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.
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