AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 10 Mensuration Ex 10.1 Textbook Questions and Answers.

## AP State Syllabus SSC 10th Class Maths Solutions 10th Lesson Mensuration Exercise 10.1

### 10th Class Maths 10th Lesson Mensuration Ex 10.1 Textbook Questions and Answers

Question 1.

A joker’s cap is in the form of right circular cone whose base radius is 7 cm and height is 24 cm. Find the area of the sheet required to make 10 such caps.

Answer:

Radius of the cap (r) = 7 cm

Height of the cap (h) = 24 cm

Slant height of the cap (l) = \(\sqrt{r^{2}+h^{2}}\)

= \(\sqrt{7^{2}+24^{2}}\)

= \(\sqrt{49+576}\)

= √625

= 25

∴ l = 25 cm.

Lateral surface area of the cap = Cone = πrl

L.S.A. = \(\frac{22}{7}\) × 7 × 25 = 550 cm^{2}.

∴ Area of the sheet required for 10 caps = 10 x 550 = 5500 cm^{2}.

Question 2.

A sports company was ordered to prepare 100 paper cylinders without caps for shuttle cocks. The required dimensions of the cylinder are 35 cm length / height and its radius is 7 cm. Find the required area of thin paper sheet needed to make 100 cylinders.

Answer:

Radius of the cylinder, r = 7 cm

Height of the cylinder, h = 35 cm

T.S.A. of the cylinder with lids at both ends = 2πr(r+h)

= 2 × \(\frac{22}{7}\) × 7 × (7 + 35)

= 2 × \(\frac{22}{7}\) × 7 × 42 = 1848 cm^{2}.

Area of thin paper required for 100 cylinders = 100 × 1848

= 184800 cm^{2}

= \(\frac{184800}{100 \times 100}\) m^{2}

= 18.48 m^{2}.

Question 3.

Find the volume of right circular cone with radius 6 cm. and height 7 cm.

Answer:

Base radius of the cone (r) = 6 cm.

Height of the cone (h) = 7 cm

Volume of the cone = \(\frac{1}{3}\)πr^{2}h

= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 6 × 6 × 7

= 264 c.c. (Cubic centimeters)

∴ Volume of the right circular cone = 264 c.c.

Question 4.

The lateral surface area of a cylinder is equal to the curved surface area of a cone. If their base be the same, find the ratio of the height of the cylinder to slant height of the cone.

Answer:

Base of cylinder and cone be the same.

CSA / LSA of cylinder = 2πrh

CSA of cone = πrl

The lateral surface area of a cylinder is equal to the curved surface area of cone.

∴ 2πrh = πrl

⇒ \(\frac{h}{l}=\frac{\pi r}{2 \pi r}\)

⇒ \(\frac{h}{l}\) = \(\frac{1}{2}\)

∴ h : l = 1 : 2

Question 5.

A self help group wants to manufacture joker’s caps (conical caps) of 3 cm radius and 4 cm height. If the available colour paper sheet is 1000 cm^{2}, then how many caps can be manufactured from that paper sheet?

Answer:

Radius of the cap (conical cap) (r) = 3 cm

Height of the cap (h) = 4 cm

Slant height l = \(\sqrt{r^{2}+h^{2}}\)

(by Pythagoras theorem)

= \(\sqrt{3^{2}+4^{2}}\)

= \(\sqrt{9+16}\)

= √25

= 5 cm

C.S.A. of the cap = πrl

= \(\frac{22}{7}\) × 3 × 5

≃ 47.14 cm^{2}

Number of caps that can be made out of 1000 cm^{2} = \(\frac{1000}{47.14}\) ≃ 21.27

∴ Number of caps = 21.

Question 6.

A cylinder and cone have bases of equal radii and are of equal heights. Show that their volumes are in the ratio of 3 : 1.

Answer:

Given dimensions are:

Cone:

Radius = r

Height = h

Volume (V) = \(\frac{1}{3}\)πr^{2}h

Cylinder:

Radius = r

Height = h

Volume (V) = πr^{2}h

Ratio of volumes of cylinder and cone = πr^{2}h : \(\frac{1}{3}\)πr^{2}h

= 1 : \(\frac{1}{3}\)

= 3 : 1

Hence, their volumes are in the ratio = 3 : 1.

Question 7.

A solid iron rod has cylindrical shape. Its height is 11 cm. and base diameter is 7 cm. Then find the total volume of 50 rods?

Answer:

Diameter of the cylinder (d) = 7 cm

Radius of the base (r) = \(\frac{7}{2}\) = 3.5 cm

Height of the cylinder (h) = 11 cm

Volume of the cylinder V = πr^{2}h

= \(\frac{22}{7}\) × 3.5 × 3.5 × 11 = 423.5 cm^{3}

∴ Total volume of 50 rods = 50 × 423.5 cm^{3} = 21175 cm^{3}.

Question 8.

A heap of rice is in the form of a cone of diameter 12 m. and height 8 m. Find its volume? How much canvas cloth is required to cover the heap? (Use π = 3.14)

Answer:

Diameter of the heap (conical) (d) = 12 cm

∴ Radius = \(\frac{d}{2}\) = \(\frac{12}{2}\) = 6 cm

Height of the cone (h) = 8 m

Volume of the cone, V = \(\frac{1}{3}\)πr^{2}h

= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 6 × 6 × 8

= 301.71 m^{3}.

Question 9.

The curved surface area of a cone is 4070 cm^{2} and its diameter is 70 cm. What is its slant height?

Answer:

C.S.A. of a cone = πrl = 4070 cm^{2}

Diameter of the cone (d) = 70 cm

Radius of the cone = r = \(\frac{d}{2}\) = \(\frac{70}{2}\) = 35 cm

Let its slant height be ‘l’.

By problem,

πrl = 4070 cm^{2}

\(\frac{22}{7}\) × 35 × l = 4070

110 l = 4070

l = \(\frac{4070}{110}\) = 37 cm

∴ Its slant height = 37 cm.