Question

⭐ A right circular cylinder of height 10 cm and radius 4.5 cm is formed by melting some coins of thickness 0.2 cm and diameter 1.5 cm. Find the number of coins required ?

⭐ Name 5 shapes with their areas ?​

Answer 1

Answer :

• 450

Given :

• Height, radius, thickness and diameter.

To Find :

• the number of coins required.

Solution :

$\small\rm Diameter \:of\: a \:coins\: (d_1)=1.5 cm \\ \small\rm \: \: \: Radius\: of\: coin\: (r_1)=\dfrac{1.5}{2}=0.75 cm \\ \small \rm \: Thickness\: or\: length \:(h_1)= 0.2 cm$

$\therefore \sf Volume \: of \: coin = \pi r_1^2h \\ \\\sf = \pi \times (0.75)^2 x 0.2 cu.cm \\ \\ \sf Again \: diameter \: of \: cylinder (d_2) =4.5 cm \\ \\\sf Radius \: of \: cylinder (r_2)= \frac{4.5}{2}=2.25 cm \\ \\ \sf Height \: of \: cylinder (h_2) = 10 cm \\ \\\therefore \sf Volume \: of \: cylinder \: = \pi r_2^2h_2 \\ \\ \sf = \pi \times (2.25)^2 \times10 cu.cm \\ \\\therefore \sf Number \: of \: coin \: required = \tiny \dfrac{volume \: of \: cylinder}{Volume \: of \: coin} \\ \\=\dfrac{ \pi \times r_1^2 \times h_1}{ \pi \times r_2^2 \times h_2}=\dfrac{(2.25)^2 \times 10}{(0.75)^2 \times 0.2} \\ \\ = \dfrac{(2.25) \times(2.25) \times10}{(0.75) \times(0.75) \times0.2} \\ \\ = 3 \times 3 \times 50 \\ \\ \bf= 450$

5 shapes with their areas :

$\large \boxed{ \bf {Mensuration} } \\ \small \boxed {\begin{array}{r | l} \sf Rectangle & length \times \: breadth \\ &= (l \times b ) \\& \\ \sf Square & side \times side \\ & \: = (side) {}^{2} \\ & = a \times a = {a}^{2} \\ \\ \sf Triangle & \frac{1}{2} \times base \times height \\ & = \frac{1}{2} \times b \times h \\ \\ \sf Circle &\pi \times (radius) {}^{2} \\ & = \pi \times {r}^{2} \\ \\ \sf Rhombus& \frac{1}{2} \times (product \: of \: the \: diagonals) \\& = \frac{1}{2} \times d_1d_2 \\ \end{array}}$