Question

From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the

same height and same diameter is hollowed out. Find the total surface area of the

remaining solid to the nearest cm2.​

Answer 1

Answer:

Given:

Height (h) of the conical part = Height (h) of the cylindrical part =2.4 cm

Diameter of the cylindrical part =1.4 cm

Radius = Diameter

2

Radius(r) of the cylindrical part =0.7 cm

Slant height (l) of conical part = √r²+h²

= √0.7²+2.4²

=√ 0.49+5.76 = √6.25 =2.5

Total surface area of the remaining solid = CSA of cylindrical part + CSA of conical part + Area of cylindrical base

=2πrh + πrl + πr²

=2×22 × 0.7×2.4+22×0.7×2.5+×0.7×0.7

7. 7

=4.4×2.4+2.2×2.+2.2×0.7

=10.56+5.50+1.54=17.60² cm

The total surface area of the remaining solid to the nearest cm² is 18² cm

Step-by-step explanation:

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Answer 2

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From the question we know the following:

The diameter of the cylinder = diameter of conical cavity = 1.4 cm

So, the radius of the cylinder = radius of the conical cavity = 1.4/2 = 0.7

Also, the height of the cylinder = height of the conical cavity = 2.4 cm

Now, the TSA of remaining solid = surface area of conical cavity + TSA of the cylinder

= πrl+(2πrh+πr2)

= πr(l+2h+r)

= (22/7)× 0.7(2.5+4.8+0.7)

= 2.2×8 = 17.6 cm2

So, the total surface area of the remaining solid is 17.6 cm2