find R: 휋푅 = 16휋; 푅 16; 푅 = 4 because of 1:3 ratio 푟 = ⁄3. 4 8휋 퐶 = 2휋푟 = 2휋( ⁄3) = ⁄3 퐵 = 휋푟2 = 9휋; 푟2 = 9; 푟 = 3 using the given ℎ = 9 fill in sum of volume of cylinder and volume of hemisphere and add to sum of the surface area (minus one base) of the Cylinder and the surface area of the hemisphere. 18. D 2휋푟3 2 푉푐푦푙 + 푉ℎ푒푚 + 푆퐴푐푦푙 − 퐵 + 푆퐴ℎ푒푚 = 퐵ℎ + ⁄3 + 퐵 + 2휋푟ℎ + 2휋푟 2휋(27) = 9휋(9) + ⁄3 + 9휋 + 2휋(3)(9) + 2휋(9) = 81휋 + 18휋 + 9휋 + 54휋 + 18휋 = 180휋 The surface of a regular octahedron is composed of 8 equilateral triangles, so the surface area equals 2 푎 √3⁄ 2 2 19. A 8 times the area of an equilateral triangle. 푆퐴 = 8 ( 4) = 2푎 √3 = 200√3 ; 푎 = 100; a=10. 3 3 푎 √2⁄ 10 √2⁄ 1000√2⁄ The volume formula for a regular octahedron given side a: 푉 = 3 = 3 = 3 ∘ 10휋 5휋 The circumference of the circle is 10휋 so the 60 arc has length ⁄6 = ⁄3. The chord of the 20. B 60∘ arc is a side of the equilateral triangle formed by two radii and the chord. Therefore, the chord has length 5 and will not be included in the perimeter of the bounded area. Therefore, the 5휋 5휋 perimeter of the area inside the square and outside the circle is 32 − 5 + ⁄3 = 27 + ⁄3 The slant height of the cone, the hypotenuse of a triangle with legs 5 and 7, has length 푙 = √74 21. D 퐿퐴푐표푛푒 = 휋푟푙 = 5휋√74 퐿퐴푐푦푙 = 2휋푟ℎ = 70휋 퐿퐴푐표푛푒: 퐿퐴푐푦푙 = 5휋√74: 70휋 = √74: 14 Two congruent cones having only a base in common is the result of the rotation. The surface area of the solid is then two times the lateral area of either of the cones. Since the area of the square is 36, the side length is 6 and the length of a diagonal, the hypotenuse of 45-45-90 triangle with legs 6, is 22. D 6√2. Half the length of a diagonal, 3√2, is the length of the radius and also the height of a cone. The slant height of the cone is the hypotenuse of legs radius and height and is therefore, 6. 푆퐴푐표푛푒 = 휋푟푙 = 휋(3√2)(6) = 18휋√2 푆퐴푠표푙푑 = 2(18휋√2) = 36휋√2 The length of the semicircle, 휋푟 = 6휋, is the circumference of the cone; since 6휋 = 2휋푟, 푟푐표푛푒 = 3. The radius of the semicircle is the slant height, ′푙′, of the cone and the hypotenuse of a triangle 2 2 23. C formed by the 푟푐표푛푒, ℎ푐표푛푒, 푎푛푑 푙푐표푛푒. Therefore: ℎ푐표푛푒 = √6 − 3 = √36 − 9 = √27 = 3√3. 2 휋푟2ℎ 휋(3 )(3√3)⁄ 푉 = ⁄3 = 3 = 9휋√3 MAΘ Theta CPAV Solutions 2018 2 2 푉푐푦푙 = 휋푟 ℎ = 휋2 (6) = 24휋 which is the volume of the new cone. 휋푟2 ℎ 9휋(ℎ) 푉 = 푐표푛푒 ⁄ ; 24휋 = ⁄ ; 24휋 = 3휋ℎ; ℎ = 8. 24. E 푐표푛푒 3 3 푐표푛푒 Slant height is hypotenuse for legs 8 & 3, therefore, 푙 = √64 + 9 = √73. Surface Area of Cone: 푆퐴 = 휋푟2 + 휋푟푙 = 9휋 + 3휋√73 The height and diameter of the cylinder are equal to 6, the side length of the cube. 25. A 3 2 3 2 푉푐푢푏푒 − 푉푐푦푙 = 푎 − 휋푟 ℎ = 6 − 휋(3 )6 = 216 − 54휋 The slant height is the hypotenuse to legs apothem and height of the pyramid. The apothem of the square base (and the radius of the cone) is 3. 26. B Therefore, height of pyramid and cone: ℎ = √92 − 32 = √72 = 6√2. 2 휋푟2ℎ 휋(3 )(6√2)⁄ 푉푐표푛푒 = ⁄3 = 3 = 18휋√2 The first information needed is the area a segment of a circular cross-section, perpendicular to the axis, that represents the liquid in the tank. Since the diameter is 8, the radius is 4 and the sector of the circle that contains the needed segment is composed of an isosceles triangle with legs 4 and height 2. The height divides the triangle into two 30-60-90 triangles with the height (2) as the short leg and half the chord of the segment as the long leg (2√3); across from the long leg with vertex at the center of the circle is the 60∘ angle and since there are two 30-60-90 triangles, the central angle 27. D of the sector is 120∘. Area of segment = area of sector – area of triangle 퐴 = 휋푟2 푏ℎ 16휋 (4√3)(2)⁄ 16휋 ⁄3 − ⁄2 = ⁄3 − 2 = ⁄3 − 4√3. All that is needed now to find the volume of the liquid is to multiply the segment area times the 16휋 320휋 (320휋 − 240√3)⁄ length of the tank. 푉 = ( ⁄3 − 4√3)20 = ⁄3 − 80√3 = 3 퐿퐴 = 2휋푟ℎ; 2휋푟2 = 2휋푟ℎ; 푟 = ℎ = 2; 푑 = 4 Hypotenuse of right triangle is √42 + 22 = 2√5 28. A 푐푦푙 Perimeter of right triangle: 푑푎푚푒푡푒푟 + ℎ푒ℎ푡 + ℎ푦푝표푡푒푛푢푠푒 = 4 + 2 + 2√5 = 6 + 2√5 Regular hexahedron is a cube. The diagonal (3√2)of the cube face is the diameter of the cylinder, 3 2 the radius is then √ ⁄ 표푟 3⁄ .
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