In this chapter, we provide RD Sharma Class 6 ex 20.5 Solutions Chapter 20 Mensuration for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 6 ex 20.5 Solutions Chapter 20 Mensuration Maths pdf, Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 6 |

Subject | Maths |

Chapter | Chapter 20 |

Chapter Name | Mensuration |

Exercise | Ex 20.5 |

Table of Contents

**RD Sharma Solutions for Class 6 Chapter 20 Mensuration** **Ex 20.5 Download PDF**

**Chapter 20: Mensuration Exercise 20.5**

**Question: 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

**Solution:**

(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

**Question: 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

**Solution:**

(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

**Question: 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:**

(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

**Question: 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

**Solution:**

(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 × Side of the square

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

**Question: 5**

The length of a rectangle is three times of its width. If the length of the diagonal is 8√10 m, then the perimeter of the rectangle is

(a) 15√10 m

(b) 16√10 m

(c) 24√10 m

(d) 64 m

**Solution:**

(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

Now, AC is the diagonal of rectangle.

In right angled triangle ABC.

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

640 = 9x^{2} + x^{2}

640 = 10x^{2}

x^{2} = 640/10 = 64

x = 64 = 8 m

Thus, breadth of the rectangle = x = 8 m

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

Perimeter of the rectangle = 2 (Length + Breadth)

= 2 (24 + 8)

= 2 × 32 = 64 m

**Question: 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) √3:1

(c) √2:1

(d) 2√2:1

**Solution:**

(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}

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

(3x)^{2} = (x)^{2} + (y)^{2}

9x^{2} = x^{2} + y^{2}

8x^{2} = y^{2}

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

**Question: 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:**

(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

PR = 2 AC

Now, area of the square = (diagonal)^{2}/2

Area of PQRS = (PR)^{2})/2

Similarly, area of ABCD = (AC)^{2}/2

According to the question:

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

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

**Question: 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:**

(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 = x^{2} / y^{2}

⟹ x^{2}/y^{2 }= 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

**Question: 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

**Solution:**

(c) becomes one fourth

Explanation:

Let the side of the square be x.

Then, area = (Side × Side) = (x × x) = x^{2}

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

Now, new area = (x/2)^{2}

= (x^{2})/4

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

**Question: 10**

A rectangular carpet has area 120 m^{2} and perimeter 46 metres. The length of its diagonal is

(a) 15 m

(b) 16 m

(c) 17 m

(d) 20 m

**Solution:**

(c) 17 m

Explanation:

Area of the rectangle = 120 m^{2}

Perimeter = 46 m

Let the sides of the rectangle be l and b.

Therefore

Area = lb = 120 m^{2} …(1)

Perimeter = 2 (l + b) = 46

Or, (l + b) = 46 / 2 =23 m …(2)

Now, length of the diagonal of the rectangle = l^{2} + b^{2}

So, we first find the value of (l^{2} + b^{2})

Using identity:

(l^{2} + b^{2}) = (l + b)^{2} – 2 (lb) [From (1) and (2)]

Therefore

(l^{2} + b^{2}) = (23)^{2} – 2 (120)

= 529 – 240 = 289

Thus, length of the diagonal of the rectangle = l^{2} + b^{2} = 289 = 17 m

**Question: 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:**

(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.

**Question: 12**

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

(a) 10√2 cm

(b) 40 cm

(c) 40√2 cm

(d) 200 cm

**Solution:**

(c) 40√2 cm

Explanation:

Length of diagonal = 20 cm

**All Chapter RD Sharma Solutions For Class 6 Maths**

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