SEBA Class 6 Mathematics Chapter 8 Perimeter and Area

SEBA Class 6 Mathematics Chapter 8 Perimeter and Area Solutions, SEBA Class 6 Maths Notes in English Medium, SEBA Class 6 Mathematics Chapter 8 Perimeter and Area Notes to each chapter is provided in the list so that you can easily browse throughout different chapter Assam Board SEBA Class 6 Mathematics Chapter 8 Perimeter and Area Notes and select needs one.

SEBA Class 6 Mathematics Chapter 8 Perimeter and Area

Also, you can read the SCERT book online in these sections Solutions by Expert Teachers as per SCERT (CBSE) Book guidelines. SEBA Class 6 General Mathematics Textual Question Answer. These solutions are part of SCERT All Subject Solutions. Here we have given SEBA Class 6 Mathematics Chapter 8 Perimeter and Area Solutions for All Subject, You can practice these here.

Perimeter and Area

Chapter – 8

Exercise – 8 (A)

1. Measure of each side of the following figures are given. Find their perimeter (Draw the figures on your note book)

Ans: Perimeter: 4 + 4 + 2 + 2 + 4 + 4 + 2 = 22 cm

Ans: Perimeter: 5 + 11 + 5 + 3 + 4 + 3 + 2 + 3 + 3 + 4 + 2 + 3 = 48 cm

2. If the length and breadth of a rectangle are 60 cm and 20 cm respectively, then find its perimeter.

Ans: Perimeter of a rectangle = 2 (L+ B)

= 2 (60 + 20) cm

= 2 × 80 cm

= 160 cm

3. If the perimeter of a rectangular floor is 400 m and its length is 150 m, find the breadth of the floor.

Ans: Perimeter of a rectangle floor = 400 m and length (L) = 150 m

∴ Perimeter = 2 (L+B)

⇒ 2 (150+ B) = 140

⇒ 1500 + B = 400/2 = 200

⇒ B = 200 – 150

⇒ B = 50

∴ The reqd. Breadth = 50m.

4. If the length of a rectangular park is 200 metre, its breadth is 150 metre, then find the cost of fencing at the rate of ₹20 per metre.

Ans: Length of a rectangular park = 200 m and breadth = 150 m 

∴ Perimeter = 2 (L + B) = 2(200 + 150)m 

= (2 × 350)m = 700m 

∴ The cost of the fencing at the rate of Rs. 20 

∴ Perimeter = Rs. (700 × 20) 

= Rs 14,000 

5. If the perimeter of a regular triangle as well as regular quadrilateral is 36 cm. Find the length of their sides in each case.

Ans: Length of a regular triangle = 36/3 = 12 cm. 

and length of a regular quadrilateral = 36/4 = 9cm

6. Perimeter of a regular pentagon is 125 cm. Find the measure of the length of each of its sides.

Ans: Perimeter of a regular pentagon = 125 cm. 

∴ The length of each of its sides 125/5 = cm

= 25 cm

7. Total cost of fencing of a squared playground at the rate of 35 per meters is 4480. Find the measure of each side of the playground.

Ans: 

৪. Length of a rectangle is twice its breadth. If the breadth of the rectangle is 17 cm, then find its perimeter.

Ans: Length of a rectangle = 2 × breadth 

∴ Length = (2 × 17) cm = 34 cm  

∴ Perimeter = 2 (L+ B) 

= 2 (34 + 17) cm 

= (2 × 51) cm 

= 102 cm

9. Ajay runs around a square field of side 60 meters. Bimla runs around a field which is the shape of a regular pentagon of side 50 meters. Who covers more distance and by how much?

Ans: Ajay runs around a square field of side 60 metres. 

Perimeter of squared field = (4 × 60)m = 240m. 

Bimla runs around a field which is the shape of a regular pentagon of side 50m.

∴ Perimeter = (5 × 50)m = 250m

∴ Bimla covers (250 – 240)m = 10 metres 

∴ Bimla covers more distance and the distance is 10 metres.

10. A piece of string is 60 cm long. What will be the length of each side if the string is used to form-

(a) a square.

Ans: Length of each side of a square = 60/4 cm = 15 cm

(b) an equilateral triangle.

Ans: Length of each side of an equilateral triangle = 60/3 cm = 20 cm.

(c) a regular pentagon.

Ans: Length of each side of a regular Pentagon = 60/5 cm = 12 cm.

(d) a regular hexagon?

Ans: Length of each side of a Hexagon 60/6 = cm 10 cm.

Exercise – 8 (B)

1. Find the area of each of the following parallelograms.

(a)

Ans: 5 × 7 sq.cm

= 35 sq.cm

Ans: 7 × 4 sq.cm

= 28 sq.cm

2. Find the area of each of the following triangles.

(a) 

Ans: 1/2 × 4 × 6 = 12 sq.cm.

Ans: 1/2 × 4.4 × 5.2 

= 11.44 sq.cm.

(c) 

Ans: 1/2 × 4 × 5.8 

= 11.6 sq.cm.

3. For the parallelogram ABCD, AB = 8 cm, AE = 4 cm, DE = 3cm, Find area of the coloured region.

Ans: Area of shaded region = Area of the parallelogram

= (4 × 8) – 1/2 × 3 × 4 sq.cm

= (32-6) sq.cm

= 26 sq.cm

4. Find the area of the triangle whose length of the base is 40 cm and the height is 12 cm.

Ans: Area of triangle = 1/2 base × height

= 1/2 (40cm × 12cm)

= 20cm × 12cm = 240 sq.cm

5.  Find the area of the coloured portions.

(i) 

Ans: 

(ii) 

Ans: 

6. Choose different lengths and breadths of some rectangles having same perimeter 16 cm. Also find the area of those rectangles.

Length (cm)	Breadth (cm)	Area (sqcm)

Ans: 

Length (cm)	Breadth (cm)	Area (sq cm)
7	1	7
6	2	12
5	3	15

7. Area of the rectangle ABCD is 225 sq cm and its breadth is 10 cm. Find its length.

Ans: Area = 225 sq.cm., Altitude = 10cm

∴ Base 225/10 cm = 22.5 cm

8. Parallelogram ABCD has AB = 8 cm, AD = 5 cm, DE = 7 cm, then find BF.

Ans: Area of a parallelogram 

= AB × DE

= (8 × 7) sq.cm

= 56 A sq. cm

9. Find the perimeter of a square whose area is 64 cm²

Ans: The formula for the area of a square is:

Area = side²

Given that the area is 64 cm², we find the side length:

Side = √64 = 8 cm

The perimeter of a square is given by:

Perimeter = 4 × side

= 4 × 8 = 32 cm

∴ The perimeter of the square is 32 cm.

10. How many squares with the side of 2 cm cover the surface of a rectangle with a length of 14 cm and breadth of 8 cm?

Ans: Area = Length × Breadth = 14 x 8 = 112 cm²

Area Side² = 2 x 2 = 4 cm²

Number of squares = Area of rectangle/Area of one square = 112/4 = 28

∴ 28 squares of side 2 cm are required to cover the rectangle.