Title MSTE 11 Solutions.pdf ✅

Vfr = πh 3 [(1 m) 2 + (0.7 m) 2 + (1 m)(0.7 m)] Since the volumes should be equal, 4π m3 = πh 3 [(1 m) 2 + (0.7 m) 2 + (1 m)(0.7 m)] h = 5.479 m ▣ 4. The base of a truncated prism is an equilateral triangle with each side 6 m long. The lengths of its lateral edges are 3, 4, and 4.5 m long. Find the volume of the prism. [SOLUTION] The volume of a truncated prism V = Bh V = ( (6 m) 2√3 4 ) ( 3 m + 4 m + 4.5 m 3 ) V = 59.7558 m3 ▣ 5. A sphere of radius 5 cm and a right circular cone of base radius 5 cm and altitude of 10 cm stand on the same horizontal plane. How far from this plane is the plane, parallel to the former, which will cut the two solids into two equal circular sections? [SOLUTION] From the sphere, r 2 = 5 2 − (5 − h) 2 From the cone, r 10 − h = 5 10 → r = 1 2 (10 − h) From the equations, [ 1 2 (10 − h)] 2 = 5 2 − (5 − h) 2

( r2 r1 ) 3 = V2 V1 ( r2 r1 ) 3 = 1.2V1 V1 r2 r1 = 1.06266 SA2 SA1 = ( r2 r1 ) 2 SA2 SA1 = (1.06266) 2 SA2 SA1 = 1.1292 Therefore, there is an increase of 12.92% . ▣ 9. A closed cylindrical tank measures 12 ft long and 5 ft in diameter. It has to contain water with a depth of 3 ft when the tank is lying in horizontal position. Find the height of the water when the tank is in its vertical position. [SOLUTION] cos θ = 0.5 2.5 → θ = 78.463° Solve for the cross-sectional area A = 360° − 2(78.463°) 360° π(2.5 ft) 2 + 1 2 (2.5) 2 sin(2 × 78.463°) = 12.3007 ft 2 The volume of water is V = (12.3007 ft 2 )(12 ft) = 147.6085 ft 3 When it is in vertical position, the water is of a cylindrical shape. 147.6085ft 3 = π(2.5 ft) 2h h = 7.518 ft