From a solid cylinder whose height is 2.4cm and diameter 1.4cm, a conical cavity of the same height and the same diameter is hollowed out. The total surface area of the remaining solid to the nearest c m 2 is ____. - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

Solution:

According to the given information, a height of 2.4 cm of a solid cylinder with a diameter of 1.4 cm, where the same conical cavity is hollowed out at the same height and diameter
So the dimensions we know are
Height of full cylinder (v) = 2.4 cm
Diameter of solid cylinder (d) = 1.4 cm
Therefore, the radius of the solid cylinder (r) = 0.7 cm
As we know, the formula of the total surface of the remaining solid is equal to the curved surface of the cylinder + the area of the base of the cylinder + the curved surface of the cone
Now we know that the curved surface of the cylinder =

2 π r h

Area of the base of the cylinder

π r 2

and Curved surface area of the cone

π r l

.
Where

l

slant height which is given as

l = r 2 + h 2

Therefore, Total surface area of the remaining solid =

2 π r h + π r 2 + π r r 2 + h 2

Now substituting the values in the above formula, we get

2 × 227 × 0.7 × 2.4 + 227 × (0.7) 2 × (2.4) 2 + (0.7) 2 ⇒ 10.56 + 1.54 + 5.5 c m 2 ⇒ 17.6 c m 2

So, Total surface area of the remaining solid is equal to

17.6 c m 2

From a solid cylinder whose height is 2.4cm and diameter 1.4cm, a conical cavity of the same height and the same diameter is hollowed out. The total surface area of the remaining solid to the nearest $c m 2$ is ____.

Solution:

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