Question

8 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 cm?
.

Answer 1

Answer:

17.6  cm²

Step-by-step explanation:

since the cylinder is solid it has a base

also whose area is to be calculated

T.S.A of remaining solid= C.S.A of cylinder+Area of cylinder base+C.S.A of cone

C.S.A of cylinder

diameter=1.4 cm

radius=1.4/2=0.7 cm

height=2.4 cm

C.S.A of cylinder=2πrh

=2×22/7×0.7×2.4

=2×22×0.1×2.4

= 10.56 cm²

Area of the cylinder base

The base of the cylinder is a circle with

radius= 0.7 cm

so,

area of base=πr²

=22/7×0.7×0.7

=22×0.1×0.7

=1.54 cm²

C.S.A of cone= πrl

radius= 0.7 cm

height= 2.4 cm

now, we find the slant height

we know that

l²=r²+h²

l²=(2.4)²+(0.7)²

l²=5.76+0.49

l²=6.25

l=√6.25

l= 2.5 cm

C.S.A of cone=πrl

=22/7×0.7×2.5

=22×0.1×2.5

=5.5 cm²

hence T.S.A of remaining solid= C.S.A of cylinder+Area of the cylinder base+C.S.A of cone

=10.56+1.54+5.5

=17.6 cm²

hence the area of the remaining solid is 17.6 cm²

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