Potwe 15 Fc Gm 02 s

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WWW.CEMC.UWATERLOO.CA | The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Problem of the Week Problem E and Solution Sphere Pressure Problem A cube rests inside a sphere so that each vertex touches the sphere. The radius of the sphere is 6 cm. Determine the volume of the cube. Solution Label four vertices of the cube A,B,C,D as shown in the diagram. Let x represent the side length of the cube. Then AB = BC = CD = x. The diagonals of a cube intersect in a point such that the distance from the intersection point to each vertex is equal. Since each vertex of the cube touches the sphere, the diagonal of the cube, AD, is equal in length to the diameter of the sphere. Therefore AD = 2(6) = 12 cm. Each face of a cube is a square so ABC = 90 . Using Pythagoras’ Theorem, AC 2 = AB 2 + BC 2 = x 2 + x 2 =2x 2 . In a cube the sides are perpendicular to the base. In particular, DC is perpendicular to the base and it follows that DC AC . Therefore 4DCA is a right angled triangle. Using Pythagoras’ Theorem, AD 2 = AC 2 + CD 2 =2x 2 + x 2 =3x 2 . But AD = 12 so AD 2 = 144. Then 3x 2 = 144, x 2 = 48 and x =4 3 since x> 0. The volume of the cube is x 3 = (4 3) 3 = 192 3 cm 3 . the volume of the cube is 192 3 cm 3 .

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Transcript of Potwe 15 Fc Gm 02 s

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WWW.CEMC.UWATERLOO.CA | The CENTRE for EDUCATION in MATHEMATICS and COMPUTING

Problem of the WeekProblem E and Solution

Sphere Pressure

ProblemA cube rests inside a sphere so that each vertex touches the sphere. The radius of the sphere is6 cm. Determine the volume of the cube.

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SolutionLabel four vertices of the cube A,B,C,D as shown in the diagram. Let xrepresent the side length of the cube. Then AB = BC = CD = x.

The diagonals of a cube intersect in a point such that the distance from theintersection point to each vertex is equal. Since each vertex of the cube touchesthe sphere, the diagonal of the cube, AD, is equal in length to the diameter ofthe sphere. Therefore AD = 2(6) = 12 cm.

Each face of a cube is a square so ∠ABC = 90◦. Using Pythagoras’ Theorem,

AC2 = AB2 +BC2 = x2 + x2 = 2x2.

In a cube the sides are perpendicular to the base. In particular, DC isperpendicular to the base and it follows that DC ⊥ AC. Therefore 4DCA is aright angled triangle. Using Pythagoras’ Theorem,

AD2 = AC2 + CD2 = 2x2 + x2 = 3x2.

But AD = 12 so AD2 = 144. Then 3x2 = 144, x2 = 48 and x = 4√3 since

x > 0. The volume of the cube is x3 = (4√3)3 = 192

√3 cm3.

∴ the volume of the cube is 192√3 cm3.