Question

a sheet of metal has dimensions 56cm by 33 cm . It is melted down and recast into discs of the same thickness and radius 7 cm . how many discs will be cast​

Answer 1

Answer:

12 discs

Step-by-step explanation:

Area of sheet:-

56 x 33 = 1848cm²

Area of disc:-

$\frac{22}{7}7^2$

= 22 x 7 = 154cm²

Number of discs:-

1848/154 = 12 discs

Answer 2

Answer:

• Total 12 discs will be cast

Step-by-step explanation:

Given,

• A sheet of metal has dimensions 56cm by 33 cm .
• It is melted down and recast into discs of the same thickness and radius 7 cm

To find:

• How many discs will be cast?

Solution:

As, here we are given a sheet of metal whose dimensions are 56 cm and 33cm and a disc of radius 7cm. So, we will Find area of both a sheet of metal and disc.

Area of Rectangular sheet = Length × Breadth

Where,

• Length = 56 cm
• Breadth = 33 cm

$\sf \mapsto56 \times 33$

$\sf \mapsto1848 \: {cm}^{2}$

• Area of metal sheet is 1848 cm²

Now,

• Area of disc = πr²

$\sf \mapsto \dfrac{22}{7} \times {(7)}^{2}$

$\sf \mapsto \dfrac{22} { \cancel{7}} \times \cancel{ 7 }\times 7$

$\sf \mapsto22 \times 7$

$\sf \mapsto154 \: {cm}^{2}$

• Area of disc is 154 cm²

We have get the required area of Metal sheet and discs:

Number of discs that will be make:

$\sf \mapsto \dfrac{1848}{154}$

$\sf\mapsto 12 \: discs$

Hence,

• Total 12 discs will be cast.