Students can practice TS SCERT Class 6 Maths Solutions Chapter 10 Perimeter and Area Ex 10.1 to get the best methods of solving problems.
TS 6th Class Maths Solutions Chapter 10 Perimeter and Area Exercise 10.1
Question 1.
Find the perimeter of each of the following shapes :
Answer:
(i) Shape – 1 :
Given that AB = 40 cm ; BC = 50 cm
CD = 35 cm ; DE = 60 cm
and EA = 45 cm
Perimeter of shape – 1 = AB + BC + CD + DE + EA
= (40 + 50 + 35 + 60 + 45) cm
= 230 cm
(ii) Shape – 2 : Given that
AB = 8 cm; i.e., AB = HG = 8 cm;
GF = 2 cm; i.e., GF = HI = 2 cm;
ED = 5 cm; i.e., ED = JK = 5 cm;
IJ = 3 cm; i.e., IJ = LK = 3 cm;
AL = 3 cm; i.e., AL = BC = 3 cm
CD = EF = 3 cm
Perimeter of shape – 2 = AB + BC + CD + DE + EF + FG + GH + HI + IJ + JK + KL + LA
= (8 + 3 + 3 + 5 + 3 + 2 + 8 + 2 + 3 + 5 + 3+ 3)cm
= 48 cm
(iii) Shape – 3 : Given that
AB = 6 cm ; BC = 2 cm ; CD = 2 cm ;
DE = 2 cm ; EF = 2 cm ; FG = 2 cm ;
GH = 2 cm ; and HA = 6 cm
Perimeter of shape – 3
= AB + BC + CD + DE + EF + FG + GH + HA
= (6 + 2 + 2 + 2 + 2 + 2 + 2 + 6) cm
= 24 cm
(iv) Shape – 4 : Given that
AL = 4 cm (i.e.) AL = BC = 4 cm
AB = 2 cm (i.e.) AB = HG = 2 cm
JK = 2 cm (i.e.) JK = ED = 2 cm
GF = 4 cm (i.e.) GF = HI = 4 cm
KL = 4 cm (i.e.) KL = IJ = 4 cm
EF = 4 cm (i.e.) EF = CD = 4 cm
Perimeter of shape – 4
= AB + BC + CD + DE + EF + FG + GH + HI + IJ + JK + KL + LA
= 2 + 4 + 4 + 2 + 4 + 4 + 2 + 4 + 4 + 2 + 4 + 4
= 40 cm
Question 2.
Find the perimeter of each of the following figures.
What would be cost of putting a wire around each of these shapes given that 1 cm wire costs ₹ 15 ?
Answer:
(i)
Perimeter = AB + BC + CA
= (30 + 55 + 35) cm
= 120 cm
Cost of putting a wire around the given shape = 120 × Rs.15 = Rs. 1800
(ii)
ABCD is a rectangle so opposite sides are equal.
AB = CD = 40 cm
BC = AD = 20 cm
Perimeter = AB + BC + CD + DA
= 40 + 20 + 40 + 20
= 120 cm
Cost of putting a wire around the given shape at Rs.15 per 1 cm = 120 × Rs.15
= Rs. 1800
(iii)
ABCD is a square, so all the sides are equal.
∴ AB = BC = CD = DA = 30 cm
Perimeter = AB + BC + CD + DA
= 30 + 30 + 30 + 30
= 120 cm
Cost of putting a wire around the given shape = 120 × Rs.15
= Rs. 1800
(iv)
Given that the length of one side = 24 cm
ABCDEF is a regular hexagon (i.e.) it contains 6 equal sides.
Perimeter = 6 × side
= 6 × 24cm = 144 cm
Cost of putting a wire around the given shape = 144 × Rs.15 = Rs. 2160
Question 3.
How many different rectangles can you make with a 24 cm long string with integral sides and what are the sides of those rectangles in cm ?
Answer:
The length of string = 24 cm
Perimeter of the rectangle = 2(l + b) =24 cm
l + b = \(\frac{24}{2}\)cm = 12 cm
We can make different rectangles with 24 cm. long string as follows.
| S.No. | Length in cm | Breadth in cm | Perimeter of the rectangle = 24 cm |
| 1 | 1 | 11 | 2(1 + 11) = 2 × 12 = 24 |
| 2 | 2 | 10 | 2(2 + 10) = 2 × 12 = 24 |
| 3 | 3 | 9 | 2(3 + 9) = 2 × 12 = 24 |
| 4 | 4 | 8 | 2(4 + 8) = 2 × 12 = 24 |
| 5 | 5 | 7 | 2(5 + 7) = 2 × 12 = 24 |
| 6 | 7 | 5 | 2 (7 + 5) = 2 × 12 = 24 |
| 7 | 8 | 4 | 2(8 + 4) = 2 × 12 = 24 |
| 8 | 9 | 3 | 2(9 + 3) = 2 × 12 = 24 |
Question 4.
A flower bed is in the shape of a square with a side 3.5 m. Each side is to be fenced with 4 rows of ropes. Find the cost of rope required at ₹ 15 per meter.
Answer:
The shape of a flower bed is a square.
The length of the side of the square
= 3.5 m
Length of 4 rows of ropes on each side = 4 × 3.5 m = 14 m
Length of the rope on 4 sides
= 14 m × 4 = 56 m
Cost of rope required at Rs. 15 per meter = Rs. 56 × 15 = Rs. 840
Question 5.
A piece of wire is 60 cm long. What will be the length of each side if the string is used to form :
(i) an equilateral triangle
(ii) a square
(iii) a regular hexagon
(iv) a regular pentagon
Answer:
(i) Length of the wire = 60 cm
The perimeter of an equilateral triangle with side x cm = 3 × x = 3x
By problem,
3x = 60 cm
∴ x = \(\frac{60}{3}\) cm = 20 cm
Length of each side of the equilateral triangle = 20 cm
(ii) Length of the wire = 60 cm
The perimeter of the square with side ‘x’ cm = 4 × x = 4x
By problem,
4x = 60 cm
x= \(\frac{60}{4}\) cm = 15 cm
Length of each side of the square =15 cm
(iii) Length of the wire = 60 cm
The perimeter of the regular hexagon with side x’ cm = 6 × x = 6x
By problem,
6x = 60 cm
∴ x = \(\frac{60}{5}\) cm = 10 cm 6
Length of each side of the regular hexagon = 10 cm
(iv) Length of the wire = 60 cm
The perimeter of the regular pentagon with side ‘x’ cm = 5 × x = 5x
By problem,
5x = 60 cm
x = \(\frac{60}{5}\) cm = 12 cm
Length of each side of the regular pentagon = 12 cm.
Question 6.
Bunty and Bubly go for jogging every morning. Bunty goes around a square park of side 80 m. Bubly goes around a rectangular park with length 00 m and breadth 60 m. If they both fake 3 rounds, who covers more distance and by how much ?
Answer:
Bunty goes around a square park.
The length of the side of square park
= 80 m.
The perimeter of the square park
= 4 × 80 = 320 m.
Distance covered by Bunty in 3 rounds = 320m × 3 = 960 m.
Bubly goes around a rectangular park.
The length and breadth of the park are 90 m and 60 m respectively.
The perimeter of the rectangular park = 2 (length + breadth)
= 2[90 + 60]
= 2 × 150
= 300 m
Distance covered by Bubly in 3 rounds = 300m × 3 = 900 m.
Bunty covers greater distance by 60 m. (∵ 960 – 900 = 60 m)
Question 7.
The length of a rectangle is twice of its breadth. If its perimeter is 48 cm, find the dimensions of the rectangle.
Answer:
The perimeter of the rectangle = 48 cm Let the breadth of the rectangle be x cm The length of the rectangle .
= 2 × x = 2x cm Perimeter of the rectangle
= 2 (length + breadth)
= 2 (2x + x)cm
= 2 × 3x = 6x
By problem,
6x = 48
∴ x = \(\frac{48}{6}\) = 8
Breadth of the rectangle = 8 cm
Length of the rectangle = 2 × 8 = 16 cm
Question 8.
Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is the length of third side ?
Answer:
Two sides of a triangle are 12 cm and 14 cm.
Length of third side = x cm .
Perimeter of the triangle
= sum of the three sides = 12 + 14 + x = (26 + x) cm
By problem,
(26 + x) cm = 36 cm
∴ x = 36 – 26 = 10 cm
Length of third side of the triangle
= 10 cm.
Question 9.
Find the perimeter of each of the following shapes:
(i) A triangle of sides 3 cm., 4 cm. and 5 cm.
(ii) An equilateral triangle of side 9 cm.
(iii) An isosceles triangle with equal sides 8 cm each and third side of 6 cm.
Answer:
(i) The sides of a triangle are 3 cm, 4 cm, 5 cm
Perimeter of the given triangle
= sum of the three sides = (3 + 4 + 5)cm
= 12 cm
(ii) Side of the equilateral triangle = 9 cm
Perimeter of the equilateral triangle = (9 + 9 + 9)cm
= 27 cm (∵ All the 3 sides are equal)
(iii) The length of one of the equal sides of the isosceles triangle is 8 cm.
Length of the third side is 6 cm.
Perimeter of the given triangle
= (8 + 8 + 6) cm
= 22 cm