Question

a solid toy is in the form of a hemisphere surmounted by a right circular cone of the height of the cone is 4cm and diameter of the base is 6 cm calculate the volume of the toy,surface area of the toy​

Answer 1

Given :

• Height of the right-circular cone = 4 cm.

• Diameter of the right-circular cone = 6 cm.

• Radius of cone = Radius of hemisphere

To find :

• Volume of the toy.

• Total surface area of the cone.

Solution :

Since the toy is made from the right-circular cone and the hemisphere , sum of their volumes will give the volume of the toy and the sum of their total surface area will give us the total surface area of the toy.

Let us find the Radius of the cone and hemisphere :

→ Radius of cone = Radius of hemisphere = Diameter/2

==> R = 6/2

==> R = 3 cm

Hence, the radius is 3 cm.

To find the volume of the cone :

Let the volume of toy be $V_{t}$

According to the Question,

$\small{\bf{Volume\:of\:toy = Volume\:of\:cone + Volume\:of\:hemisphere}}$

$:\implies \bf{V_{t} = \dfrac{1}{3}\pi r^{2}h + \dfrac{2}{3}\pi r^{3}}$

Where :-

• r = Radius
• h = Height

$\boxed{\begin{minipage}{7 cm} Volume of a Cone = \bf{\dfrac{1}{3}\pi r^{2}h} and \\ \\ Volume of a hemisphere = \bf{\dfrac{2}{3}\pi r^{3}} \end{minipage}}$

By using the Equation for volume of the toy and substituting in it , we get :

$:\implies \bf{V_{t} = \dfrac{1}{3}\pi r^{2}h + \dfrac{2}{3}\pi r^{3}}$

$:\implies \bf{V_{t} = \bigg(\dfrac{1}{3} \times \dfrac{22}{7} \times 3^{2} \times 4\bigg) + \bigg(\dfrac{2}{3} \times \dfrac{22}{7} \times 3^{3}\bigg)}$

$:\implies \bf{V_{t} = \bigg(\dfrac{1}{3} \times \dfrac{22}{7} \times 9 \times 4\bigg) + \bigg(\dfrac{2}{3} \times \dfrac{22}{7} \times 27\bigg)}$

$:\implies \bf{V_{t} = \bigg(\dfrac{1}{\not{3}} \times \dfrac{22}{7} \times \not{9} \times 4\bigg) + \bigg(\dfrac{2}{\not{3}} \times \dfrac{22}{7} \times \not{27}\bigg)}$

$:\implies \bf{V_{t} = \bigg(\dfrac{22}{7} \times 3 \times 4\bigg) + \bigg(2 \times \dfrac{22}{7} \times 9\bigg)}$

$:\implies \bf{V_{t} = \dfrac{264}{7} + \dfrac{396}{7}}$

$:\implies \bf{V_{t} = \dfrac{760}{7}}$

$:\implies \bf{V_{t} = 108.57}$

$\boxed{\therefore \bf{Volume\:of\:toy = 108.57\:cm^{3}}}$

Hence, the volume of the toy is 108.57 cm³.

To find the Total surface area of the toy :

Let the TSA of toy be $T_{t}$

According to the Question,

$\small{\bf{TSA\:of\:toy = TSA\:of\:cone + TSA\:of\:hemisphere}}$

$:\implies \bf{T_{t} = \pi r(l + r) + 3\pi r^{2}}$

Where :-

• r = Radius

• h = Height

• l = Slant height

$\boxed{\begin{minipage}{7 cm} TSA of a Cone = \bf{\pi r(l + r)} and \\ \\ TSA of a hemisphere = \bf{3\pi r^{2}} \\ \\ Slant height = \bf{l = \sqrt{r^{2} + h^{2}}} \end{minipage}}$

By using the Equation for TSA of the toy and substituting in it , we get :

$:\implies \bf{T_{t} = \pi r\big[\big(\sqrt{r^{2} + h^{2}}\big) + r\big] + \big(3\pi r^{2}\big)}\:\:\:\big(\because \bf{l = \sqrt{r^{2} + h^{2}}}\big)$

$:\implies \bf{T_{t} = \dfrac{22}{7} \times 3\big[\big(\sqrt{3^{2} + 4^{2}}\big) + 3\big] + \bigg(3 \times \dfrac{22}{7} \times 3^{2}\bigg)}$

$:\implies \bf{T_{t} = \dfrac{22}{7} \times 3\big[\big(\sqrt{9 + 16}\big) + 3\big] + \bigg(3 \times \dfrac{22}{7} \times 9\bigg)}$

$:\implies \bf{T_{t} = \dfrac{22}{7} \times 3\big[\big(\sqrt{25}\big) + 3\big] + \bigg(\not{3} \times \dfrac{22}{7} \times \not{9}\bigg)}$

$:\implies \bf{T_{t} = \dfrac{22}{7} \times 3\big[5 + 3\big] + \bigg(\dfrac{22}{7} \times 3\bigg)}$

$:\implies \bf{T_{t} = \dfrac{22}{7} \times 3 \times 8 + \dfrac{66}{7}}$

$:\implies \bf{T_{t} = \dfrac{22}{7} \times 24 + \dfrac{66}{7}}$

$:\implies \bf{T_{t} = \dfrac{528}{7} + \dfrac{66}{7}}$

$:\implies \bf{T_{t} = \dfrac{528 + 66}{7}}$

$:\implies \bf{T_{t} = \dfrac{594}{7}}$

$:\implies \bf{T_{t} = 84.86}$

$\boxed{\therefore \bf{TSA\:of\:the\:toy = 84.86\:cm^{2}}}$

Hence, the total surface area of the toy is 84.86 cm².