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

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

**Exercise 17A**

**Question 1:**

**More Resources**

**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

Hope given RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures are helpful to complete your math homework.

If you have any doubts, please comment below. A Plus Topper try to provide online math tutoring for you.

Ambarish Sharma says

In Sum No.6 How is (s-a) coming 29? It’s 270-250=20.

Please explain.

Regards