# 10th class maths mensuration exercise 10.3 solutions

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## NCERT Solutions for Class 6 Maths Exercise 10.3 Chapter 10 Mensuration

NCERT Solutions For Class 6 Maths Chapter 10 Mensuration Exercise 10.3 is provided here. Click here to download Free PDFs of NCERT Solutions prepared by expert teachers at BYJU'S.

NCERT SolutionsNCERT Class 6NCERT 6 MathsChapter 10: Measurement And MensurationExercise 10.3

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## NCERT Solutions For Class 6 Maths Chapter 10 Mensuration Exercise 10.3

NCERT Solutions For Class 6 Maths Chapter 10 Mensuration Exercise 10.3 discusses the key topics including the important formulas and steps to be followed while determining the area of rectangle and square. We also have a set of examples which help students analyse the type of problems that would appear in the Class 6 exam. These NCERT Solutions Class 6 Maths help students improve their problem solving abilities which are important from the exam point of view.

Chapter 10 Measurement …

Exercise 10.3

## NCERT Solutions for Class 6 Chapter 10: Mensuration Exercise 10.3 Download PDF

### Access NCERT Solutions for Class 6 Chapter 10: Mensuration Exercise 10.3

**1. Find the area of the rectangles whose sides are:**

**(a) 3 cm and 4 cm**

**(b) 12 m and 21 m**

**(c) 2 km and 3 km**

**(d) 2 m and 70 cm**

**Solutions:**

We know that

Area of rectangle = Length × Breadth

(a) l = 3 cm and b = 4 cm

Area = l × b = 3 × 4

= 12 cm2

(b) l = 12 m and b = 21 m

Area = l × b = 12 × 21

= 252 m2

(c) l = 2 km and b = 3 km

Area = l × b = 2 × 3

= 6 km2

(d) l = 2 m and b = 70 cm = 0.70 m

Area = l × b = 2 × 0.70

= 1.40 m2

**2. Find the areas of the squares whose sides are:**

**(a) 10 cm**

**(b) 14 cm**

**(c) 5 m**

**Solutions:**

**(a)**Area of square = side2

= 102 = 100 cm2

(b) Area of square = side2

= 142 = 196 cm2

(c) Area of square = side2

= 52 =25 cm2

**3. The length and breadth of three rectangles are as given below:**

**(a) 9 m and 6 m**

**(b) 17 m and 3 m**

**(c) 4 m and 14 m**

**Which one has the largest area and which one has the smallest?**

**Solutions:**

**(a)**Area of rectangle = l × b

= 9 × 6 = 54 m2

(b) Area of rectangle = l × b

= 17 × 3 = 51 m2

(c) Area of rectangle = l × b

= 4 × 14 = 56 m2

Area of rectangle 56 m2 i.e (c) is the largest area and area of rectangle 51 m2 i.e (b) is the smallest area

**4. The area of a rectangular garden 50 m long is 300 sq m. Find the width of the garden.**

**Solutions:**

Area of rectangle = length × width

300 = 50 × width width = 300 / 50 width = 6 m

∴ The width of the garden is 6 m

**5. What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of ₹ 8 per hundred sq m.?**

**Solutions:**

Area of land = length × breadth

= 500 × 200 = 1,00,000 m2

∴ Cost of tiling 1,00,000 sq m of land = (8 × 1,00,000) / 100

= ₹ 8000

**6. A table top measures 2 m by 1 m 50 cm. What is its area in square metres?**

**Solutions:**

Given l = 2m

b = 1m 50 cm = 1.50 m

Area = l × b = 2 × 1.50

= 3 m2

**7. A room is 4 m long and 3 m 50 cm wide. Howe many square metres of carpet is needed to cover the floor of the room?**

**Solutions:**

Given l = 4m

b = 3 m 50 cm = 3.50 m

Area = l × b = 4 × 3.50

=14 m2

**8. A floor is 5 m long and 4 m wide. A square carpet of sides 3 m is laid on the floor. Find the area of the floor that is not carpeted.**

**Solutions:**

Area of floor = l × b = 5 × 4

= 20 m2

Area of square carpet = 3 × 3

= 9 m2

Area of floor that is not carpeted = 20 – 9

= 11 m2

∴ Area of the floor that is not carpeted is 11 m2

**9. Five square flower beds each of sides 1 m are dug on a piece of land 5 m long and 4 m wide. What is the area of the remaining part of the land?**

**Solutions:**

Area of flower square bed = 1 × 1

= 1 m2

Area of 5 square bed = 1 × 5

= 5 m2

Area of land = 5 × 4

= 20 m2

Remaining part of the land = Area of land – Area of 5 square bed

= 20 – 5 = 15 m2

∴ Remaining part of the land is 15 m2

**10. By splitting the following figures into rectangles, find their areas (The measures are given in centimetres).**

**Solutions:**

**(a)**

Area of yellow region = 3 × 3

= 9 cm2

Area of orange region = 1× 2

= 2 cm2

Area of grey region = 3 × 3

= 9 cm2

Area of brown region = 2 × 4

= 8 cm2

Total area = 9 + 2 + 9 + 8

= 28 cm2

∴ Total area is 28 cm2

(b)

## AP SSC 10th Class Maths Solutions Chapter 10 Mensuration Ex 10.3 – AP Board Solutions

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 10 Mensuration Ex 10.3 Textbook Questions and Answers. AP State Syllabus SSC 10th Class Maths Solutions 10th Lesson Mensuration Exercise 10.3 10th Class Maths

## AP SSC 10th Class Maths Solutions Chapter 10 Mensuration Ex 10.3

March 5, 2021 by Mahesh

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 10 Mensuration Ex 10.3 Textbook Questions and Answers.

## AP State Syllabus SSC 10th Class Maths Solutions 10th Lesson Mensuration Exercise 10.3

### 10th Class Maths 10th Lesson Mensuration Ex 10.3 Textbook Questions and Answers

Question 1.

An iron pillar consists of a Cylindrical portion of 2.8 m. height and 20 cm. in diameter and a cone of 42 cm. height surmounting it. Find the weight of the pillar if 1 cm3 of iron weighs 7.5 g.

Answer:

Volume of the iron pillar = Volume of the cylinder + Volume of the cone

Cylinder:

Radius = d 2 = 20 2 = 10 cm

Height = 2.8 m = 280 cm

Volume = πr2h = 22 7 × 10 × 10 × 280 = 88000 cm3 Cone: Radius ‘r’ = d 2 = 20 2 = 10 cm height ‘h’ = 42 cm Volume = 1 3 πr2h = 1 3 × 22 7 × 10 × 10 × 42 = 4400 cm3

∴ Total volume = 88000 + 4400 = 92400 cm3

∴ Total weight of the pillar at a weight of 7.5 g per 1 cm3 = 92400 × 7.5

= 693000 gms = 693000 1000 kg = 693 kg. Question 2.

A toy is made in the form of hemisphere surmounted by a right cone whose circular base is joined with the plane surface of the hemisphere. The radius of the base of the cone is 7 cm. and its volume is 3/2 of the hemisphere. Calculate the height of the cone and the surface area of the toy correct to 2 places of decimal.

(Take π = 3 1 7 ) Answer:

Given r = 7 cm and

Volume of the cone =

3 2

volume of the hemisphere

1 3 πr2h = 3 2 × 2 3 × πr3 ∴ h = 3r = 3 × 7 = 21 cm

Surface area of the toy = C.S.A. of the cone + C.S.A. of hemisphere

Cone: Radius (r) = 7 cm Height (h) = 21 cm Slant height l = r 2 + h 2 − − − − − − √ = 7 2 + 21 2 − − − − − − − √ = 49+441 − − − − − − − √ = √490 = 22.135 cm. ∴ C.S.A. = πrl = 22 7

× 7 × 22.135 = 486.990 cm2

Hemisphere: Radius (r) = 7 cm C.S.A. = 2πr2 = 2 × 22 7 × 7 × 7 = 308 cm2

C.S.A. of the toy = 486.990 + 308 = 794.990 cm2

Question 3.

Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 7 cm.

Answer:

Radius of the cone with the largest volume that can be cut out from a cube of edge 7 cm =

7 2 cm

Height of the cone = edge of the cube = 7 cm

∴ Volume of the cone V =

1 3 πr2h = 1 3 × 22 7 × 7 2 × 7 2 × 7 = 89.83 cm3. Question 4.

A cylindrical tub of radius 5 cm and length 9.8 cm is full of water. A solid in the form of right circular cone mounted on a hemisphere is immersed into the tub. The radius of the hemi¬sphere is 3.5 cm and height of cone outside the hemisphere is 5 cm. Find the volume of water left in the tub. (Take π =

22 7 ) Answer:

The tub is in the shape of a cylinder, thus

Radius of the cylinder (r) = 5 cm

Length of the cylinder (h) = 9.8 cm

Volume of the cylinder (V) = πr2h

= 22 7 × 5 × 5 × 9.8

Volume of the tub = 770 cm3.

Radius of the hemisphere (r) = 3.5 cm

Volume of the hemisphere =

2 3 πr3 = 2 3 × 22 7 × 3.5 × 3.5 × 3.5 = 22×12.25 3 = 269.5 3

Radius of the cone (r) = 3.5 cm

Height of the Cone (h) = 5 cm

Volume of the cone V =

1 3 πr2h = 1 3 × 22 7 × 3.5 × 3.5 × 5 = 192.5 3

Volume of the solid = Volume of the hemisphere + Volume of the cone

= 269.5 3 + 192.5 3 = 462 3 = 154 cm3

Now, when the solid is immersed in the tub, it replaces the water whose volume is equal to volume of the solid itself.

Thus the volume of the water replaced = 154 cm3.

∴ Volume of the water left in the tub = Volume of the tub – Volume of the solid = 770 – 154 = 616 cm3.

Question 5.

In the adjacent figure, the height of a solid cylinder is 10 cm and diameter 7 cm. Two equal conical holes of radius 3 cm and height 4 cm are cut off as shown in the figure. Find the volume of the remaining solid.

## Telangana SCERT Class 10 Maths Solution Chapter 10 Mensuration Exercise 10.3

Telangana SCERT Class 10 Maths Chapter 10 Mensuration Exercise 10.3 Maths Problems and Solution Here in this Post. Telangana SCERT Class 10 Maths Solution Chapter 10 Mensuration Exercise 10.3 .

## Telangana SCERT Class 10 Maths Solution Chapter 10 Mensuration Exercise 10.3

Mihir July 2, 2021

Telangana SCERT Solution

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Advertisement **Telangana SCERT Solution Class X (10) Maths Chapter 10 Mensuration Exercise 10.3**

**Exercise 10.3 **

**1) An Iron Pillar consists of a cylindrical Portion of 2.8m height and 20cm in diameter and a cone of 42cm height surmounting it. Find the weight of the Pillar if 1cm3 of iron weights 7.5 kg**

**Ans:**

Cone: Diameter d = 20cm

Radius = r = d/2= 20/2 = 10cm

Height h = 42 cm

Volume V = 1/3 π r2h

= 1/3 x 22/7 x 10 x 10 x 42

= 1/3 x 22 x 10 x 10 x 6

= 22 x 10 x 10 x 2 = 22 x 100 x 2 = 22 x 200 = 4400cm3 Cylinder: Radius r = 10cm Height h = 2.8 cm = 2.8 x 100 h = 280 cm Volume V = π r2h

= 22/7 x 10 x 10 x 280

= 22 x 10 x 10 x 40 = 22 x 100 x 40 = 22 x 4000 = 88000 cm3

Volume of the Iron Pillar = Volume of Cone + Volume of Cylinder

= 4400cm3 + 88000 cm3

= 92400 cm3

Weight of the iron Pillar

1cm3 …..> 7.5 gr 92400 cm3……..> 7

92400 x 7.5 = 693000 gr

= 693000/1000 = 693 kg

**2) A toy is made in the form of hemisphere Surmounted by a right cone whose circular base is joined with the plane surface of the hemisphere. The radius of the base of the cone is 3/2 of the hemisphere. Calculate the height of the cone and the surface of the area of the toy correct to 2 places of decimal**

**Ans:**

hemisphere Radius r = 7cm Volume = 2/3 π r3

= 2/3 x π x 7 x 7 x 7cm3

Cone: Radius r = 7cm Height = h

Volume V = 1/3 π r2h

= 1/3 π x 7 x 7 x h

**Accordingly to the Problem**

Volume of Cone = 3/2 Volume of Hemisphere

1/3 π x 7 x7 x h = 3/2 x 2/3 x π x 7 x 7 x 7

h = 7 x 3 h = 21 cm

Surface Area of the toy = CSA of Hemisphere + CSA of Cone

= 2 π r2 + π r l

= 2 x 22/7 x 7 x 7 + 22/7 x 7 x √r2 + h2

= 2 x 22/7 x 7 x 7 + 22/7 x 7 x √(7)2 + (21)2

= 2 x 22 x 7 + 22 x √ 49 + 441

= 308 + 22 x √ 490

= 308 + 22 x √ 49 x 10

= 308 + 22 x 7 √10

= 308 + 154 + (3.16)

= 306 + 48664 = 794.64cm2

**3) Find the Volume of the largest right circular cone that can be cut out of a cube whose edge is 7cm**

Ans:

Diameter d =7cm Radius r = 7/2 cm Height h = 7 cm Volume = 1/3 π r2 h

= 1/3 x 22/7 x 7/2 x 7/2 x 7

= 11 X 49 / 6 = 539 / 6 = 89.83 cm3

Therefore, Volume of the cone is 89.83

**4)**

**A Cylinder tub of height is 9.8cm is fall of water. A solid in the form of right circular cone mounted on a hemisphere is immersed into the mug. The radius of the hemisphere is 3.5cm and height of Conical part 5cm. Find the Volume of Water left in the Tub**

Ans:

Cylindrical Tubs Radius r = 5cm height (h) 9.8 cm Volume v = π r2 h

= 22/7 x 5 x 5 x 9.8

= 22 x 5 x 5 x 98/10

= 22 x 5 x 5 x 14/10

= 22 x 5 x 14/2 = 770 cm3

Volume of Hemisphere + Volume of Cone

= 2/3 π r3 + 1/3 π r2 h

= 1/3 π r2 (2r + h)

= 1/3 x 22/7 x 3.5 x 3.5 x (7+5)

= 1/3 x 22/7 x 3.5 x 3.5 x 12

= 22/7 x 3.5 x 3.5 x 4

= 22 x 0.5 x 3.5 x 4

= 154 cm3

Volume of the Water left in the tub = Volume of the Cylinder – Volume of Solid

= 770 cm3 – 154 cm3 = 616 cm3

**5) In the adjacent figure, the height of a solid Cylinder is 10 cm and Diameter is 7cm . Two Equal conical holes of radius 3cm and height 4cm are cut off as shown the figure. Find the volume of the remaining solid**

Ans:

Cylindrical Solid’s diameter d = 7cm

radius r = d/2 = 7/2 cm

height h = 10cm volume v = π r2 h

= 22/7 x 7/2 x 7/2 x 10

= 11 x 7/2 x 10 = 11 x 7 x 5 = 385cm3 In Cone’s Radius r = 3cm Height h = 4cm

Volume V = 1/3 π r2 h

= 1/3 x 22/7 x 3 x 3 x 4

= 22/7 x 3 x 4 = 264/7 = 37.71 cm3

If two equal conical holes are cut off then volume of the remaining solid

= Volume of the Cylinder – 2 x volume of the cable

= 385 – 2(37.71) = 385 – 75.42 = 309.58 cm3

**6) Spherical Marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7cm, which contains some water, Find the number of marbles that Should be dropped into the beaker, so that water level rises by 5.6 cm**

**Ans:**

In Spherical marbles,

Guys, does anyone know the answer?