Solve the question---24 The diagram shows a frustum of a cone, and a sphere. The frustum, shown shaded in the diagram, is made by removing the small cone from the large cone. The small cone and the large cone are similar. [Diagram 1: Frustum of a cone and a sphere] Description: Left part: A frustum of a cone is shown shaded. Above the frustum, a dashed outline of a small cone removed from the top is shown, completing a large cone. Vertical arrow with label "Small cone h cm" pointing down from the apex of the complete large cone to the top circular base of the frustum. Vertical arrow with label "Large cone kh cm" pointing down from the apex of the complete large cone to the bottom circular base of the frustum. The top circular base of the frustum has a radius labeled "r cm". The bottom circular base of the frustum has a radius labeled "kr cm". A horizontal dashed line segment across the top base circle is labeled "r cm" with arrows pointing to the circumference. A horizontal dashed line segment across the bottom base circle is labeled "kr cm" with arrows pointing to the circumference. Right part: A sphere is shown. A dashed line segment from the center to the surface is labeled "r cm", representing the radius. Annotation: Diagram NOT accurately drawn. The height of the small cone is h cm and the radius of the base of the small cone is r cm. The height of the large cone is kh cm and the radius of the base of the large cone is kr cm. The radius of the sphere is r cm. The sphere is divided into two hemispheres, each of radius r cm. Solid A is formed by joining one of the hemispheres to the frustum. The plane face of the hemisphere coincides with the upper plane face of the frustum, as shown in the diagram below. Solid B is formed by joining the other hemisphere to the small cone that was removed from the large cone. The plane face of the hemisphere coincides with the plane face of the base of the small cone, as shown in the diagram below. [Diagram 2: Solid A and Solid B] Description: Left part labeled "A": A frustum with a hemisphere attached to its top circular base. The radius of the hemisphere is labeled "r cm". The bottom circular base of the frustum has radius labeled "kr cm". Right part labeled "B": A cone with a hemisphere attached to its circular base. The vertex of the cone is pointing upwards. Annotation: Diagram NOT accurately drawn. The volume of solid A is 6 times the volume of solid B. Given that k > sqrt(7) find an expression for h in terms of k and r

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We have a frustum formed by removing a small cone from a large cone. The small cone has height h and radius r, while the large cone has height kh and radius kr. We also have a sphere with radius r. We need to find an expression for h in terms of k and r, given that the volume of solid A equals 6 times the volume of solid B. Solid A is formed by joining a hemisphere to the top of the frustum. The hemisphere has radius r and coincides with the upper circular face of the frustum. Solid B is formed by joining a hemisphere to the base of the small cone that was removed. Both hemispheres have the same radius r. Let's calculate the volumes step by step. A hemisphere has volume two-thirds pi r cubed. A cone has volume one-third pi r squared h. The frustum volume equals the large cone volume minus the small cone volume. The large cone has volume one-third pi k cubed r squared h, and the small cone has volume one-third pi r squared h. Therefore, the frustum volume is one-third pi r squared h times k cubed minus 1. Now we set up the equation. Volume of solid A equals the frustum volume plus hemisphere volume. Volume of solid B equals the cone volume plus hemisphere volume. Given that volume A equals 6 times volume B, we substitute and simplify. Factoring out common terms and rearranging, we get h times k cubed minus 7 equals 10r. Therefore, h equals 10r divided by k cubed minus 7. Solid A is formed by joining a hemisphere to the top of the frustum. The hemisphere has radius r and coincides with the upper circular face of the frustum. Solid B is formed by joining a hemisphere to the base of the small cone that was removed. Both hemispheres have the same radius r. Let's calculate the volumes step by step. A hemisphere has volume two-thirds pi r cubed. A cone has volume one-third pi r squared h. The frustum volume equals the large cone volume minus the small cone volume. The large cone has volume one-third pi k cubed r squared h, and the small cone has volume one-third pi r squared h. Therefore, the frustum volume is one-third pi r squared h times k cubed minus 1. Now we set up the equation. Volume of solid A equals the frustum volume plus hemisphere volume. Volume of solid B equals the cone volume plus hemisphere volume. Given that volume A equals 6 times volume B, we substitute and simplify. Factoring out common terms and rearranging, we get h times k cubed minus 7 equals 10r. Therefore, h equals 10r divided by k cubed minus 7. Our final answer is h equals 10r divided by k cubed minus 7. We can verify this is correct because the given condition k greater than square root of 7 ensures that k cubed minus 7 is positive, making our denominator valid. This formula successfully expresses the height h in terms of the parameters k and r, completing our solution to the frustum and sphere volume problem.