Question

A circular metallic plate of radius 8cm and thickness 6mm is melted and molded into a pie of radius 16cm and thickness 4mm find the angle of the sector

Answer 1

We will first have to find the volume of the circular metallic plate. It can be found by multiplying the area of the the circle with it's thickness as:

$V_1=\pi\times radius^2\times thickness$

Now, radius = 8 cm= 0.08 m

thickness=6 mm=0.006 m

Therefore, volume, $V_1=\pi\times 0.08^2\times0.006=1.206\times 10^{-4}$ $m^3$

Now, the volume of the remolded pie will be the product of the area of the pie and the thickness of the pie. Now the area of the pie is given as:

$Area=\frac{\theta}{360^{\circ}} \times \pi \times r^2$ (where the symbols have usual meanings)

The thickness of the pie is given as: thickness=4 mm=0.004 m

Therefore, the volume of the remoulded pie will be:

$\frac{\theta^{\circ}}{360^{\circ}}\times \pi \times (0.16)^2\times 0.004=1.2064\times 10^{-4}$

$\therefore \theta\times 8.936\times 10^{-7}=1.2064\times 10^{-4}$

$\therefore \theta=\frac{1.2064\times 10^{-4}}{8.936\times 10^{-7} } \approx 135^{\circ}$

Thus, the angle of the sector is $135^{\circ}$.

Answer 2

HELLO DEAR,

the volume of the circular metallic plate. It can be found by multiplying the area of the the

circle with it's thickness as:

$\boxed{\bold{V = \pi r^2 \times thickness}}$

Now, radius = 8 cm

thickness = 6 mm = 6/10cm

Therefore, volume,

$\boxed{ = \bold{\pi (8)^2 * \frac{6}{10}}}$

Now, the volume of the remolded pie will be the product of the area of the pie and the thickness of the pie.

Now the area of the pie is

$\boxed{\bold{AREA = \frac{\Theta}{360}\pi r^2}}$

radius = 16cm

thickness = 4 mm = (4/10)m

$\boxed{\bold{AREA = \frac{\Theta}{360}\pi (16)^2 \frac{4}{10}}}$

NOW, the volume of the circular metallic plate = the volume of the remolded pie

so, $\bold{\frac{\Theta \times \pi \times 4 \times \times (16)^2}{360\times 10} = \frac{\pi \times (8)^2 \times 6}{10}}$

$\bold{\Rightarrow \Theta \times 4 \times 16 \times 16 \times \pi = 360 \times 8 \times 8 \times 6 \times \pi }$

$\bold{\Rightarrow \Theta = \frac{360 \times 8 \times 8 \times 6 \times \pi}{4 \times 16 \times 16 \times \pi }}$

$\bold{\Rightarrow \Theta = \frac{360 \times 6}{4 \times 2 \times 2}}$

$\bold{\Rightarrow \Theta = \frac{2160}{16}}$

$\bold{\Rightarrow \Theta = 135\degree}$

HENCE, the angle of the sector is 135°

I HOPE ITS HELP YOU DEAR,

THANKS