Question

A frustum of a right circular cone has a diameter of base 10cm of top 6cm and height of 5cm; find the area of its whole surface and volume?

Answer 1

Given base diameter of cone (d1) = 10 cm

Radius (r1) = 10/2 cm = 5 cm

Top diameter of Cone (d2) = 6 cm

Radius (r2) = 6/2 cm = 3 cm

Height of the cone (h) = 5 cm

Volume of the frustum right circular cone=

$\pi \div 3 (r1^{2} + r2 ^{2} + r1r2)h$

$\pi \div 3(5 ^{2} + 3^{2} + 5 \times 3)^{3}$

= Solve it and get the answer of it.....

Let 'L' be the slant height of cone

L=

$\sqrt{(r1 - r2)^{2} + h^{2} }$

L=

$\sqrt{(5 - 3)^{2} + 5^{2} }$

L=

$\sqrt{29}$

Slant height of cone = sq root of 29 cm Total surface area of the cone

=

$\pi(r1 + r2) \times l + \pi \times r1^{2} + \pi \times r2 ^{2}$

$\pi(5 + 3) \times \sqrt{29} + \pi \times 5^{2} + \pi \times 3^{2}$

Now solve it and then you will get the total surface area of the cone...........

*********************Hope it helped***************************

Answer 2

Concept:

Area of a figure is the region that a figure encloses.

Volume of an object is the capacity that an object has.

Given:

We are given that:

A frustum of right circular cone.

Diameter of base is 10 cm.

Diameter of top is 6 cm.

Height is 5 cm.

Find:

We need to find the whole surface area and volume.

Solution:

We will first find the area of the frustum:

Radius of base(R1)=5 cm

Radius of top(R2)=3 cm

Height(h)=5 cm

Slant Height (l)=√(R1-R2)²+h²=√(2)²+5²=√4+25=√29

Area=π(R1+R2)l+π(R1²+R2²)

A=π(5+3)(√29)+π(25+9)

A=π(8√29+34)

A=242.2556 cm²

Now we will find the volume of the frustum:

Volume of frustum = 1/3(πh)(R1²+R2²+R1R2)

V=1/3(22/7)(5)(5²+3²+15)

V=110/105(49)

V=51.3333 cm³

Therefore, the total surface area is 242.25556 cm² and the volume is 51.3333 cm³.

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