Surface Areas and Volumes Test

Result

Surface Areas and Volumes Test - 15

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Find the volume of a frustum cone, whose base and upper area of a circle is 6 and 6 $$cm^2$$. The height of the cone is 100 cm.

A
600 $$cm^3$$

B
2,800 $$cm^3$$

C
3,800 $$cm^3$$

D
4,800 $$cm^3$$

Solution

Volume of a frustum cone $$= \frac{1}{3} (A_1 + A_2 + \sqrt{A_1 A_2 }) h$$
$$= \frac{1}{3} (6 + 6 + \sqrt{6 \times 6}) \times 100$$
$$=\frac{1}{3} (12 + \sqrt{36}) \times 100$$
$$= \frac{1}{3} (12+6) \times 100$$
$$=\frac{1}{3} \times 18 \times 100$$
$$= \frac{1,800}{3}$$
$$= 600 cm^3$$
Question 2
1 / -0

The curved surface area of frustum cone is 612 $$mm^2$$. The diameter of a top cone is 10 and the radius of the bottom cone is 4 mm. Find the slant height. (Use $$\pi$$ = 3).

A
17.5 mm

B
22.66 mm

C
19.5 mm

D
10.5 mm

Solution

Curved surface area = $$\pi$$(R + r)s
Diameter = radius $$\times 2$$
Radius, R = 5 mm, r = 4 mm
612 = $$3 \times (5 + 4) \times s$$
612 = $$3 \times (9) \times s$$
$$612/3 \times 9 $$ = s
Slant Height = 22.66 mm
Question 3
1 / -0

Find the volume of a frustum cone, whose base and upper area of a circle is 20 and 80 $$m^2$$. The height of the cone is 45.5 m.

A
6,270 $$m^3$$

B
6,170 $$m^3$$

C
2,123.33 $$m^3$$

D
6,470 $$m^3$$

Solution

Volume of a frustum cone = $$\frac{1}{3} (A_1 + A_2 + \sqrt{A_1 A_2 }) h$$
$$= \frac{1}{3} (20 + 80 + \sqrt{20 \times 80 }) \times 45.5$$
$$= \frac{1}{3} (100 + \sqrt{1600 }) \times 45.5$$
$$= \frac{1}{3} (100 + 40) \times 45.5$$
$$= \frac{1}{3} \times 140 \times 45.5$$
$$=$$ 2,123.33 $$m^3$$
Question 4
1 / -0

The base and top radius of a cone is 36 cm and 16 cm respectively. The height of the cone is 12.6 cm. What is the volume of frustum of a cone? (Use $$\pi$$ = 3.14).

A
28,064.064 cm$$^3$$

B
27,064.064 cm$$^3$$

C
26,064.064 cm$$^3$$

D
25,064.064 cm$$^3$$

Solution

Volume of the frustum cone V = $$\displaystyle{\frac{\pi h}{3}}(R^2 + Rr + r^2)$$
$$R = 36 cm, r = 16 cm, h = 12.6 cm$$
$$V = (3.14 \times 12.6)/(3)[362 + 36 \times 16 + 162]$$
$$= 3.14 \times 4.2[1,296 + 576 + 256]$$
$$= 13.188[2,128]$$
$$= 28,064.064 cm^3$$
Question 5
1 / -0

Find the volume of the frustum cone whose base and top
radius is 21 in and 5 in respectively. The height of the cone is 300 in. (Use $$\pi$$ = 3).

A
181,300 in$$^3$$

B
171,300 in$$^3$$

C
161,300 in$$^3$$

D
151,300 in$$^3$$

Solution

Volume of the frustum cone is $$V = \displaystyle{\frac{\pi h}{3}}(R^2 + Rr + r^2)$$
$$R = 21 ft, r = 5 ft, h = 300 ft$$
$$V = (3 \times 300)/(3)[21^2 + 21 \times 5 + 5^2]$$
$$= 300[441 + 105 + 25]$$
$$= 300[571]$$
$$=$$ 171,300 in$$^3$$
Question 6
1 / -0

Find the volume of the frustum cone whose base and top radius is 12.4 ft and 4.5 ft respectively. The height of the cone is 1,200 ft. (Use $$\pi$$ = 3).

A
265,772 ft$$^3$$

B
255,772 ft$$^3$$

C
245,772 ft$$^3$$

D
275,772 ft$$^3$$

Solution

Volume of the frustum cone is $$V = \displaystyle{\frac{\pi h}{3}}(R^2 + Rr + r^2)$$
$$R = 12.4 ft, r = 4.5 ft, h = 1,200 ft$$
$$V = (3 \times 1,200)/(3)[12.4^2 + 12.4 \times 4.5 + 4.5^2]$$
$$= 1,200[153.76 + 55.8 + 20.25]$$
$$= 1,200[229.81]$$
$$=$$ 275,772 ft$$^3$$
Question 7
1 / -0

The total surface area of a frustum of cone is calculated as ________________.

A
SA = $$\pi(r + R)\sqrt{(R - r)^2 + h^2} + \pi r^2 + \pi R^2$$

B
SA = $$\pi(r - R)\sqrt{(R - r)^2 + h^2} + \pi r^2 + \pi R^2$$

C
SA = $$\pi(r + R)\sqrt{(R - r)^2 + h^2} + \pi r^2 - \pi R^2$$

D
SA = $$\pi(r + R)\sqrt{(R - r)^2 + h^2} - \pi r^2 + \pi R^2$$

Solution

Image 1
Let us assume the frustum is cut an open
Image 2
By ratio and proportion
$$\cfrac { { l }_{ 1 } }{ R } =\cfrac { l }{ R-r } \Rightarrow { l }_{ 1 }=\cfrac { Rl }{ R-r } $$
From figure,
$${ l }_{ 2 }={ l }_{ 1 }-l$$
$${ l }_{ 2 }=\cfrac { Rl }{ R-r } -l\Rightarrow \cfrac { Rl(R-r)l }{ R-r } $$
$${ l }_{ 2 }=\cfrac { rl }{ R-r } $$
The length of the arc is circumference of base
$${ S }_{ 1 }=2\pi r,{ S }_{ 2 }=2\pi r$$
From figure,
$$A=\cfrac { 1 }{ 2 } { S }_{ 1 }{ l }_{ 1 }-\cfrac { 1 }{ 2 } { S }_{ 2 }{ l }_{ 2 }$$$$=\cfrac { 1 }{ 2 } \left( 2\pi R \right) \left( \cfrac { Rl }{ R-r } \right) -\cfrac { 1 }{ 2 } \left( 2\pi r \right) \left( \cfrac { rl }{ R-r } \right) $$
$$\Rightarrow \cfrac { \pi { R }^{ 2 }l-\pi { r }^{ 2 }l }{ R-r } $$
$$A=\cfrac { \pi \left( { R }^{ 2 }-{ r }^{ 2 } \right) l }{ R-r } $$
$$A=\cfrac { \pi (R-r)\left( R+r \right) l }{ \left( R-r \right) } $$
$$A=\pi \left( R+r \right) l$$
Now, $$l=\sqrt { { \left( R-r \right) }^{ 2 }+{ h }^{ 2 } } $$
$$A=\pi \left( R+r \right) \sqrt { { \left( R-r \right) }^{ 2 }+{ h }^{ 2 } } $$ (Lateral surface area)
Total area
$$A=\pi \left( R+r \right) \sqrt { { \left( R-r \right) }^{ 2 }+{ h }^{ 2 } } +\pi { r }^{ 2 }+\pi { R }^{ 2 }$$
Question 8
1 / -0

Find the curved surface area of a frustum cone whose larger and smaller radius is 12.2 and 8.8 ft. The slant height is 5.2 ft. (Use $$\pi$$ = 3.14)

A
342.888 ft$$^2$$

B
348.75 ft$$^2$$

C
338.75 ft$$^2$$

D
328.75 ft$$^2$$

Solution

Curved surface area of a frustum cone = $$\pi$$(r + R)s
= 3.14 $$\times$$ (12.2 + 8.8) $$\times$$ 5.2
= 3.14 $$\times$$ (21) $$\times$$ 5.2
Curved surface area of a frustum cone = 342.888 ft$$^2$$
Question 9
1 / -0

A flower pot in the shape of a frustum with the top and bottom circles of radii $$20$$ in and $$10$$ in. Its depth is $$30$$ in. Find its volume.

A
$$7,100\pi\ {in}^2$$

B
$$7,000\pi\ {in}^3$$

C
$$7,0000\pi\ {in}$$

D
$$8,500\pi\ {in}^3$$

Solution

Volume of the frustum cone $$=\displaystyle \frac{\pi h}{3} [R^2 + Rr + r^2] $$

$$=\displaystyle \frac{\pi \times 30}{3} [{20}^2 + 20 \times 10 + {10}^2] $$

$$=10 \pi (400 + 200 + 100)$$
$$=7,000\pi\ in^3$$
Question 10
1 / -0

A flower pot in the shape of a frustum with the top and bottom circles of radii $$15$$ cm and $$10$$ cm. Its depth is $$36$$ cm. Find the surface area.

A
$$1,233.5$$ $$\pi cm^2$$

B
$$1,23.5$$ $$ cm^2$$

C
$$1,24.5$$ $$m^2$$

D
$$1,33.5$$ $$\pi cm^2$$

Solution

Surface area = $$ \pi (R + r) \sqrt{(R - r)^2 + h^2} + \pi R^2 + \pi r^2$$
= $$ \pi (15 + 10) \sqrt{(15 - 10)^2 + {36}^2} + \pi {10}^2 + \pi {15}^2$$
= $$25 \pi \sqrt{1,321} + 325 \pi $$
= $$25 \pi \times 36.34 + 325 \pi$$
= $$1,233.5 \pi cm^2$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Surface Areas and Volumes Test - 15

Result

Surface Areas and Volumes Test - 15

Score

-

Rank

-

SHARING IS CARING

Question 1

600 $$cm^3$$

2,800 $$cm^3$$

3,800 $$cm^3$$

4,800 $$cm^3$$

Solution

Question 2

17.5 mm

22.66 mm

19.5 mm

10.5 mm

Solution

Question 3

6,270 $$m^3$$

6,170 $$m^3$$

2,123.33 $$m^3$$

6,470 $$m^3$$

Solution

Question 4

28,064.064 cm$$^3$$

27,064.064 cm$$^3$$

26,064.064 cm$$^3$$

25,064.064 cm$$^3$$

Solution

Question 5

181,300 in$$^3$$

171,300 in$$^3$$

161,300 in$$^3$$

151,300 in$$^3$$

Solution

Question 6

265,772 ft$$^3$$

255,772 ft$$^3$$

245,772 ft$$^3$$

275,772 ft$$^3$$

Solution

Question 7

SA = $$\pi(r + R)\sqrt{(R - r)^2 + h^2} + \pi r^2 + \pi R^2$$

SA = $$\pi(r - R)\sqrt{(R - r)^2 + h^2} + \pi r^2 + \pi R^2$$

SA = $$\pi(r + R)\sqrt{(R - r)^2 + h^2} + \pi r^2 - \pi R^2$$

SA = $$\pi(r + R)\sqrt{(R - r)^2 + h^2} - \pi r^2 + \pi R^2$$

Solution

Question 8

342.888 ft$$^2$$

348.75 ft$$^2$$

338.75 ft$$^2$$

328.75 ft$$^2$$

Solution

Question 9

$$7,100\pi\ {in}^2$$

$$7,000\pi\ {in}^3$$

$$7,0000\pi\ {in}$$

$$8,500\pi\ {in}^3$$

Solution

Question 10

$$1,233.5$$ $$\pi cm^2$$

$$1,23.5$$ $$ cm^2$$

$$1,24.5$$ $$m^2$$

$$1,33.5$$ $$\pi cm^2$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.