Question

$\huge\mathtt{Question }$
The dimensions of a metallic cuboid are 44 cm x 42 cm x 21 cm. It is melted and recast into a sphere. Find the surface area of the sphere.

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Answer 1

Answer:

5544 cm² is the required surface area of sphere .

Step-by-step explanation:

According to the Question

It is given that dimensions of metallic cuboid

• Length ,l = 44 cm
• Breadth ,b = 42cm
• Height ,h = 21 cm

The metallic cuboid are melted and recast into a sphere . So, the volume is constant here.

Firstly we calculate the radius of sphere

• Volume of Cuboid = Volume of Sphere

Substitute the value we get

➻ 44 × 42 × 21 = 4/3 *π *r³

➻ 44×42×21 × 3/4 = 22/7 * r³

➻ 44×42×21×3×7 /4×22 = r³

➻ 21×21×21 = r³

➻ r = 21 cm

So ,the radius of the sphere is 21 cm

Now, calculating the surface area of the sphere.

• Surface Area of Sphere = 4πr²

substitute the value we get

➻ Surface Area of Sphere = 4×22/7× 21×21

➻ Surface Area of Sphere = 88×3×21

➻ Surface Area of Sphere = 5544cm²

• Hence, the surface area of the sphere is 5544cm².

Answer 2

$\huge\mathcal\fcolorbox{orange}{black}{Answer}$

Step-by-step explanation:

The volume of cuboid = l × b × h

= (44 × 42 × 21) cm³

= 38808 cm³

Since the metallic cuboid is melted and into a sphere,

volume of sphere = volume of cuboid

Therefore, volume of sphere = 38808 cm³

$but \: volume \: of \: sphere \: = \frac{4}{3}\pi \: {r}^{3} = 38808$

Therefore,

${r}^{3} = 38808 \: \times \frac{3}{4} \: \times \frac{7}{22}$

${r}^{3} = 21 \times 21 \times 21$

$r = 21cm$

$surface \: area \: of \: sphere = 4\pi \: {r}^{2}$

$= 4 \times \frac{22}{7} \times {21}^{2}$

$= 5544 {cm}^{2}$

Therefore, the surface area of the sphere is 5544 cm²