This exercise contains objective type questions at the end covering topics from the entire chapter. The solutions contain brief explanations in order to make them easier for the students to understand from the exam point of view. Here, important formulas and topics are highlighted to help students grasp the concepts. The students can gain knowledge on how much they have understood across all concepts. To know more about the important concepts which are covered here, RD Sharma Solutions Class 6 Maths Chapter 20 Mensuration Objective Type Questions are available for free download.

## RD Sharma Solutions for Class 6 Chapter 20: Mensuration Objective Type Questions Download PDF

### Access RD Sharma Solutions for Class 6 Chapter 20: Mensuration Objective Type Questions

**Mark the correct alternative in each of the following:**

**1. The side of a rectangle are in the ratio 5: 4. If its perimeter is 72 cm, then its length is (a) 40 cm (b) 20 cm(c) 30 cm (d) 60 cm**

**Solution:**

The option (b) is the correct answer.

Consider the sides of the rectangle asÂ 5x*Â *and 4x.Â

We know that, perimeter of rectangle =Â 2 (Length + Breadth)

By substituting the values

72 = 2 (5x*Â *+ 4x)

On further calculation

72 = 2Â Ã—Â 9x

So we get

72 = 18x

By division

x*Â *=Â 72/18Â = 4

Hence, the length of the rectangle = 5x*Â *= 5Â Ã—Â 4 = 20 cm

**2. The cost of fencing a rectangular field 34 m long and 18 m wide at Rs 2.25 per metre is (a) Rs 243(b) Rs 234(c) Rs 240(d) Rs 334**

**Solution:**

The option (b) is the correct answer.

We must find the perimeter of the rectangle for fencing the field.

The dimensions of the rectangle are

Length = 34m

Breadth =Â 18m

We know that Perimeter = 2 (Length + Breadth)

By substituting the values

Perimeter of the rectangle = 2 (34 + 18) = 2Â Ã—Â 52 = 104 m

So the cost of fencing the field at the rate of Rs. 2.25 per meter = 104Â Ã—Â 2.25 = Rs. 234

**3. If the cost of fencing a rectangular field at Rs. 7.50 per metre is Rs. 600, and the length of the field is 24 m, then the breadth of the field is(a) 8 m(b) 18 m(c) 24 m(d) 16 m**

**Solution:**

The option (d) is the correct answer.

It is given thatÂ cost of fencing the rectangular field = Rs. 600

So the rate of fencing the field = Rs. 7.50 per m

We know that perimeter of the field =Â CostÂ ofÂ fencing/RateÂ ofÂ fencingÂ

By substituting the values

Perimeter of the field =Â 600/7.50Â = 80 m

Length of the field = 24 m

So we get breadth of the field =Â Perimeter/2-Â Length =Â 80/2-Â 24 = 16 m

**4. The cost of putting a fence around a square field at Rs 2.50 per metre is Rs 200. The length of each side of the field is(a) 80 m(b) 40 m(c) 20 m(d) None of these**

**Solution:**

The option (c) is the correct answer.

It is given that cost of fencing the square field = Rs. 200

So the rate of fencing the field = Rs. 2.50

We know that, perimeter of the square field = CostÂ ofÂ fencing/RateÂ ofÂ fencingÂ

By substituting the values

Perimeter of the square field =Â 200/2.50Â = 80 m

Perimeter of square = 4Â Ã—Â Side of the square

It can be written as

Side of the square =Â Perimeter/4Â =Â 80/4Â = 20 m

**5. The length of a rectangle is three times of its width. If the length of the diagonal isÂ 8âˆš10Â m, then the perimeter of the rectangle is (a)Â 15âˆš10Â m (b)Â 16âˆš10Â m(c)Â 24âˆš10Â m (d) 64 m**

**Solution:**

The option (d) is the correct answer.

Consider ABCD as a rectangle.

Assume that the width of the rectangle BC =Â xÂ m

We know that the length is three times width of the rectangle.

So, length of the rectangle AB = 3xÂ m

AC is the diagonal of rectangle

Consider ABC as a right angled triangle.

AC^{2}Â =Â AB^{2}Â +Â BC^{2}Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

By substituting the values

640 = 9x^{2}Â +Â x^{2}

We get

640 = 10x^{2}

On further calculation

x^{2}Â =Â 640/10Â =Â 64

xÂ = âˆš64Â = 8 m

So the breadth of the rectangleÂ x*Â *= 8 m

Length of the rectangle 3xÂ = 3Â Ã—Â 8 = 24 m

Perimeter = 2 (Length + Breadth)

By substituting the values

Perimeter = 2 (24 + 8) = 2Â Ã—Â 32 = 64 m

**6. If a diagonal of a rectangle is thrice its smaller side, then its length and breadth are in the ratio (a) 3: 1(b)Â âˆš3:Â 1(c)Â âˆš2Â :Â 1(d)Â 2âˆš2:Â 1**

**Solution:**

The option (d) is the correct answer.

Assume that the length of the smaller side of the rectangle BC =Â x

Length of the larger side AB =Â y

We know that the length of the diagonal is three times that of the smaller side.

Diagonal of the rectangle 3xÂ = AC

By using Pythagoras theorem

(AC) ^{2}Â = (AB) ^{2Â Â Â }+ (BC) ^{2}

By substituting the values

(3x) ^{2}Â = (x) ^{2}Â + (y) ^{2}

On further calculation

9x^{2}Â =Â x^{2}Â +Â y^{2}

We get

8x^{2}Â =Â y^{2}

By taking square roots of both sides,

2âˆš2Â x*Â *=Â y

Hence, the ratio of the larger side to the smaller side is 2âˆš2:Â 1.

**7. The ratio of the areas of two squares, one having its diagonal double than the other, is (a) 1: 2(b) 2: 3(c) 3: 1(d) 4: 1**

**Solution:**

The option (d) is the correct answer.

Consider ABCD and PQRS as the two squares. We know that, the diagonal of square PQRS is twice the diagonal of squareÂ ABCD.

PRÂ = 2 AC

Area of the square =Â Diagonal^{2}/2

Area of PQRS =Â PR^{2}/2Â

In the same way, area of ABCDÂ =Â AC^{2}/2

From the question, we know that:Â

If AC =Â xÂ units,Â we get PR = 2xÂ units

AreaÂ ofÂ PQRS/AreaÂ ofÂ ABCDÂ = (PR^{2}Ã—2)/ (2Ã—AC^{2})Â

By substituting the value

AreaÂ ofÂ PQRS/AreaÂ ofÂ ABCDÂ =Â [(2x)^{ 2}Ã—2]/ [2Ã—x^{2}]Â =Â 4/1

Hence, the ratio of the areas of squares PQRS and ABCD is 4: 1.

**8. If the ratio of areas of two squares is 225 : 256, then the ratio of their perimeters is(a) 225 : 256(b) 256 : 225(c) 15 : 16(d) 16 : 15**

**Solution:**

The option (c) is the correct answer.

ConsiderÂ ABCD and PQRS as the two squares.

Let the lengths of each side of ABCDÂ and PQRS beÂ xÂ andÂ y.

We know that

AreaÂ ofÂ sq.Â ABCD/AreaÂ ofÂ sq.Â PQRSÂ =Â x^{2}/y^{2}

So we get x^{2}/y^{2}=Â 225/256

By taking square roots on both sides,

x/yÂ =Â 15/16

By taking the ratio of their perimeters, we get

PerimeterÂ ofÂ sq.Â ABCD/PrimeterÂ ofÂ sq.Â PQRSÂ =Â (4Â Ã—Â sideÂ ofÂ sq.Â ABCD)/ (4Â Ã— sideÂ ofÂ sq.Â PQRS)Â =Â 4x/4y

By removing common terms in numerator and denominator

PerimeterÂ ofÂ sq.Â ABCD/PerimeterÂ ofÂ sq.Â PQRSÂ =Â x/ y

So perimeterÂ ofÂ sq.Â ABCD/PerimeterÂ ofÂ sq.Â PQRSÂ =Â 15/16

Hence, the ratio of their perimeters = 15: 16

**9. If the sides of a square are halved, the its area (a) remains same (b) becomes half(c) becomes one fourth (d) becomes double**

**Solution:**

The option (c) is the correct answer.

ConsiderÂ x as the side of the square.

We know that area of a square = SideÂ Ã—Â Side = xÂ Ã—Â x =Â x^{2}

If the sides are halved, we get new side =Â x/2

So the new areaÂ =Â (x/2)^{2}=Â x^{2}/4

From this we know that the area has become one fourth of its previous value.

**10. A rectangular carpet has area 120 m ^{2}Â and perimeter 46 metres. The length of its diagonal is (a) 15 m (b) 16 m(c) 17 m (d) 20 m**

**Solution:**

The option (c) is the correct answer.

It is given that area of the rectangle = 120 m^{2}

Perimeter of the rectangle = 46 m

ConsiderÂ lÂ andÂ b as the length and breadth.

Area of the rectangle =Â l Ã— bÂ = 120 m^{2}Â Â Â Â Â Â Â Â Â

Perimeter of the rectangle = 2 (l*Â *+Â b) = 46

So we get

(l*Â *+Â b) =Â 46/2Â = 23 mÂ Â Â Â Â Â Â

Length of the diagonal of the rectangle =Â âˆšl^{2}Â +Â b^{2}

It can be written as

(l^{2}Â +Â b^{2}) = (l*Â *+Â b) ^{2}Â – 2 (l Ã— b)Â Â Â Â Â Â Â Â Â Â Â Â

By substituting the values

(l^{2}Â +Â b^{2}) = (23)^{2}Â – 2 (120) = 529Â – 240 = 289

Hence, length of the diagonal of the rectangle =Â âˆšl^{2}Â +Â b^{2}Â =Â âˆš289Â = 17 m

**11. If the ratio between the length and the perimeter of a rectangular plot is 1: 3, then the ratio between the length and breadth of the plot is (a) 1: 2(b) 2: 1(c) 3: 2(d) 2: 3**

**Solution:**

The option (b) is the correct answer.

We know that LengthÂ ofÂ rectangle/PerimeterÂ ofÂ rectangleÂ =Â 1/3

So we get

l/ (2lÂ +Â 2b)Â =Â 1/3

By cross multiplication, we get:

3lÂ = 2l + 2b

On further calculation

l = 2b

We get

l/ b=Â 2/1

Hence, the ratio of the length and the breadth is 2: 1.

**12. If the length of the diagonal of a square is 20 cm, then its perimeter is (a)Â 10âˆš2cm (b) 40 cm(c)Â 40âˆš2cm (d) 200 cm**

**Solution:**

The option (c) is the correct answer.

It is given that length of the diagonal = 20 cm

So the length of the side of a square =Â LengthÂ ofÂ Diagonal/âˆš2Â =Â 20/âˆš2Â =Â (2 Ã— 10)/âˆš2

We get

Length of the side of a square =Â (âˆš2 Ã— âˆš2 Ã— 10)/âˆš2Â = 10âˆš2Â cm

Hence, perimeter of the square = 4Â Ã—Â Side = 4Â Ã—10âˆš2Â = 40âˆš2Â cm