RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b.

Other Exercises

Question 1.
Solution:
Let the length of plot be x meters
Its perimeter = 2 [length + breadth]
= 2(x + 16) = (2x + 32) meters
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 1
Length of the rectangle is 21. 5 meter
Area of the rectangular plot = length × breadth
= (16 × 21.5)m2
= 344 m2
The length = 21.5 m and the area = 344 m2

Question 2.
Solution:
The length of a rectangular park is twice its breadth
and its perimeter is 840 m.
Find the area of the park.
Let the breadth of a rectangular park = x m
Then, length of a rectangular park = 2x m
Perimeter of a rectangular park = 840 m
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 2
∴ Area of a rectangular park = length x breadth = 140x 280 = 39200 m2

Question 3.
Solution:
Let ABCD be the rectangle in which AB = 12 cm and AC = 37 m
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 3
Thus, length = 35 cm and breadth = 12 cm
Area of rectangle = (12 × 35) = 420
Hence, the other side = 35 cm and the area = 420 cm2

Question 4.
Solution:
Let the breadth of the plot be x meter
Area = Length × Breadth = (28 × x) meter
= 28x m2
Breadth of plot is = 16. 5 m
Perimeter of the plot is = 2(length + breadth)
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 4

Question 5.
Solution:
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 5

Question 6.
Solution:
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 6

Question 7.
Solution:
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 7

Question 8.
Solution:
Area of verandah = (36 × 15)m2 = 540 m2
Area of stone = (0.6 × 0.5)m2 [10 dm = 1 m]
Number of stones required = \(\frac { Area\quad of\quad verandah }{ Area\quad of\quad stone } =\frac { 540 }{ 0.3 } =1800\)
Hence, 1800 stones are required to pave the verandah

Question 9.
Solution:
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 8

Question 10.
Solution:
Length of the park = 35 m
Breadth of the park = 18 m
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 9
Area of the park = (35 18)m2 = 630 m2
Length of the park with grass =(35 – 5) = 30 m
Breadth of the park with grass = (18 – 5) m = 13 m
Area of park with grass = (30 × 13) m= 390 m2
Area of path without grass = Area of the whole park area of park with grass
= 630 m2 – 390 m= 240 m2
Hence, area of the park to be laid with grass = 240 m2

Question 11.
Solution:
Length of the plot = 125 m
Breadth of the plot = 78 m
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 10
Area of plot ABCD = (125 × 78)m2 = 9750 m2
Length of the plot including the path= (125 + 3 + 3) m = 131 m
Breadth of the plot including the path = (78 + 3 + 3) m = 84 m
Area of plot PQRS including the path
= (131 × 84) m= 11004 m2
Area of path = Area of plot PQRS Area of plot ABCD
= (11004 – 9750)m2 = 1254 m2
Cost of gravelling = Rs 75 per m2
Cost of gravelling the whole path = Rs. (1254 × 75) = Rs. 94050
Hence, cost of gravelling the path = Rs. 94050.

Question 12.
Solution:
Area of rectangular field including the foot path = (54 × 35)m2
Let the width of the path be x m
Then, area of rectangle plot excluding the path = (54 × 2x) (35 × 2x)
Area of path = (54 × 35) (54 × 2x) (35 × 2x) = 54 ×35) (54 × 2x) (35 × 2x) = 420
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 11

Question 13.
Solution:
Let the length and breadth of a rectangular garden be 9x and 5x.
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 12
Then, area of garden = (9x 5x)m = 45xm2
Length of park excluding the path = (9x 7) m
Breadth of the park excluding the path = (5x 7) m
Area of the park excluding the path = (9x 7)(5x 7) m2
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 13

Question 14.
Solution:
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 14

Question 15.
Solution:
Let the width of the carpet = x meter
Area of floor ABCD = (8 ×5) m2
Area of floor PQRS without border
= (8 × 2x)(5 × 2x)
= 40 × 16 x 10x + 4x2
= 40 × 26x + 4x2
Area of border = Area of floor ABCD Area of floor PQRS
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 15

Question 16.
Solution:
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 16

Question 17.
Solution:
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 17

Question 18.
Solution:
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 18

Question 19.
Solution:
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 19

Question 20.
Solution:
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 20

Question 21.
Solution:
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 21

Question 22.
Solution:
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 22

Question 23.
Solution:
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 23

Question 24.
Solution:
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 24

Question 25.
Solution:
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 25
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 26

Question 26.
Solution:
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 27
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 28

Question 27.
Solution:
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 29
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 30

Question 28.
Solution:
Area of the ||gm = (base height) sq. unit
= (25 × 16.8) cm= 420 cm2

Question 29.
Solution:
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 31

Question 30.
Solution:
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 32

Question 31.
Solution:
Area of parallelogram = 2 area of DABC
Opposite sides of parallelogram are equal
AD = BC = 20 cm
And AB = DC = 34 cm
In ∆ABC we have
a = AC = 42 cm
b = AB = 34 cm
c = BC = 20 cm
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 33

Question 32.
Solution:
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 34
We know that the diagonals of a rhombus, bisect each other at right angles
OA = OC = 15 cm,
And OB = OD = 8 cm
And ∠AOB = 900
∴By Pythagoras theorem, we have
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 35

Question 33.
Solution:
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 36
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 37

Question 34.
Solution:
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 38
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 39

Question 35.
Solution:
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 40

Question 36.
Solution:
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 41

Question 37.
Solution:
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 42
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b 43

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