**1. **Given: The side of a square = 60 m

And the length of rectangular field = 80 m According to question,

Perimeter of rectangular field = Perimeter of square field

⇒ 2 (l + b) = 4 × side

⇒ 2 (80 + b) = 4 × 60 ⇒ (160 + 21) = 240

⇒ 2h = 240 – 160 ⇒ 2h =80

⇒ b* = *40 m

Now Area of Square field = (Side)^{2} = (60)^{2} = 3600 m^{2}

And Area of Rectangular field = length × breadth = 80 × 40 = 3200 m^{2}

Hence, area of square field is larger.

**2. ** Side of a square plot = 25 m

∴ Area of square plot = (Side)^{2} = (25)^{2} = 625 m^{2}

Length of the house = 20 m and

Breadth of the house = 15 m

∴ Area of the house = length × breadth = 20 × 15 = 300 m^{2}

Area of garden = Area of square plot – Area of house = 625 – 300 = 325 m^{2}

∵ Cost of developing the garden per sq. m = 55

∴ Cost of developing the garden 325 sq. m = 55 × 325 = 17,875

Hence total cost of developing a garden around is ₹ 17,875.

**3. ** Given: Total length = 20 m

Diameter of semi circle = 7 m

∴ Radius of semi circle = 3.5 m

Length of rectangular field = 20 – (3.5 + 3.5) = 20 – 7 = 13 m

Breadth of the rectangular field = 7 m

∴ Area of rectangular field = lxb *= *13 x 7 = 91 m^{2}

Area of garden = 91 + 38.5 = 129.5 m^{2}

Now Perimeter of two semi circles

And Perimeter of garden = 22 + 13 + 13 = 48 m

**4. **Given: Base of flooring tile = 24 cm = 0.24 m

Corresponding height of a flooring tile = 10 cm = 0.10 m

Now Area of flooring tile = Base × Altitude = 0.24 × 0.10 = 0.024 m^{2}

∴ Number of tiles required to cover the floor

Hence 45000 tiles are required to cover the floor.

Total distance covered by the ant = Circumference of semi circle + Diameter = 4.4 + 2.8 = 7.2 cm

(b) Diameter of semi circle = 2.8 cm

(c) Diameter of semi circle = 2.8 cm

Total distance covered by the ant = 2 + 2 + 4.4 = 8.4 cm

Hence for figure (b) food piece, the ant would take a longer round.

**6. ** Here one parallel side of the trapezium *(a) = 1 *m

And second side *(b) = *1.2 m and height *(h) = *0.8 m

Area of top surface of the table

Hence surface area of the table is 0.88 m^{2}

**7. **Let the length of the other parallel side be

Length of one parallel side *(a) = *10 am and height *(h) = *4 cm

⇒ 34 = 20+ 2b ⇒ 34 – 20 = 2b

⇒ 14 = 2b ⇒ 7 = b

⇒ b = 7

Hence another required parallel side is 7 cm.

**8. **Given: BC = 48 m, CD = 17 m, AD = 40 m and perimeter = 120 m

∵ Perimeter of trapezium ABCD = AB + BC + CD + DA

⇒ 120 = AB + 48 + 17 + 40

⇒ 120 = AB = 105

⇒ AB = 120 – 105 = 15 m

Hence area of the field ABCD is 660 m^{2}

**9.** Here h_{1} = 13 m, h_{2} = 8 m and AC = 24 m

Area of quadrilateral ABCD = Area of Δ* *ABC + Area of Δ ADC

Hence required area of the field is 252 m^{2}

**10. ** Given: d, =7.5 cm and d_{2} = 12 cm

We know that,

Hence area of rhombus is 45 cm^{2}.

**11. **Since rhombus is also a kind of parallelogram.

Area of rhombus = Base × Altitude = 6 × 4 = 24 cm^{2}

Hence the length of the other diagonal is 60 cm.

**12. ** Here, d_{1} = 45 cm and d_{2} = 30 cm

∴ Area of 3000 tiles = 675 × 3000 = 2025000 cm^{2}

∵ Cost of polishing the floor per sq. meter = ₹ 4

∴ Cost of polishing the floor per 202.50 sq. meter = 4 × 202.50 = ₹ 810

Hence the total cost of polishing the floor is ₹ 810.

**13. **Given: Perpendicular distance *(h) = *100 m

Area of the trapezium shaped field = 10500 m^{2}

Let side along the road be x m and side along the river = 2 × m

Hence the side along the river = 2x = 2 × 70 = 140 m.

**14. **Given: Octagon having eight equal sides, each 5 m.

Construction: Divided the octagon in 3 figures, two trapeziums whose parallel and perpendicular sides are 11 m and 4 m respectively and third figure is rectangle having length and breadth 11 m and 5 m respectively.

And Area of rectangle = length × breadth = 11 × 5 = 55 m^{2}

∴ Total area of octagon = 64 + 55 = 119 m^{2}

**15.** First way: By Jyoti’s diagram,

Area of pentagon = Area of trapezium ABCP + Area of trapezium AEDP

Second way: By Kavita’s diagram

Here, a perpendicular AM drawn to BE.

AM = 30 – 15 = 15 m

Area of pentagon = Area of Δ ABE + Area of square BCDE

Hence total area of pentagon shaped park = 337.5 m^{2}.

**16. **Here two of given figures (I) and (II) are similar in dimensions.

And also figures (III) and (IV) are similar in dimensions.

Area of figure (I) = Area of trapezium

Also Area of figure (II) = 96 cm^{2}

Now Area of figure (III) = Area of trapezium

Also Area of figure (IV) = 80 cm^{2}

**17. **(a) Given: Length of cuboidal box *(l) = *60 cm

Breadth of cuboidal box *(b) = *40 cm Height of cuboidal box *(h) = *50 cm

Total surface area of cuboidal box = 2* (lb + bh + hl)*

= 2 (60 × 40 + 40 × 50 + 50 × 60)

= 2 (2400 + 2000 + 3000)

= 2 × 7400 = 14800 cm^{2.}

(b) Given: Length of cuboidal box *(l) = *50 cm

Breadth of cuboidal box *(b) = *50 cm Height of cuboidal box *(h) = *50 cm

Total surface area of cuboidal box = 2 *(lb + bh **+ hl)*

= 2(50 × 50+ 50 × 50 + 50 × 50)

= 2 (2500 + 2500 + 2500)

= 2 × 7500 = 15000 cm^{2}

Hence cuboidal box *(a) *requires the lesser amount of materal to make, since surface area of box *(a) *is less than that of box *(b).*

**18. **Given: Length of suitcase box *(l) = *80 cm, Breadth of suitcase box *(b) = *48 cm

And Height of cuboidal box *(h) = *24 cm

Total surface area of suitcase box = 2 *(lb + bh **+ hl)*

= 2 (80 × 48 + 48 × 24 + 24 × 80)

= 2 (3840 + 1152 + 1920)

= 2 × 6912 = 13824 cm^{2}

Area of Tarpaulin cloth = Surface area of suitcase

Required tarpaulin for 100 suitcases = 144 × 100 = 14400 cm = 144 m

Hence tarpaulin cloth required to cover 100 suitcases is 144 m.

**19. **Here Surface area of cube = 600 cm^{2}

⇒ 6l^{2} = 600 ⇒ l^{2} = 100 ⇒ l = 10 cm

Hence the side of cube is 10 cm

**20. **Here, Length of cabinet *(1) = *2 m, Breadth of cabinet *(b) = *1 m

And Height of cabinet *(h) = *1.5 m

∴ Surface area of cabinet = *lb + *2 *(bh + **hl)*

= 2 × 1 + 2 (1 × 1.5 + 1.5 × 2)

= 2 + 2 (1.5 + 3.0)

= 2 + 9.0

= 11 m^{2}

Hence required surface area of cabinet is 11 m^{2}.

**21. **Here, Length of wall *(l) = *15 m, Breadth of wall *(b) = *10 m And Height of wall *(h) = 7 *m

Total Surface area of classroom = *lb + *2 *(bh + hl)*

= 15 × 10 + 2 (10 × 7 + 7 × 15)

= 150 + 2 (70 + 105)

= 150 + 350

= 500 m^{2}

Hence 5 cans are required to paint the room.

**22.** Given: Diameter of cylinder = 7 cm

Radius of cylinder (r) = 2 – 7 cm

And Height of cylinder (h) = 7 cm

Lateral surface area of cylinder

Now lateral surface area of cube = 4l^{2} = 4 × (7)^{2} = 4 × 49 = 196 cm^{2}

Hence the cube has larger lateral surface area.

**23. ** Given: Radius of cylindrical tank *(r) = *7 m

Height of cylindrical tank *(h) = *3 m

Total surface area of cylindrical tank

Hence 440 m^{2} metal sheet is required.

**24. **Given: Lateral surface area of hollow cylinder = 4224 cm^{2}

And Height of hollow cylinder = 33 cm

Curved surface area of hollow cylinder = 2π*rh*

Now Length of rectangular sheet = 2π*r*

Perimeter of rectangular sheet = 2 (*l* + *b)*

= 2 (128 + 33) = 2 × 161 = 322 cm

Hence perimeter of rectangular sheet is 322 cm.

**25. **Given: Diameter of road roller = 84 cm

Length of road roller *(h) = *1 m = 100 cm

Curved surface area of road roller

∵ Area covered by road roller in 750 revolutions = 26400 × 750 = 1,98,00,000 cm^{2}

= 1980 m^{2} [∵ 1 m^{2} = 10,000 cm^{2}]

Hence the area of the road is 1980 m^{2}.

**26.** Given: Diameter of cylindrical container = 14 cm

∴ Radius of cylindrical container

Height of cylindrical container = 20 cm

Height of the label *(h)= *20 – 2 – 2 = 16 cm

Hence the area of the label of 704 cm^{2}.

**27. **We find area when a region covered by a boundary, such as outer and inner surface area of a cylinder, a cone, a sphere and surface of wall or floor.

When the amount of space occupied by an object such as water, milk, coffee, tea, etc., then we have to find out volume of the object.

(a) Volume (b) Surface are (c) Volume

**28. ** Yes, we can say that volume of cylinder B is greater, since radius of cylinder B is greater than that of cylinder A (and square of radius gives more value than previous).

Diameter of cylinder A = 7 cm ⇒ Radius of cylinder A = cm

And Height of cylinder A = 14 cm

= 539 cm^{3}

Now Diameter of cylinder B = 14 cm

And Height of cylinder B = 7 cm

∴ Volume of cylinder A = = 1078 cm^{3}

Total surface area of cylinder A = π*r (*2*h + r) *[∵* *It is open from top]

Total surface area of cylinder B = π*r (*2*h + r) ** * [∵ It is open from top]

Yes, cylinder with greater volume also has greater surface area.

**29.** Given: Base area of cuboid = 180 cm^{2} and Volume of cuboid = 900 cm^{3}

We know that,

Volume of cuboid = *l* *× **b × h [•.• Base area = lxb= 180 (given)] *

⇒ 900 = 180 *× **h*

Hence the height of cuboid is 5 m.

**30.** Given: Length of cuboid *(l)* = 60 cm, Breadth of cuboid *(b) = *54 cm and Height of cuboid *(h) = *30 cm

We know that, Volume of cuboid = *l × b × h = *60 *×* 54 *×* 30 cm^{3}

And Volume of cube = (Side)^{3} = 6 *×* 6 *×* 6 cm^{3}

Hence required cubes are 450.

**31. **Given: Volume of cylinder = 1.54 m^{3} and Diameter of cylinder = 140 cm

Volume of cylinder = π*r ^{2}h*

Hence height of the cylinder is 1 m.

**32. **Given: Radius of cylindrical tank *(r) = *5 m and Height of cylindrical tank *(h) = *7 m Volume of cylindrical tank = π*r ^{2}h*

*= *×1.5 × 1.5 × 7 = 49.5 cm^{3}

= 49.5 × 1000 liters [∵ m^{3} = 1000 liters]

= 49500 liters

Hence required quantity of milk is 49500 liters.

**33. **(i) Let the edge of cube be

Since, Surface area of the cube (A) = 6*l*^{2} When edge of cube is doubled, then

Surface area of the cube (A’)= 6 (2*l*)^{2} = 6 × 41^{2} = 4 × 61^{2}

A’ = 4 × A

Hence surface area will increase four times.

(ii) Volume of cube (V) = *l*^{3}

When edge of cube is doubled, then Volume of cube (V’)= (2*l)*^{3} = 8*l ^{3}*

V’ = 8 × V

Hence volume will increase 8 times.

**34. **Given: volume of reservoir = 108 m^{3}

Rate of pouring water into cuboidal reservoir = 60 liters/minute

∵ m^{3 }water filled in reservoir will take = 1 hour

∴ 1 m^{3} water filled in reservoir will take

∴ 108 m^{3 }water filled in reservoir will take hours = 30 hours

It will take 30 hours to fill the reservoir.