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

Volume = Area of cross section × height

A circular metallic plate of some thickness is a cylinder

Area of a sector

$[tex] \frac{\theta}{360}\pi r^{2}=\pi r^{2} h$$\frac{\theta}{360}\pi r^{2}=\pi r^{2} h$} [/tex]

Volume of cylinder

$\pi r^{2}h$

Now both the volumes must be equal

So we have

$\frac{\theta}{360}\pi 16^2(\frac{4}{10})=\pi 8^2(\frac{6}{10})$

Solving we get

$\frac{1024\theta}{3600} \pi =\frac{384}{10} \pi$

Solving for theta

we get

$\theta=\frac{384(3600)}{1024(10)} =135$

Hence the angle of the sector is 135°.

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