The radio of the ends of a frustum of a cone 45cm high are 28 CM and 7CM .Find It's Volume,The curved Surface area And total surface Area.
(Take π =22/7)
Answers
MATHS
The radii of the ends of frustum of a cone 45 cm high are 28 cm and 7 cm. Find its volume, the curved surface area and the total surface area
ANSWER
Let 'ℓ' & R be the radii of the bottom and top circular ends of a frustum of a cone, respectively and 'h' be height of cone.
Volume of the frustum of the cone =31×722×45(282+72+28×7)cm3
=722×15(784+49+196)cm3
=722×15×1029cm3=22×15×147cm3
=330×147cm3
Now, l=h2+(R−r2)=(45)2+(28−7)2=2025+441=2466
=2466=49.65cm
curved surface area of the frustum of the cone
=π(R+r)ℓ=[722∗(28+7)∗49.65]cm2
=[22×5×49.65]cm2
5461.5cm2
total surface area of frustum of the cone
π(R+r)ℓ+πR2+πr2
=[5461.5+722×(28)2+722×(7)2]cm2
=[5261.5+722×(28)2+(7)2]cm2
=[5461.5+22×1+9]cm2
=8079.5cm2
Answer:
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