Question

solve q 6 and 7..........​

Answer 1

SOLUTION:6

Let r cm be the radius of the sphere.

Given, r= 4.2cm

Now,

We know that, volume of the sphere;

$= > \frac{4}{3} \pi {r}^{3} \\ \\ = > v = ( \frac{4}{3} \times \pi \times (4.2) \times (4.2) \times (4.2)) {cm}^{3}$

Let R cm be the radius & h cm be the height at the cylinder.

So,

Volume= πr²h

=) (π × 6× 6 ×h)cm³

Therefore,

Sphere is recast into the shape of a cylinder. So, volume remains same

=) 4/3πr³ = πr² h

$= > \frac{4}{3} \times \pi \times 4.2 \times 4.2 \times 4.2 = \pi \times 6 \times 6 \times h \\ \\ = > h = \frac{4 \times 4.2 \times 4.2 \times 4.2}{6 \times 6 \times 3} \\ \\ = > h = \frac{296.35}{108} \\ \\ = > h = 2.74cm$

SOLUTION:7

Here,

r= 21cm

& theta= 60°

We know that area of circle= πr²

$= > ( \frac{22}{7} \times 21 \times 21) {cm}^{2} \\ \\ = >( 22 \times 3 \times 21) {cm}^{2} \\ \\ = > 1386 {cm}^{2}$

Area of ∆ (AOB);

$= > \frac{1}{2} {r}^{2} sin \theta \\ \\ = > \frac{1}{2} {r}^{2} sin60 \degree \\ = > (\frac{1}{2} \times 21 \times 21 \times \frac{ \sqrt{3} }{2} ) \\ \\ = > (\frac{21 \times 21 \times \sqrt{3} }{4} ) \\ \\ = > (\frac{441 \times 1.732}{4} ) \\ \\ = > \frac{763.81}{4} {cm}^{2} \\ \\ = > 190.95 {cm}^{2}$

Now,

Area of minor segment (ACBA);

=) Area of circle - area of ∆(AOB)

=) (1386 -190.95)cm²

=) 1195.05cm²

Hope it helps ☺️