Question

A wall clock is equally divided into 12 sections. If the clock reads 8:00 and has a diameter of 12.5 inches, find the area of the smaller sector formed by the minute and hour hands.

Answer 1

Answer: Our required area would be 11.93 sq. inches.

Step-by-step explanation:

Since we have given that

Number of sections = 12

Angle formed in every 5 minutes is given by

$\dfrac{360}{12}=30^\circ$

Diameter = 12.5 inches

Radius = $\dfrac{12.5}{2}=6.25\ inches$

Area of the smaller sector formed by the minute and hour hands would be

$\dfrac{\theta}{360^\cir}\times \pi r^2\\\\=\dfrac{30}{360}\times \dfrac{22}{7}\times 6.75\times 6.75\\\\=11.93\ sq.\ inches$

Hence, our required area would be 11.93 sq. inches.

Answer 2

Answer:

The area of the sector is 81.8 square inches.

Step-by-step explanation:

The wall clock has handas that move 360°. So, if it's equally distributed then, we can divide

$\frac{360\°}{12}= 30\°$

Thich means that during each interval of hour, the hands move 30°.

So, if the wall clock is showing 8 o'clock, than one hand is pointing 12, which is gonna be the reference point 0°. The hour hand would be at number 8, which forms an angle of

$8(30\°)=240 \°$

Now, to find the circular sector area we used the following formula

$A= \frac{\theta}{360 \°} \times \pi r^{2}$

Where $\theta = 240 \°$ and $r=\frac{12.5 in}{2}= 6.25 in$, because the radius is defined as half the diameter.

Replacing all these values and $\pi =3.14$, we have

$A= \frac{240\°}{360 \°} \times (3.14) (6.25in)^{2}=81.8 in^{2}$

Therefore, the area of the smaller sector formed by the minute and hour hands is 81.8 square inches.