TS 6th Class Maths Solutions Chapter 10 Perimeter and Area InText Questions

Students can practice TS SCERT Class 6 Maths Solutions Chapter 10 Perimeter and Area InText Questions to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 10 Perimeter and Area InText Questions

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Question 1.
Give five examples of situations where you need to know the perimeter.
Answer:

  • To construct the house.
  • To pave the tiles of a room.
  • To find the perimeter of a flat.
  • To determine the perimeter of a field.
  • To construct a well around the school ground. For the above cases the knowledge of concept of perimeter is required.

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Question 1.
What would be the perimeter of these shapes ? Fill in the blanks given and in each case start from the point A.
TS 6th Class Maths Solutions Chapter 10 Perimeter and Area InText Questions 1
Answer:
(i) Perimeter = AB + BC + CD + DA
= 10m + 40m + 10m + 40m
= 100 m

TS 6th Class Maths Solutions Chapter 10 Perimeter and Area InText Questions 2
= 100m + 120m + 90m + 45m + 60m + 80m
= 495 m

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Find the perimeter of the following :
Question 1.
A table with sides equal to 30 cm, 15 cm, 30 cm, 15 cm respectively.
Answer:
Perimeter of table = 2 (l + b)
= 2 (30cm + 15cm) = 2 (45)cm = 90 cms.

Question 2.
Measure the length of the sides of your text book cover. What is the perimeter ?
Answer:
The length of the sides of the text book (Maths) is 30 cm, 20 cm.
Its perimeter = 2 (l + b)
= 2 (30 + 20)cm
= 2 × 50cm
= 100 cms.

TS 6th Class Maths Solutions Chapter 10 Perimeter and Area InText Questions

Question 3.
Around a rectangular park of sides 100 meters and 70 meters a wire has to be put. The cost of the wire is ₹ 20 per meter. What is the total cost of the wire ?
Answer:
Given that length of a park = 100 m
breadth of a park = 70 m
∴ Perimeter of a rectangular park = 2 (l + b)m
= 2 (100 + 70)m = 2 × 170m = 340 m.
The total cost of the wire at the rate of ₹ 20 perimeter = 340m × 20 = ₹ 6,800
Perimeter = AB + BC + £D + DA = 10m + 40m + 10m + 40m = 100 m

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Question 1.
Find the perimeter of the following rectangles.
TS 6th Class Maths Solutions Chapter 10 Perimeter and Area InText Questions 3
Answer:
TS 6th Class Maths Solutions Chapter 10 Perimeter and Area InText Questions 4

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Question 1.
A square picture frame has sides of 0. 75 mts. If the cost of a coloured paper is ₹ 20 per meter, what is the cost of putting coloured paper around the frame ?
Answer:
The side of sqaured frame = 0.75 m
Its perimeter =4 × s = 4 × 0.75 = 3m
∴ The cost of a coloured paper around the square frame at the rate of ₹ 20 per meter = 3 × 20 = ₹ 60.

Question 2.
There is a string of length 44 cm. How many different rectangles with positive integers as length and breadth can be made with this string ?
Answer:
The length and breadth of rectangle as follows whose perimeter is 44 cm.

S.No.LengthBreadthPerimeter fin cm)
1.101244
2.91344
3.81444
4.71544
5.121044

Question 3.
If I have a string 41 cm long can I make a rectangle using the string completely ? Give reasons.
Answer:

Length (cm)Breadth (cm)Perimeter (cm)
1010.541
911.541
119.541
137.541
146.541

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Question 1.
Find the perimeter of the following squares. Figures are drawn on 1 cm grids.
TS 6th Class Maths Solutions Chapter 10 Perimeter and Area InText Questions 5
Answer:
i) Perimeter of square ABCD = 4 × s = 4 × 4 = 16 cm.
ii) Perimeter of square PQRS = 4 × s = 4 × 6 = 24 cm.
iii) Perimeter of square EFGH = 4 × s = 4 × 5 = 20 cm.
iv) Perimeter of square WXYZ = 4 × s = 4 × 2 = 8 cm.

Question 2.
Find various objects from your surroundings which have regular shapes and their perimeters.
Answer:
The objects which are having the regular shapes are
1) A ‘4’ size squared paper.
2) The pentagon building in U.S.A.
3) A square shaped cake.
4) A square shaped field.

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Question 1.
Find the perimeter of a regular pentagon of side 8 cm.
Answer:
Perimeter of a regular polygon = n × length of its side
∴ Perimeter of a pentagon = 5 × 8 cm = 40 cm

TS 6th Class Maths Solutions Chapter 10 Perimeter and Area InText Questions

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Question 1.
Find the areas of the following figures by counting squares.
Area of each square is 1 sq.cm
TS 6th Class Maths Solutions Chapter 10 Perimeter and Area InText Questions 6
Answer:

FigureIts Area (in sq. cms)
(i)3 × 3 = 9
(ii)7 × 1 = 7
(iii)9 × 1 = 9
(iv)8 × 1 = 8
(v)6 × 1 = 6
(vi)4 × 1 = 4
(vii)4 × 1 = 4
(viii)7 × 1 = 7

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Question 1.
Trace shapes of leaves, flower petals and other such objects on the graph paper and find their area approximately.
Answer:
Student Activity

Question 2.
Draw any line diagram on a graph sheet. Count the squares and use them to estimate the area of the region.
Answer:
Student Activity

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Question 1.
Draw two different rectangles having the same perimeter. Compare their areas. Are they same ? Can you draw two different squares having the same perimeter ?
Answer:
i) If the perimeters of two rectangles are equal then the areas of the rectangles need , not to be equal.
ii) No. We can’t draw two different squares having same perimeter. ;

TS 6th Class Maths Solutions Chapter 10 Perimeter and Area InText Questions

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Question 1.
Find the area of:
(i) The floor of your classroom.
(ii) A door in your house
(iii) The black board in your classroom.
Answer:
(i) Length of our class room floor = 15m
Breadth of the class room floor = 10m
Area of the floor = lb
= (15m) (10m)
= 150 sq.m

(ii) Length of the door in our house = 5m
Breadth of the door = 1.5 m
Area of the door = lb
= (5m) (1.5m)
= 7.5 sq.m

(iii) Length of the black board = 2.5m
Breadth of the black board = 2m
Area of the Black board = lb = 5sq.m

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Question 1.
The length of one side of few squares are given. Find their areas using graph papers also find side × side. What do you notice from the result obtained ?
i) 4 cm
ii) 6 cm
iii) 2 cm
iv) 8 cm
Answer:

Side of square (cms)Its area by graph paper (sq. cm)Side × Side
i) 4 cm4 + 4 + 4 + 4 = 16S × S = 4 × 4 = 16
ii) 6 cm6 + 6 + 6 + 6 + 6 + 6 = 36S × S = 6 × 6 = 36
iii) 2 cm1 + 1 + 1 + 1 = 4S × S = 2 × 2 = 4
iv) 8 cm8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 = 64S × S = 8 × 8 = 64

We notice that two results are equal.

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