**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 m^{2}

If the cost of sowing is Re. 1, area = m^{2}

If the cost of sowing is Rs. 783, area = m^{2}

Area of the field = 135000 m^{2}

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 =

Length of the second side =

Length of third side =

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 m^{2}

**Question 6:**

On dividing 540 m in ratio 25 : 17 : 12, we get

Length of one side =

Length of second side =

Length of third side =

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.

The cost of ploughing 9000 area = Rs.

= 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 cm^{2}

**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 = Sq.unit

(ii) Height of the triangle= 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 =

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 cm^{2}

**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 cm^{2}

**Question 17:**

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

Hence, area of the triangle = 83.1 cm^{2}.

**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 =

perimeter = (2a + b) cm

Hence each side = 12 cm and Base = = 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 cm^{2}.

**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 cm^{2} 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 ) m^{2} = 344 m^{2}

The length = 21.5 m and the area = 344 m^{2}

**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 ) m^{2} = 39200 m^{2}

Hence, area of the park = 39200 m^{2}

**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) cm^{2} = 420 cm^{2}

Hence, the other side = 35 cm and the area = 420 cm^{2}

**Question 4:**

Let the breadth of the plot be x meter

Area = Length × Breadth = (28 × x) meter

= 28x m^{2}

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) m^{2} = (15x^{2}) m^{2}

Area of lawn = 3375 m^{2}

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) m^{2}

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) m^{2}

Area of carpet = Length × Breadth

= (2.5 x 0.8) m^{2}

Number of carpets =

= 216

Hence the number of carpet pieces required = 216

**Question 9:**

Area of verandah = (36 × 15) m^{2} = 540 m^{2}

Area of stone = (0.6 × 0.5) m^{2 } [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 m^{2}

l × (28 – l) = 192

28l – l^{2 }= 192

l^{2} – 28l + 192 = 0

l^{2} – 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) m^{2} = 630 m^{2}

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^{2} = 390 m^{2}

Area of path without grass = Area of the whole park – area of park with grass

= 630 – 390 = 240 m^{2}

Hence, area of the park to be laid with grass = 240 m^{2}

**Question 12:**

Length of the plot = 125 m

Breadth of the plot = 78 m

Area of plot ABCD = (125 × 78) m^{2} = 9750 m^{2}

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^{2} = 11004 m^{2}

Area of path = Area of plot PQRS – Area of plot ABCD

= (11004 – 9750) m^{2}

= 1254 m^{2}

Cost of gravelling = Rs 75 per m^{2}

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) m^{2}

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 – 4x^{2} = 420

178x – 4x^{2} = 420

4x^{2} – 178x + 420 = 0

2x^{2} – 89x + 210 = 0

2x^{2} – 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) m^{2} = 45 x^{2}m^{2}

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) m^{2}

= 4.4 × 3.0 = 13.2 m^{2}

Length of the carpet = = 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) m^{2}

Area of floor PQRS without border

= (8 – 2x)(5 – 2x)

= 40 – 16x – 10x + 4x^{2}

= 40 – 26x + 4x^{2}

Area of border = Area of floor ABCD – Area of floor PQRS

= [40 – (40 – 26x + 4x^{2 })] m^{2}

=[40 – 40 + 26x – 4x^{2 }] m^{2}

= (26x – 4x^{2 }) m^{2}

**Question 17:**

Area of road ABCD

= ( 80 × 5 ) m^{2}

= 400 m^{2}

Area of road EFGH

= ( 64 × 5 ) m^{2}

= 320 m^{2}

Area of common road PQRS

= ( 5 × 5 ) m^{2}

= 25 m^{2}

Area of the road to be gravelled

=(400 + 320 – 25) m^{2} = 695 m^{2}

Cost of gravelling the roads

=Rs. (695 × 24) m^{2 }= Rs. 16680

**Question 18:**

Area of four walls of room = 2(l + b) × h

= 2(14 + 10) × 6.5 = 2 × 24 × 6.5

= 312 m^{2}

Area of two doors = 2 × (2.5 × 1.2) m^{2} = 6 m^{2}

Area of four windows = 4 (1.5 × 1) m^{2 }= 6 m^{2}

Area of four walls to be painted = [Area of 4 walls – Area of two doors – Area of two windows]

= [312 – 6 – 6] m^{2 }= 300 m^{2}

Cost of painting the walls = Rs 38 per m^{2}

Cost of painting 300 m^{2} of walls = Rs 38 x 300

= Rs. 11400

**Question 19:**

Cost of papering the wall at the cost of Rs. 30 m^{2} 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 m^{2} is Rs. 1620

**Question 20:**

(i) Area of the square = Sq.unit

(ii) Side of the square = = 16.97 m

Perimeter of the square = (4 × side) units

= (4 × 16.97)m

= 67.88 m

**Question 21:**

Area of the square = Sq.unit

Let diagonal of square be x

Length of diagonal = 16 m

Side of square = = 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 = Sq.unit = 80000 m^{2} (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 m^{2}

Re 1 is the cost of harvesting an area = m^{2}

Rs. 1620 is the cost of harvesting an area = m^{2}

Area = 90000 m^{2}

Area of square = (side)^{2} = 90000 m^{2}

side = = 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 = m = 2000 m

Perimeter = 4 × side = 2000

side = 500 m

Area of a square = (side)^{2} = (500)^{2} m

= 250000 m^{2}

Cost of mowing the lawn = Rs. = Rs. 135000

**Question 25:**

Largest possible size of square tile = HCF of 525 cm and 378 cm

= 21 cm

Number of tiles =

= cm^{2}

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 ) cm^{2} = 114 cm^{2}

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) cm^{2} = 412.37 cm^{2}

Perimeter ABCD = AD + AB + BC + CD

= 24 cm + 10 cm + 26 cm + 26 cm

= 86 cm

We hope the given Solutions of **RS Aggarwal Solutions Class 10 Chapter 17** are helpful to complete your math homework. If you have any doubts, please comment below. We will try to provide online math tutoring for you.

**RS Aggarwal Solutions Class 10 **

- Real Numbers
- Polynomials
- Linear equations in two variables
- Triangles
- Trigonometric Ratios
- T-Ratios of Some Particular Angles
- Trigonometric Identities
- Trigonometric Ratios of Complementary Angles
- Mean, Median, Mode of Grouped Data, of Grouped Data Ex.9A Ex.9B Ex.9C and 9D Ex.9E
- Quadratic Equations Ex.10A Ex.10B Ex.10C Ex.10D
- Arithmetic Progressions
- Circles
- Constructions
- Height and Distance
- Probability
- Coordinate Geometry
- Perimeter and Areas of Plane Figures
- Areas of Circle, Sector and Segment
- Volume and Surface Areas of Solids