ML Aggarwal Mensuration Check Your Progress Class 6 ICSE Maths Solutions

ML Aggarwal Mensuration Check Your Progress Class 6 ICSE Maths Solutions. We Provide Step by Step Answer of  Check Your Progress Questions for Mensuration as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-6.

ML Aggarwal Mensuration Check Your Progress Class 6 ICSE Maths Solutions

Board ICSE
Publications Avichal Publishig Company (APC)
Subject Maths
Class 6th
Chapter-14 Mensuration
Writer ML Aggarwal
Book Name Understanding
Topics Solution of Check Your Progress Questions
Edition 2023-2024

Mensuration Check Your Progress

ML Aggarwal Class 6 ICSE Maths Solutions

Page-298

Question 1. The perimeter of a square ABCD is twice the perimeter of ∆PQR. Find the area of the square ABCD.

Question 1. The perimeter of a square ABCD is twice the perimeter of ∆PQR. Find the area of the square ABCD.

Answer:

Perimeter of ∆PQR = 6 cm + 5 cm + 7 cm = 18 cm
Perimeter of square = 2 × 18 = 36 cm
Side of square = ?
4 × side = 36
⇒ Side = 36/4 = 9
Area of square = (side)2 = (9)= 9 × 9 = 81 cm2

Question 2. The perimeter of an equilateral triangle is 42 cm. Find the perimeter of a square, each side of which is double the side of the triangle.

Answer:

Let the length of a side of the equilateral triangle and the square be l and L respectively.

Then, according to the given information, we have

Perimeter of equilateral triangle=42

⇒ l + l + l = 42

⇒ 3l = 42

⇒ l = 14.

So, the length of a side of the square will be

L = 2 × l = 2 × 14 = 28.

Therefore, the perimeter of the square is given by

Ps = 4L = 4 × 28 = 112

=4L = 4 × 28 = 112.

Thus, the required perimeter of the square is 112 cm.

Question 3. A wire of length 60 cm is cut into two pieces. One piece is used to form a rectangle of length 10 cm and width 8 cm. the other piece is bent into the shape of a regular hexagon. What is the length of each side of hexagon ?

Answer:

Length of wire = 60cm

After dividing it in to piece, one piece is used for making a rectangle.

Length of rectangle = 10cm, width = 8cm

Perimeter of rectangle = 2(L + B)

= 2(10+8)

= 2 x 18

= 36cm

Therefore, to make the rectangle, 36cm of wire is used of 60cm wire.

The remaining wire will be used to make a hexagon.

Therefore, Remaining wire = Total length – Used for rectangle

= 60 – 36

= 24

Therefore, 24cm wire will be used to make the hexagon.

To determine the side of hexagon,

Perimeter of hexagon = Remaining wire = 6 x side

24 = 6 x side

24 / 6 = side

4 = side

Therefore, if the string is to be bent to make a hexagon, the side should be 4cm.

Question 4. A wire is in the shape of a square of side 10 cm. If the wire is rebent into a rectangle of length 12 cm, find its breadth. Which encloses more area, the square or the rectangle and by how much?

Answer:

Side of square = 10 cm
Perimeter of square = 4 × 10 cm = 40 cm
According to question,
Perimeter of rectangular = 2(l + b)
⇒ 40 cm = 2 (12 cm + b)
⇒ 40 cm = 24 cm + 2b
⇒ 2b = 40 cm – 24 cm
⇒ 2b = 16 cm
⇒ b = 16/2 cm = 8 cm
Area of square = (a)= (10 cm)2 = 100 cm2
Area of rectangular = (l × b) = 12 × 8 = 96 cm2
Area of square is more and it is = 100 cm2 – 96 cm2 = 4 cm2

Question 5. A rectangular room is 9 m long and 6 m wide. Find the cost of covering the floor with carpet 2 m wide at ₹35 per metre.

Answer:

Length of room = 9m
Width of room = 6m
Area of room = l × b = 9 × 6 = 54 m2
Width of carpet = 2m
Area of carpet = 54 m2
Length = ?
Length of carpet = 54/2 = 27 m
Cost of covering = ₹35 × 27 m = ₹945

Question 6. If the cost of fencing a square plot at the rate of ₹ 2.50 per metre is ₹ 200, then find the length of each side of the field.

Answer:

Total cost of fencing a square plot = ₹200
Rate of fencing = ₹2.50
∴ Perimeter of square = Total cost/rate = 200/2.5 = 80m
Since, we know that,
Perimeter of square = 4a
⇒ 80 m = 4a
⇒ a = 20 m
∴ Length of a square plot = 20 m

Question 7. If the cost of fencing a rectangular park at the rate of ₹7.50 per metre is ₹600 and the length of the park is 24 m, find the breadth of the park.

Answer:

Cost of fencing a rectangular park = ₹600
Rate of fencing = ₹7.50
Perimeter of a park = Total cost/rate  = 600/7.5 = 80m
Length of the park = 24 cm
Let breadth of the park = b
∴ Perimeter of a square = 80 m
⇒ 2(l + b) = 80 m
⇒ 2(24 + b) = 80 m
⇒ 24 + b = 80 m
⇒ b = 40 – 24 m
∴ b = 16 m

Question 8. By splitting the following figures into rectangle, find their areas (The measures are given in centimetres).

Question 8. By splitting the following figures into rectangle, find their areas (The measures are given in centimetres).

Answer:

(a) Area of the figure,
= (3 × 1 + 3 × 1 + 3 × 1) sq m
= (3 + 3 + 3) sq m
= 9 sq m

Question 8. By splitting the following figures into rectangle, find their areas (The measures are given in centimetres).

(b) Area of the figure,
= (3 × 3 + 1 × 2 + 3 × 3 + 4 × 2) sq cm
= (9 + 2 + 9 + 8) sq cm
= 28 sq cm

Question 8. By splitting the following figures into rectangle, find their areas (The measures are given in centimetres).

—  : End of ML Aggarwal Mensuration Check Your Progress Class 6 ICSE Maths Solutions :–

Return to   ML Aggarwal Maths Solutions for ICSE Class -6

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