Question

From a solid cylinder of height 12 centimetre and diameter 10 cm a conical cavity of same height and same diameter is hollowed out find the total surface area of remaining solid?​

Answer 1

Answer:

Volume of remaining solid = Volume of cylinder - Volume of Cone

In the given case,

r = 5cm

h = 12cm

Volume of Cylinder = pi * r * r * h = 3.14 * 5 * 5 * 12 = 942 cm^3

Volume of Cone = (1/3) * pi * r * r * h = (1/3) * 3.14 * 5 * 5 * 12 = 314 cm^3

Remaining Volume = 942 - 314 = 628cm^3

Total Surface of Remaining Solid = Conical surface area of Cone + Area of Circle + Cylindrical area of Cylinder

Conical surface area of Cone = pi * r * ((r^2 + h^2)^(1/2)) = 282.74cm^2

Area of Circle = pi * r * r = 78.54cm^2

Cylindrical area of Cylinder = 2 * pi * r * h = 376.8cm^2

Total Surface Area = 738.08cm^2

Answer 2

Answer:

Given that:-

➤Volume of remaining solid = volume of cylinder - volume of cone

r = 5cm

h = 12cm

Then:-

➤ Volume of cylinder is

➠ $\bf \pi {r}^{2} h$

➠ 3.14 × 5² × 12

➠ 942cm²

➤Volume of cone is

➠$\bf \frac{1}{3} \pi {r}^{2} h$

➠$\bf \frac{1}{3} \times 3.14 \times {5}^{2} \times 12$

➠314cm³

➤Remaining volume = 942 - 314

= 628cm³

Solution:-

Total surface of remaining solid = conical surface area of cone + Area of circle + cylindrical area of cylinder

➤Conical surface area of cone is

⇒$\bf \pi \: r \: \sqrt{ {r}^{2} } + {h}^{2}$

⇒204.203cm²

➤Conical of circle is

⇒$\bf \pi {r}^{2}$

⇒78.54cm²

➤Cylindrical area of cylinder is

⇒$\bf 2\pi \: rh$

⇒376.8cm²

$\therefore therefore$

⇨Total surface area is 659.543cm²

HOPE IT HELPS YOU BRO..☺