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

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RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures is a part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures.

## Exercise 17A

Question 1:

Question 2:
If the cost of sowing the field is Rs. 58, then area = 10000 m2
If the cost of sowing is Re. 1, area =  $\frac { 10000 }{ 58 }$ m2
If the cost of sowing is Rs. 783, area =  $(\frac { 10000 }{ 58 } \times 783)$ m2
Area of the field = 135000 m2
Let the attitude of the field be x meters
Then, Base = 3x meter
Area of the field =

Hence, the altitude = 300m and the base = 900 m

Question 3:
Let a = 42 cm, b = 34 cm and c = 20 cm

(i) Area of triangle =

(ii) Let base = 42 cm and corresponding height = h cm
Then area of triangle =

Hence, the height corresponding to the longest side = 16 cm

Question 4:
Let a = 18 cm, b = 24 cm, c = 30 cm
Then, 2s = (18 + 24 + 30) cm = 72 cm
s = 36 cm
(s – a) = 18cm, (s – b) = 12 cm and (s – c) = 6 cm
(i) Area of triangle =

(ii) Let base = 18 cm and altitude = x cm
Then, area of triangle =

Hence, altitude corresponding to the smallest side = 24 cm

Question 5:
On dividing 150 m in the ratio 5 : 12 : 13, we get
Length of one side =  $(150\times \frac { 5 }{ 30 } )m=25m$
Length of the second side =  $(150\times \frac { 12 }{ 30 } )m=60m$
Length of third side =  $(150\times \frac { 13 }{ 30 } )m=65m$
Let a = 25 m, b = 60 m, c = 65 m

(s – a) = 50 cm, (s – b) = 15 cm, and (s – c) = 10 cm

Hence, area of the triangle = 750 m2

Question 6:
On dividing 540 m in ratio 25 : 17 : 12, we get
Length of one side =  $(540\times \frac { 25 }{ 54 } )m=250m$
Length of second side =  $(540\times \frac { 17 }{ 54 } )m=170m$
Length of third side =  $(540\times \frac { 12 }{ 54 } )m=120m$
Let a = 250m, b = 170 m and c = 120 m

Then, (s – a) = 29 m, (s – b) = 100 m, and (s – c) = 150m

The cost of ploughing 100 area is = Rs. 18. 80
The cost of ploughing 1 is =  Rs. $\frac { 18. 80 }{ 100 }$
The cost of ploughing 9000 area = Rs. $(\frac { 18. 80 }{ 100 } \times 9000)$
= Rs. 1692
Hence, cost of ploughing = Rs 1692.

Question 7:
Let the length of one side be x cm
Then the length of other side = {40 × (17 + x)} cm = (23 – x) cm
Hypotenuse = 17 cm
Applying Pythagoras theorem, we get

Hence, area of the triangle = 60 cm2

Question 8:
Let the sides containing the right angle be x cm and (x × 7) cm

One side = 15 cm and other = (15 × 7) cm = 8 cm

perimeter of triangle (15 + 8 + 17) cm = 40 cm

Question 9:
Let the sides containing the right angle be x and (x × 2) cm

One side = 8 cm, and other (8 × 2) cm = 6 cm
= 10 cm

Therefore, perimeter of the triangle = 8 + 6 + 10 = 24 cm

Question 10:
(i) Here a = 8 cm
Area of the triangle =  $(\frac { \sqrt { 3 } }{ 4 } \times { a }^{ 2 })$ Sq.unit

(ii) Height of the triangle=  $(\frac { \sqrt { 3 } }{ 4 } \times { a })$ Sq.unit

Hence, area = 27.71 cm2 and height = 6.93 cm

Question 11:
Let each side of the equilateral triangle be a cm

Question 12:
Let each side of the equilateral triangle be a cm

Perimeter of equilateral triangle = 3a = (3 × 12) cm = 36 cm

Question 13:
Let each side of the equilateral triangle be a cm
area of equilateral triangle =  $\frac { \sqrt { 3 } }{ 4 } { a }^{ 2 }$

Height of equilateral triangle

Question 14:
Base of right angled triangle = 48 cm
Height of the right angled triangle =

Question 15:
Let the hypotenuse of right angle triangle = 6.5 m
Base = 6 cm

Hence, perpendicular = 2.5 cm and area of the triangle =7.5 cm2

Question 16:
The circumcentre of a right triangle is the midpoint of the hypotenuse

Hypotenuse = 2 ×(radius of circumcircle)
= (2 × 8) cm = 16 cm
Base = 16 cm, height = 6 cm
Area of right angled triangle

Hence, area of the triangle= 48 cm2

Question 17:
Let each side a = 13 cm and the base b = 20 cm

Hence, area of the triangle = 83.1 cm2.

Question 18:
Let each equal side be a cm in length.
Then,

Hence, hypotenuse = 28.28 cm and perimeter = 68.28 cm

Question 19:
Let each equal side be a cm and base = 80 cm

perimeter of triangle = (2a + b) cm
= (2 ×41 + 80) cm
= (82 + 80) cm = 162 cm
Hence, perimeter of the triangle = 162 cm

Question 20:
Perimeter of an isosceles triangle = 42 cm
(i) Let each side be a cm, then base =  $\frac { 3 }{ 2 } { a }$
perimeter = (2a + b) cm

Hence each side = 12 cm and Base = $\frac { 3 }{ 2 } { 12 }$ = 18cm
(ii) Area of triangle =

(iii) Height of the triangle =

Question 21:
Let the height be h cm, then a= (h + 2) cm and b = 12 cm

Squaring both sides,

Therefore, a = h + 2 = (8 + 2)cm = 10 cm

Hence, area of the triangle = 48 cm2.

Question 22:
Perimeter of triangle = 324 cm
(i) Length of third side = (324 – 85 – 154) m = 85 m
Let a = 85 m, b = 154 m, c = 85 m

(ii) The base = 154 cm and let the perpendicular = h cm

Hence, required length of the perpendicular of the triangle is 36 m.

Question 23:

Area of shaded region = Area of ∆ABC – Area of ∆DBC
First we find area of ∆ABC

Second we find area of ∆DBC which is right angled

Area of shaded region = Area of ∆ABC – Area of ∆DBC
= (43.30 – 24) = 19. 30
Area of shaded region = 19.3

Question 24:
Let ∆ABC is a isosceles triangle. Let AC, BC be the equal sides
Then AC = BC = 10cm. Let AB be the base of ∆ABC right angle at C.

Area of right isosceles triangle ABC

Hence, area = 50 cm2 and perimeter = 34.14 cm

## Exercise 17B

Question 1:
Let the length of plot be x meters
Its perimeter = 2 [length + breadth]
=2(x + 16) = (2x + 32) meters

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:
Let the breadth of a rectangular park be x meter
Then, its length = 2x meter
∴ perimeter = 2(length + breadth)
=2(2x + x) = 6x meters
∴ 6x = 840 m [ ∵ 1 km = 1000 m]
⇒ x = 140 m
Then, breadth = 140 m and length = 280 m
Area of rectangular park = (length × breadth) = ( 140 × 280 ) m2 = 39200 m2
Hence, area of the park = 39200 m2

Question 3:
Let ABCD be the rectangle in which AB = 12 cm and AC = 37 m

By Pythagoras theorem, we have

Thus, length = 35 cm and breadth = 12 cm
Area of rectangle = (12 × 35) cm2 = 420 cm2
Hence, the other side = 35 cm and the area = 420 cm2

Question 4:
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)
= 2 (28 + 16.5 ) m = 2 ( 44.5) m = 89 m

Question 5:
Let the breadth of rectangular hall be x m
Then, Length = (x + 5) m

Breadth = 25 m and length = (25 + 5) m = 30 m
Perimeter of rectangular hall = 2(length + breadth)
= 2(30 + 25)m = (2 × 55) m = 110 m

Question 6:
Let the length of lawn be 5x m and breadth of the lawn be 3x m
Area of rectangular lawn = (5x × 3x) m2 = (15x2) m2
Area of lawn = 3375 m2

Length = 5 × 15 = 75
Breadth = (3 × 15)m = 45 m
Perimeter of lawn = 2(length + breadth)
=2 (75 + 45)m = 240 m
Cost of fencing the lawn per meter = Rs. 8.50 per meter
Cost of fencing the lawn = Rs 8. 50 × 240 = Rs. 2040

Question 7:
Length of the floor = 16 m
Breadth of the floor = 13.5 m
Area of floor = (16 x 13.5) m2

Cost of carpet = Rs. 15 per meter
Cost of 288 meters of carpet = Rs. (15 × 288) = Rs. 4320

Question 8:
Area of floor = Length × Breadth
= (24 x 18) m2
Area of carpet = Length × Breadth
= (2.5 x 0.8) m2
Number of carpets =
= 216
Hence the number of carpet pieces required = 216

Question 9:
Area of verandah = (36 × 15) m2 = 540 m2
Area of stone = (0.6 × 0.5) m2   [10 dm = 1 m]
Number of stones required =
Hence, 1800 stones are required to pave the verandah.

Question 10:
Perimeter of rectangle = 2(l + b)
2(l + b) = 56 ⇒ l + b = 28 cm
b = (28 – l) cm
Area of rectangle = 192 m2
l × (28 – l) = 192
28l – l= 192
l2 – 28l + 192 = 0
l2 – 16l – 12l + 192 = 0
l(l – 16) – 12(l – 16) = 0
(l – 16) (l – 12) = 0
l = 16 or l = 12
Therefore, length = 16 cm and breadth = 12 cm

Question 11:
Length of the park = 35 m
Breadth of the park = 18 m

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) m2 = 390 m2
Area of path without grass = Area of the whole park – area of park with grass
= 630 – 390 = 240 m2
Hence, area of the park to be laid with grass = 240 m2

Question 12:
Length of the plot = 125 m
Breadth of the plot = 78 m

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) m2 = 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 13:
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
1890 – 1890 + 108x + 70x – 4x2 = 420
178x – 4x2 = 420
4x2 – 178x + 420 = 0
2x2 – 89x + 210 = 0
2x2 – 84x – 5x + 210 = 0
2x(x – 42) – 5(x – 42) = 0
(x – 42) (2x – 5) = 0

Question 14:
Let the length and breadth of a rectangular garden be 9x and 5x.

Then, area of garden = (9x × 5x) m2 = 45 x2m2
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)
Area of the path =

(98x – 49) = 1911
98x = 1911 + 49

Length = 9x = 9 × 20 = 180 m
Breadth = 5x = 5 × 20 = 100 m
Hence, length = 180 m and breadth = 100 m

Question 15:
Area of carpet = (4.9 – 0.5) (3.5 – 0.5) m2
= 4.4 × 3.0 = 13.2 m2
Length of the carpet = $(\frac { 13.2 }{ 0.80 })m$  = 16.5m
Cost of carpet = Rs. 40 per meter
Cost of 16.5 m carpet = Rs. (40 × 16.5) = Rs. 660

Question 16:
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 – 16x – 10x + 4x2
= 40 – 26x + 4x2
Area of border = Area of floor ABCD – Area of floor PQRS
= [40 – (40 – 26x + 4x)] m2
=[40 – 40 + 26x – 4x] m2
= (26x – 4x) m2

Question 17:
= ( 80 × 5 ) m2
= 400 m2

= ( 64 × 5 ) m2
= 320 m2
= ( 5 × 5 ) m2
= 25 m2
Area of the road to be gravelled
=(400 + 320 – 25) m2 = 695 m2
=Rs. (695 × 24) m= Rs. 16680

Question 18:
Area of four walls of room = 2(l + b) × h
= 2(14 + 10) × 6.5 = 2 × 24 × 6.5
= 312 m2
Area of two doors = 2 × (2.5 × 1.2) m2 = 6 m2
Area of four windows = 4 (1.5 × 1) m= 6 m2
Area of four walls to be painted = [Area of 4 walls – Area of two doors – Area of two windows]
= [312 – 6 – 6] m= 300 m2
Cost of painting the walls = Rs 38 per m2
Cost of painting 300 m2 of walls = Rs 38 x 300
= Rs. 11400

Question 19:
Cost of papering the wall at the cost of Rs. 30 m2 per in Rs. 7560

Let h meter be the height and b m be the breadth of the room
Length of the room = 12 m
Area of four walls = 2 ×(12 + b) × h
2(12 + b) × h = 252
Or (12 + b) h = 126 —–(1)
The cost of covering the floor with mat at the cost of Rs. 15 per m2 is Rs. 1620

Question 20:
(i) Area of the square =  $\frac { 1 }{ 2 } { (diagonal) }^{ 2 }$ Sq.unit

(ii) Side of the square = $\sqrt { 288 } m$ = 16.97 m
Perimeter of the square = (4 × side) units
= (4 × 16.97)m
= 67.88 m

Question 21:
Area of the square =  $\frac { 1 }{ 2 } { (diagonal) }^{ 2 }$ Sq.unit
Let diagonal of square be x

Length of diagonal = 16 m
Side of square =  $\sqrt { 128 } m$ = 11.31 m
Perimeter of square = [4 × side] sq. units
=[ 4 × 11.31] cm = 45.24 cm

Question 22:
Let d meter be the length of diagonal
Area of square field =  $\frac { 1 }{ 2 } { (diagonal) }^{ 2 }$ Sq.unit = 80000 m2 (given)

Time taken to cross the field along the diagonal

Hence, man will take 6 min to cross the field diagonally.

Question 23:
Rs. 180 is the cost of harvesting an area = 1 hectare = 10000 m2
Re 1 is the cost of harvesting an area = $\frac { 10000 }{ 180 }$ m2
Rs. 1620 is the cost of harvesting an area = $(\frac { 10000 }{ 180 } \times 1620)$ m2
Area = 90000 m2
Area of square = (side)2 = 90000 m2
side = $\sqrt { 90000 } m$ = 300 m
Perimeter of square = 4 × side = 4 × 300 = 1200 m
Cost of fencing = Rs 6.75 per meter.
Cost of fencing 1200 m long border = 1200 × Rs 6.75 = Rs. 8100

Question 24:
Rs. 14 is the cost of fencing a length = 1m
Rs. 28000 is the cost of fencing the length =  $\frac { 28000 }{ 14 }$ m = 2000 m
Perimeter = 4 × side = 2000
side = 500 m
Area of a square = (side)2  = (500)2  m
= 250000 m2
Cost of mowing the lawn = Rs. $(250000\times \frac { 54 }{ 100 } )$ = Rs. 135000

Question 25:
Largest possible size of square tile = HCF of 525 cm and 378 cm
= 21 cm
Number of tiles =  $\frac { Area\quad of\quad rectangle }{ Area\quad of\quad square\quad tiles }$
=  $\frac { (525\times 378) }{ (21\times 21) }$ cm2
Number of tiles = 450

Question 26:
Area of quad. ABCD = Area of ∆ABD + Area of ∆DBC
For area of ∆ABD
Let a = 42 cm, b = 34 cm, and c = 20 cm

For area of ∆DBC
a = 29 cm, b = 21 cm, c = 20 cm

Question 27:

Area of quad. ABCD = Area of ∆ABC + Area of ∆ACD

Now, we find area of a ∆ACD

Area of quad. ABCD = Area of ∆ABC + Area of ∆ACD
= ( 60+54 ) cm2 = 114 cm2
Perimeter of quad. ABCD = AB + BC + CD + AD
=(17 + 8 + 12 + 9) cm
= 46 cm
Perimeter of quad. ABCD = 46 cm

Question 28:
ABCD be the given quadrilateral in which AD = 24 cm, BD = 26 cm, DC = 26 cm and BC = 26 cm
By Pythagoras theorem

For area of equilateral ∆DBC, we have
a = 26 cm

Area of quad. ABCD = Area of ∆ABD + Area of ∆DBC
= (120 + 292.37) cm2 = 412.37 cm2
Perimeter ABCD = AD + AB + BC + CD
= 24 cm + 10 cm + 26 cm + 26 cm
= 86 cm

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