Question

The answer of this problem

Answer 1

1.a. HG has length 26 m (same as the base of the cuboid), so the length of GM is 13 m. If OG = 30 m, then by the Pythagorean theorem, we have

OM² + (13 m)² = (30 m)²   ===>   OM = √(30² - 13²) m = √731 m ≈ 27.037 m

1.b. Let P be the center of the square EFGH. Then using the Pythagorean theorem again, we have

OP² + (13 m)² = (√731 m)²   ===>   OP = √(731 - 13²) m = √562 m

Then the total height of the tower is 70 m + √562 m ≈ 93.707 m

1.c. OM makes an angle θ with the plane EFGH such that

tan(θ) = OP/PM = (√562 m)/(13 m)   ===>   θ ≈ 61.261°

1.d. The given rate for cleaning doesn't quite make sense, it probably is meant to say the service charges $78 per square meter of the outside of the building. Compute the area of each exposed face (that is, excluding the base):

• Each rectangular face has length 26 m and height 70 m, hence an area of

(26 m) (70 m) = 1820 m²

• Each triangular face has height OM = √731 m and base 26 m, hence an area of

1/2 (√731 m) (26 m) = 13√731 m² ≈ 351.481 m²

Then the total area to be cleaned would be (1820 + 13√731) m², which could cost

($78/m²) ((1820 + 13√731) m²) = $(141,960 + 1014√731) ≈ $169,376.53

2.a. The volume of a sphere with radius r is 4/3 πr ³, so the volume of a hemisphere with the same radius is 2/3 πr ³. If the radius is 3 cm, then the volume of this hemisphere is

2/3 π (3 cm)³ = 2/3 (27π cm³) = 18π cm³

2.b. The volume of a right cone with base radius r and height h is 1/3 πr ²h. Note that the cone has the same radius as the hemisphere, r = 3 cm. If this volume is 2/3 as large as the hemisphere's, then solve for the height h :

1/3 π (3 cm)²h = 2/3 (18π cm³)

(3π cm²) h = 12π cm³

h = (12π cm³) / (3π cm²)

h = 4 cm

2.c. Using the Pythagorean theorem,

h ² + r ² = ℓ ²   ===>   ℓ = √((4 cm)² + (3 cm)²) = √25 cm = 5 cm

2.d. The angle θ between the slant height and the flat of the hemisphere is such that

tan(θ) = r/h = (3 cm)/(4 cm)   ===>   θ ≈ 36.87°