Class 6 Exam  >  Class 6 Notes  >  RD Sharma Solutions for Class 6 Mathematics  >  RD Sharma Solutions -Ex-20.5, Mensuration, Class 6, Maths

Ex-20.5, Mensuration, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics PDF Download

Mark the correct alternative in each of the following:

Q. 1. The sides 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

Answer: (b) 20 cm

Explanation:

Let the sides of the rectangle be 5x and 4x. (Since, they are in the ratio 5 : 4)

Now, perimeter of rectangle = 2 (Length + Breadth)

72 = 2 (5x + 4x)

72 = 2 × 9x

72 = 18x

x = 4

Thus, the length of the rectangle = 5x = 5 × 4 = 20 cm

 

Q. 2. The cost of fencing a rectangular field 34 m long and 18 m wide at As 2.25 per metre is

(a) Rs 243 (b) Rs 234 (c) Rs 240 (d) Rs 334

Answer: (b) Rs. 234

Explanation:

For fencing the rectangular field, we need to find the perimeter of the rectangle.

Length of the rectangle = 34 m

Breadth of the rectangle = 18 m

Perimeter of the rectangle = 2 (Length + Breadth) = 2 (34 + 18) m = 2 × 52 m = 104 m

Cost of fencing the field at the rate of Rs. 2.25 per meter = Rs. 104 × 2.25 = Rs. 234

 

Q. 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

Answer: (d) 16 m

Explanation:

Cost of fencing the rectangular field = Rs. 600

Rate of fencing the field = Rs. 7.50 per m

Therefore, perimeter of the field = Cost of fencing / Rate of fencing = 600 / 7.50 = 80 m

Now, length of the field = 24 m

Therefore, breadth of the field = Perimeter / 2 - Length = 80 / 2- 24 = 16 m

 

Q. 4. The cost of putting a fence around a square field at As 2.50 per metre is As 200. The length of each side of the field is

(a) 80 m (b) 40 m (c) 20 m (d) None of these

Answer: (c) 20 m

Explanation:

Cost of fencing the square field = Rs. 200

Rate of fencing the field = Rs. 2.50

Now, perimeter of the square field = Cost of fencing / Rate of fencing = 200 / 2.50 = 80 m

Perimeter of square = 4 x Side of the square

Therefore, side of the square = Perimeter / 4 = 80 / 4 = 20 m

 

Q. 5. The length of a rectangle is three times of its width. If the length of the diagonal! is Ex-20.5, Mensuration, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics, then the perimeter of the rectangle is

(a) Ex-20.5, Mensuration, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics (b) Ex-20.5, Mensuration, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics (c) Ex-20.5, Mensuration, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics (d) 64 m

Answer: (d) 64 m

Explanation:

Let us consider a rectangle ABCD.

Also, let us assume that the width of the rectangle, i.e., BC be x m.

It is given that the length is three times width of the rectangle.

Therefore, length of the rectangle, i.e., AB = 3x m

Ex-20.5, Mensuration, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

Now, AC is the diagonal of rectangle.

In right angled triangle ABC.

AC2 = AB2 + BC2

Ex-20.5, Mensuration, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics = (3x)2 + x2

640 = 9x2 + x2

640 = 10x2

x2 = 640 / 10 = 64

x = 64 = 8 m

Thus, breadth of the rectangle = x = 8 m

Similarly, length of the rectangle = 3x = 3 x 8 = 24 m

Perimeter of the rectangle = 2 (Length + Breadth)

= 2 (24 + 8)

= 2 x 32 = 64 m

 

Q. 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) Ex-20.5, Mensuration, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics (c) Ex-20.5, Mensuration, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics (d) Ex-20.5, Mensuration, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

Answer: (d) 22 : 1

Explanation:

Let us assume that the length of the smaller side of the rectangle, i.e., BC be x and length of the larger side , i.e., AB be y.

It is given that the length of the diagonal is three times that of the smaller side.

Therefore, diagonal = 3x = AC

Now, applying Pythagoras theorem, we get:

(Diagonal)2 = (Smaller side)2 + (Larger side)2

Ex-20.5, Mensuration, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

(AC)2 = (AB)2 + (BC)2

(3x)2 = (x)2 + (y)2

9x2 = x2 + y2

8x2 = y2

Now, taking square roots of both sides, we get:

22 x = y

or, y / x = 22 / 1

Thus, the ratio of the larger side to the smaller side = 22 : 1

 

Q. 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

Answer: (d) 4 : 1

Explanation:

Let the two squares be ABCD and PQRS. Further, the diagonal of square PQRS is twice the diagonal of square ABCD

Ex-20.5, Mensuration, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 MathematicsEx-20.5, Mensuration, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

PR = 2 AC

Now, area of the square = Ex-20.5, Mensuration, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

Area of PQRS = Ex-20.5, Mensuration, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

Similarly, area of ABCD = Ex-20.5, Mensuration, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

According to the question:

If AC = x units, then, PR = 2x units

Therefore, Area of PQRS / Area of ABCD = Ex-20.5, Mensuration, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 MathematicsEx-20.5, Mensuration, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

Thus, the ratio of the areas of squares PQRS and ABCD = 4 : 1

 

Q. 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

Answer: (c) 15 : 16

Explanation:

Let the two squares be ABCD and PQRS.

Further, let the lengths of each side of ABCD and PQRS be x and y, respectively.

Therefore Area of sq. ABCD / Area of sq. PQRS = x2 / y2

=> x2 / y2= 225 / 256

Taking square roots on both sides, we get:

x / y = 15 / 16

Now, the ratio of their perimeters:

Perimeter of sq. ABCD / Perimeter of sq. PQRS

= 4 × side of sq. ABCD / 4 × Side of sq. PQRS = 4x / 4y

Perimeter of sq. ABCD / Perimeter of sq. PQRS = x y

Perimeter of sq. ABCD / Perimeter of sq. PQRS = 15 / 16

Thus, the ratio of their perimeters = 15 : 16

 

Q. 9. If the sides of a square are halved, then its area

(a) remains same (b) becomes half (c) becomes one fourth

(d) becomes double

Answer: (c) becomes one fourth

Explanation:

Let the side of the square be x.

Then, area = (Side x Side) = (x × x) = x2

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

Now, new area = Ex-20.5, Mensuration, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

Ex-20.5, Mensuration, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

It is clearly visible that the area has become one-fourth of its previous value.

 

Q. 10. A rectangular carpet has area 120 m2 and perimeter 46 metres. The length of its diagonal is

(a) 15 m (b) 16 m (c) 17 m (d) 20 m

Answer: (c) 17 m

Explanation:

Area of the rectangle = 120 m2

Perimeter = 46 m

Let the sides of the rectangle be l and b.

Therefore

Area = lb = 120 m2 ...(1)

Perimeter = 2 (l + b) = 46

Or, (l + b) = 46 / 2 =23 m ...(2)

Now, length of the diagonal of the rectangle = l2 + b2

So, we first find the value of (l2 + b2)

Using identity:

(l2 + b2) = (l + b)2 - 2 (lb) [From (1) and (2)]

Therefore

(l2 + b2) = (23)2 - 2 (120)

= 529 - 240 = 289

Thus, length of the diagonal of the rectangle = l2 + b2 = 289 = 17 m

 

Q. 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

Answer: (b) 2 : 1

Explanation:

It is given that Length of rectangle / Perimeter of rectangle = 1 / 3

=> l / (2l + 2b) = 1 / 3

After cross multiplying, we get:

3l = 2l + 2b

=> l = 2b

=> l / b= 2 / 1

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

 

Q. 12. If the length of the diagonal of a square is 20 cm, then its perimeter is

(a) Ex-20.5, Mensuration, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics (b) 40 cm (c) Ex-20.5, Mensuration, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics (d) 200 cm

Answer: (c) Ex-20.5, Mensuration, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

Explanation:

Length of diagonal = 20 cm

Length of side of a square = Ex-20.5, Mensuration, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

Ex-20.5, Mensuration, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

Ex-20.5, Mensuration, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

Therefore, perimeter of the square is 4 × Side = 4 × Ex-20.5, Mensuration, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

Ex-20.5, Mensuration, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

The document Ex-20.5, Mensuration, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics is a part of the Class 6 Course RD Sharma Solutions for Class 6 Mathematics.
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FAQs on Ex-20.5, Mensuration, Class 6, Maths RD Sharma Solutions - RD Sharma Solutions for Class 6 Mathematics

1. What are the different types of mensuration covered in the RD Sharma Solutions for Class 6 Maths?
Ans. The RD Sharma Solutions for Class 6 Maths covers various types of mensuration such as perimeter, area, volume, and surface area of different geometrical shapes like squares, rectangles, circles, triangles, and cubes.
2. How can RD Sharma Solutions for Class 6 Maths help in understanding mensuration concepts?
Ans. RD Sharma Solutions for Class 6 Maths provides step-by-step solutions to the problems related to mensuration. These solutions help students in understanding the various formulas and concepts used in mensuration. By practicing these solutions, students can gain a better understanding of the topic and improve their problem-solving skills.
3. Are RD Sharma Solutions for Class 6 Maths helpful for exam preparation?
Ans. Yes, RD Sharma Solutions for Class 6 Maths are highly helpful for exam preparation. These solutions cover all the important topics and provide comprehensive explanations for each problem. By practicing these solutions, students can not only strengthen their understanding of mensuration but also enhance their problem-solving abilities, which is crucial for scoring well in exams.
4. Can RD Sharma Solutions for Class 6 Maths be used for self-study?
Ans. Absolutely! RD Sharma Solutions for Class 6 Maths are designed to be easily understandable and self-explanatory. The solutions provide detailed step-by-step explanations, making it easier for students to study independently. By working through these solutions, students can effectively learn and revise the concepts of mensuration at their own pace.
5. Are RD Sharma Solutions for Class 6 Maths available in the same language as the article title?
Ans. Yes, RD Sharma Solutions for Class 6 Maths are available in the same language as the article title. These solutions are written in a language that is easy to understand for Class 6 students. The solutions are designed to cater to the needs of students who are studying in schools where the same language is used as the medium of instruction.
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