Asked by MTHOMBENI BAFANA
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.
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.
Answers
Answered by
Reiny
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.
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.
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