Assessment
Solves routine and non-routine problems involving volume of solid figures. Do it
on page 22-23
1. Mr. Jaime Go is building an additional cylindrical aquarium to add more attraction
to his Fish Park business. How much water is needed if it has a radius of 3 m and
a height of 4 m?
2. A 4 cm - radius sphere is cut from a cylindrical solid whose volume is 992.50 cm
What is the volume of the remaining parts of the cylinder after the cut?
3. John is making a pyramid-shaped water tank as a final and unique project for his
science exhibit. If the pyramid has a base measuring 24 in x 30 in and the height is
40 in, how much water is needed to fill in the tank?
4. A resident of Brgy. Masunuron is planning to purchase a cylindrical water tank as a
storage for her family for a 1 week water shortage. How much water could be poured
into the tank if it has a radius of 3 it and a height of 8 ft?
5. Mr. Santos is buying a spherical water tank for his poultry. How much water could
be stored if the water tank has a radius of 2 meters?
6. Marian, a math wizard pupil, is drinking a lemon juice using a conical-shaped
glass. After drinking, she is asking her mother on how much juice could be poured
into her glass. She measured the glass. It has as radius of 4 cm and a height of
15 cm
7 A pyramid-shaped aquarium is to be filled with water. How much water is needed
if the aquarium has a 2 ft and 3 ft as base and a height of 3 ft?
8. Mrs. Sandra Reyes is building a spherical swimming pool as a gift for her
daughter Glenda. How much water is needed to fill in the said pool if it has a
radius of 6 meters?
9 A 2 cm-radius sphere is cut from a cylindrical solid whose volume is 645 50 cm
What is the volume of the remaining parts of the cylinder after the cut?
10. Find the volume of a right circular cone-shaped buildino with a height of 9 cm an
radius base of 7 cm.
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Answer:
1.) SA= 2πrh + 2πr²
= (2×3.14×3×4) + (2×3.14×(3)²)
= (2×3.14×12) + ( 2×3.14×9)
=(2×37.68) + (2×28.26)
=75.36 + 56.52
= 131.88
2.) unknown
3.) unknown
4.) SA= 2πrh + 2πr²
= (2×3.14×3×8) + (2×3.14×(3)²)
=(2×3.14×24) + (2×3.14×9)
=(2×75.36) + (2×28×26)
=(150.72)+(56.52)
=207.24
5.) SA=4πr²
= 4×3.14×2²
=4×12.56
=50.24
6.) unknown
7.) SA= (2Lw+2Lh+2hw)
= (2×3×2)+(2×3×3)+(2+2+3)
=12+18+12
=30+12
=42
8.) SA=4πr²
=4×3.14×6²
=12.56×36
=452.16
9.) unknown
10.) unknown
Step-by-step explanation:
i calculate them all but the other's is not because is complicated
so hope it helps