A CONE WAS CONSTRACTED FROM PLASTIC, HAVE A HEIGHT OF 20cm (PART C). THE LID IS MADE UP OF HEMISPHERE HAVE RADIUS 0F 5cm (PART A) AND PYRAMID HAVE VOLUME OF 81,667cm³. THE LID FITS EXACTLY OVER THE CONE.
A) CALC. THE VOL. OF PART A.
B) CALC. THE VOL. OF PART C.
C) CALC. THE SURFACE AREA OF PART C.
D) CALC. THE VOL. OF THE REMAINING SPACE IN PART C IF THE LID IS CLOSED. PLZ HELP.
volume of cone = (1/3)πr^2h, where r is radius and h is height,
you have the height, and the hemisphere fits exactly over the cone, so the radius of the cone is the same as that of the hemisphere.
volume of hemisphere = (1/2)(4/3)πr^3
(hemi means half)
I don't know how your pyramid fits in here, but the volume of a pyramid with square base
= (1/3)base x height.
To solve this problem, we need to break it down into different parts and calculate the volumes and surface areas separately. Let's go step by step:
A) Calculating the volume of Part A (the hemisphere lid):
The volume of a hemisphere is given by the formula (2/3)πr^3, where r is the radius. In this case, the radius is 5cm. So we can substitute the values into the formula:
Volume of Part A = (2/3)π(5cm)^3
= (2/3)π(125cm^3)
= (250/3)π cm^3
≈ 261.8 cm^3 (rounded to one decimal place)
B) Calculating the volume of Part C (the cone):
The volume of a cone is given by the formula (1/3)πr^2h, where r is the radius of the base and h is the height. In this case, the height is given as 20cm. To find the radius, we need to use the fact that the lid fits exactly over the cone, so the radius of the lid hemisphere (5cm) is also the radius of the cone. Substituting the values into the formula, we get:
Volume of Part C = (1/3)π(5cm)^2(20cm)
= (1/3)π(25cm^2)(20cm)
= (1/3)π(500cm^3)
= (500/3)π cm^3
≈ 523.6 cm^3 (rounded to one decimal place)
C) Calculating the surface area of Part C (the cone):
The surface area of a cone is given by the formula πr(r + √(r^2 + h^2)), where r is the radius of the base and h is the slant height (in this case, the height of the cone). We already know the radius is 5cm and the height is 20cm, so we can substitute the values into the formula:
Surface area of Part C = π(5cm)(5cm + √(5cm^2 + 20cm^2))
= π(5cm)(5cm + √(25cm^2 + 400cm^2))
= π(5cm)(5cm + √(425cm^2))
= π(5cm)(5cm + 20.62cm)
≈ 339.3 cm^2 (rounded to one decimal place)
D) Calculating the volume of the remaining space in Part C when the lid is closed:
To find the volume of the remaining space in Part C, we subtract the volume of the lid (Part A) from the volume of the cone (Part C):
Volume of remaining space = Volume of Part C - Volume of Part A
≈ 523.6 cm^3 - 261.8 cm^3
≈ 261.8 cm^3
So, the volume of the remaining space in Part C when the lid is closed is approximately 261.8 cm^3.