Question

the length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand from 6.00pm to 6.10pm.

Answer 1

answer=25.6cm²(approx.)

length of the min hand of the clock(r)=7cm

to find:-area swept by the min hand

sol:-angle=360/60=6°

angle made in 10 min=10×6=60°

thus, area swept by the min hand=$\pi$×r²×angle/360

>>>22/7×(7)²×60/360

>>>154/6

>>>25.6 cm²

Answer 2

Answer:

$2.564 cm^2$

Step-by-step explanation:

Length of minute hand = Radius = 7 cm

Minute hand moves from 6.00pm to 6.10pm.

It covers 10 minutes .

60 minutes= 360°

1 minute = $\frac{360}{60}$

10 minutes = $\frac{360}{60}\times 10$

                  = $60^{\circ}$

So, $\theta = 60 ^{\circ}$

Now Area of sector = $\frac{\theta}{360^{\circ}} \pi r^{2}$

So, area swept by minute hand  = $\frac{60^{\circ}}{360^{\circ}} \pi (7)^{2}$

                                                      = $\frac{1}{60} \times 3.14 \times49$

                                                      = $2.564 cm^2$

Hence the area swept by minute hand from 6.00pm to 6.10pm is $2.564 cm^2$