# Circumference and Arc Length Worksheet

Circumference and Arc Length Worksheet
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1.
The diameter of a cart wheel is 4.3 m. How much distance will it cover if it makes 10 revolutions?
 a. 49$\pi$ m b. 48$\pi$ m c. 43$\pi$ m d. 47$\pi$ m

#### Solution:

Diameter of the cart wheel is 4.3 m
[Given.]

Circumference of the wheel = 4.3π
[Circumference = πd.]

Distance covered for 10 revolutions = 10 × 4.3π = 43π m.
[Distance covered = circumference × no. of revolutions.]

2.
The radius of a bicycle wheel is 50 cm. How many revolutions does the wheel make when the bicycle travels 270 m? [Take $\pi$ = 3.]
 a. 81000 revolutions b. 180 revolutions c. 90 revolutions d. 360 revolutions

#### Solution:

Radius of the wheel is 50 cm
[Given.]

Diameter = 100 cm = 1 m

Circumference of the wheel = 1π
[Circumference = πd.]

No. of revolutions made in traveling a distance of 270 m = 2701π = 90
[Distancecircumference.]

3.
Find the circumference of a circle in which a right triangle of base ($b$) 13.5 cm and height ($h$) 18 cm is inscribed. [Take $\pi$ = 3.]

 a. 67.5 cm b. 135 cm c. 73.5 cm d. 61.5 cm

#### Solution:

Base of the right triangle = 13.5 cm.

Height of the right triangle = 18 cm.

Radius of the circle = 13.52+1822 = 11.25 cm.
[The radius of the circumcircle of a right triangle is half of the hypotenuse.]

Circumference of the circle = 2πr = 2 × π × 11.25 = 67.5 cm.

4.
Find the perimeter of the figure if $a$ = 1 cm, $b$ = 5 cm, $c$ = 4 cm.

 a. 12.85 cm b. 38.55 c. 33.85 cm d. 30 cm

#### Solution:

Perimeter of the figure = Length of the line segment AB + Length of arc BC + Length of line segment CD + length of arc DE + length of line segment EF + length of line segment FA

AB = 2 cm
[Given.]

Length of arc BC = 90360 × π × 2 = 1.57 cm
[Formula for the length of the arc = θ360 × π × radius.]

CD = 5 cm
[Given.]

Length of arc DE = π × 42 = 6.28 cm

Δ AEF is equilateral.
[AF = EF, mAFE = 60.]

AF = EF = AE = 1 + 5 + 4 = 10 cm
[AE = AC + CD + DE.]

Perimeter of the figure = 1 + 1.57 + 5 + 6.28 + 10 + 10 = 33.85 cm
[Step 1.]

5.
Select the correct statement / statements.
I. When a square is inscribed in a circle, the perimeter of the square is always less than the perimeter of the circle.
II. When a circle is inscribed in a square, the perimeter of the circle will be less than the perimeter of the square if the radius of the circle is greater than 2.
III. When a circle is inscribed in a square, the perimeter of the circle will be always less than the perimeter of the square.
 a. I only b. I, II, and III c. I, and II only d. I, and III only

#### Solution:

When the circle is inscribed in a square (figure A), perimeter of the square = 4 × 2r = 8r
[Side of the square = 2r.]

Circumference of the circle = 2 π r

2π < 8

So, in figure(A), the perimeter of the circle is always less than the perimeter of the square.

When the square is inscribed in the circle, side of the square = 2 × r
[Diagonal of the square = 2r, side = 2r2 = 2 × r.]

Perimeter of the square = 4 2 × r

Perimeter of the circle = 2πr

42 < 2 π

So, in figure(B), the Perimeter of the square is less than perimeter of the circle.

Statements I and III are correct.

6.
Arrange the arcs in the figure in the ascending order of its lengths.

 a. arc 3, arc 2, arc 1 b. arc 2, arc 3, arc 1 c. arc 3, arc 1, arc 2 d. arc 1, arc 2, arc 3

#### Solution:

Length of arc 1 = 30360 × 2π × 6 = π
[Formula for the length of arc = θ360 × 2π × r.]

Length of arc 2 = 90360 × 2π × 3 = 32 × π

Length of arc 3 = 120360 × 2π = 23 × π

The ascending order is arc 3, arc 1, arc 2.

7.
Semi circles are drawn on a line of length 7 m as shown. The first semicircle has a radius of 17 cm. Radius of each consecutive semicircle is half the radius of the preceding semi circle. What is the sum of the arc lengths of all the semicircular portions?

 a. 41.1$\pi$ m b. 7$\pi$ m c. Infinity d. $\frac{7}{2}$$\pi$ m

#### Solution:

The radius of the first semicircle is 17.

Radius of the second semi circle = 172

Radius of the third semicircle = 174

Circumference of the first semicircle = π × 17

Circumference of the second semicircle = π × 172

Circumference of the third semicircle = π × 174

Sum of the circumferences = π × 17 + π × 172 + π × 174 + ------------------

= π(17 + 172 + 174 + ---------------- )

= 2(17 + 172 + 174 + ---------------- ) = 7 m
[Given, the semicircles are drawn on a line of length 7 m.]

So, the sum of the arclengths of all the semi circular portions = π × 72 m = 7 / 2π m.
[Given, line length = 7 m.]

8.
Find the perimeter of the figure.
[Take $\alpha$ = 77o, $\beta$ = 165o, $a$ = 6 cm, $b$ = 5 cm.]

 a. 35.01 cm b. 3.16 cm c. 10.55 cm d. 6.33 cm

#### Solution:

Perimeter of the outer curve = 165360 × 2π × 5 = 14.39 cm

Perimeter of the inner curve = 77360 × 2π × 6 = 8.06 cm

Perimeter = 14.39 - 8.06 = 6.33 cm

9.
Three cyclists P, Q and R with the same speed starts from A to reach E as shown in the figure. P will follow the circular route AB-BC-CD-DE; Q will follow the route AC-CE and R will follow the route AE. Who will reach point E first? [Take $a$ = 2.]

 a. All will reach at the same time b. Q c. R d. P

#### Solution:

The distance covered by the cyclists are the circumference of the semi circles.

Distance covered by the cyclist R = Circumference of the semi circle with AE as diameter.

= 12×π×AE
[Circumference of semi circle with diameter d is 12×π×d.]

= 12×π×8 = 4π
[Substitute AE = 8.]

Distance covered by the cyclist Q = Circumference of semicircle with diameter AC + circumference of semi circle with diameter CE.

= 12×π×AC+12×π×CE

= 12×π×4+12×π×4=4π

Distance covered by the cyclist P = circumference of the semi circle with diameter AB + circumference of the semi circle with diameter BC + circumference of the semi circle with diameter CD + circumference of the semicircle with diameter DE.

= 1 / 2 × π × AB + 1 / 2 × π × BC + 1 / 2 × π × CD + 1 / 2 × π × DE

= 1 / 2 × π × 2 + 1 / 2 × π × 2 + 1 / 2 × π × 2 + 1 / 2 × π × 2 = 4π

The distances covered by the cyclists P, Q and R are same.

Therefore, three cyclists P, Q and R reach the point E in same time.

10.
Ebin bakes a christmas cake in a 30 cm diameter tin. He has a 100 cm length of decorative band to go around the outer line of the cake. Is this band long enough? [Take $\pi$ = 3.]

 a. yes b. cannot be determined c. exactly matches the required length d. no

#### Solution:

Radius of the cake = 30 / 2= 15 cm.
[Radius = 1 / 2 × diameter.]

Length of the decorative band is 100 cm.
[Given.]

Perimeter of the cake = 2π r.
[Perimeter of the circle = 2 π r.]

2 × 3 × 15 = 90 cm
[Substitute the values and simplify.]

The required length of the band is 90 cm, therefore the given band is sufficient to decorate the cake.