How do you find the Surface Area and Volume of a Cuboid
If ℓ, b and h denote respectively the length, breadth and height of a cuboid, then
- Total surface area of the cuboid = 2 (ℓb + bh + ℓh) square units.
- Volume of the cuboid = Area of the base × height = ℓ × b × h cubic units.
where base area = Breadth × length - Diagonal of the cuboid or longest rod = \(\sqrt {{\ell ^2} + {b^2} + {h^2}} \) units. and Total length of its edges = 4 (ℓ + b + h)
- Area of four walls of a room = 2 (ℓ + b) h sq. units.
Surface Area and Volume of a Cuboid Example Problems with Solutions
Example 1: The length, breadth and height of a cuboid are in the ratio 6 : 4 : 5. If the total surface area of the cuboid is 2368 cm2 ; find its dimension.
Solution: Let length (ℓ) = 6x cm, breadth (b) = 4x cm and height (h) = 5x cm,
∴ Total surface area
= 2(ℓ × b + b × h × h × ℓ)
= 2(6x × 4x + 4x × 5x + 5x × 6x)cm2
= 2(24x2 + 20x2 + 30x2) cm2
=148x2 cm2
Given : Total surface = 2368 cm2
⇒ 148x2 = 2368
⇒ x2 = \(\frac{{2368}}{{148}}\) = 16
and x = \(\sqrt {16} = 4\)
∴ length = 6x cm = 6 × 4 cm = 24 cm,
breadth = 4x cm = 4 × 4 cm = 16 cm and
height = 5x cm = 5 × 4 cm = 20 cm
Example 2: A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is to be open at the top. Ignoring the thickness of the plastic sheet, determine :
(i) The area of the sheet required for making the box.
(ii) The cost of sheet for it, if a sheet measuring 1 m2 costs Rs. 20.
Solution: Given. length (ℓ) = 1.5 m, breadth (b) = 1.25 m and depth i.e., height (h) = 65 cm = 0.65 m.
(i) Since, the box is open it has five faces in which four faces are the walls forming lateral surface area and one face is the base.
∴ The area of the sheet required for making the box.
= Lateral surface area of the box + area of its base
= 2(ℓ + b) × h + ℓ × b
= 2(1.5m + 1.25m) × 0.65m + 1.5m × 1.25 m
= 2 × 2.75 m × 0.65 m + 1.875 m2
= 3.575 m2 + 1.875 m2 = 5.45 m2
(ii) Since, a sheet measuring 1 m2 costs Rs.20
∴ The cost of sheet for the box
= 5.45 × Rs 20 = Rs. 109
Example 3: The length, breadth and height of a room are 5m, 4m and 3m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs. 7.50 per m2.
Solution: Given. ℓ = 5m, b = 4m and h = 3m
Since, the area of the walls of the room
= its lateral surface area = 2(ℓ + b) × h
And, the area of the ceiling of the room
= ℓ × b
∴ Total area to be white washed
= Area of the walls + area of the ceiling
= 2(ℓ + b) × h + ℓ × b.
= 2(5m + 4m) × 3m + 5m × 4m
= 2 × 9m × 3m + 20m2
= 54m2 + 20m2 = 74m2
Since, the rate of white washing
= Rs. 7.50 per m2
∴ Cost of white washing
= 74 × Rs.7.50 = Rs. 555
Example 4: The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of Rs. 10 per m2 is Rs : 15,000, find the height of the hall.
Solution: We know, the perimeter of a rectangle
= 2(length + breadth) = 2 (ℓ + b)
And, given the floor of a rectangular hall has perimeter 250 m
⇒ 2(ℓ + b) = 250 m
Area of the four walls of the hall
= Lateral surface area of the hall
= 2(ℓ + b) × h = 250 m × hm = 250 hm2
Since, the rate of painting four walls is
Rs. 10 per m2.
∴ Cost of painting 250 hm2 = 250 h × Rs.10
According to the given statements :
250 h × Rs. 10 = Rs. 15,000
⇒ h = \(\frac{{15,000}}{{250 \times 10}}m\) = 6 m
∴ The height of the hall = 6 m
Example 5: The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimension 22.5 cm × 10 cm × 7.5 cm can be painted out of this container.
Solution: For each brick, ℓ = 22.5 cm, b = 10 cm and h = 7.5 cm.
∴ Surface area of each brick
= 2 (ℓ × b + b × h + h × ℓ)
= 2 (22.5 × 10 + 10 × 7.5 + 7.5 × 22.5)cm2
= 2 (225 × 75 + 168.75) cm2 = 937.5 cm2
Since, the paint in the container is sufficient to paint an area.
= 9.375 m2 = 9.375 × 100 × 100 cm2
[1 m = 100 cm and 1 m2 = 100 × 100 cm2]
= 93750 cm2
∴ Number of bricks that can be painted
= \(\frac{{Total\,\,area\,\,which\,\,can\,\,be\,\,painted}}{{Surface\,\,area\,\,of\,\,the\,\,brick}}\)
= \(\frac{{\,\,\,93750\,\,c{m^2}}}{{937.5\,\,cm}} = 100\)
Example 6: A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tap. It is 30 cm long, 25 cm wide and 25 cm high.
(i) What is the area of the glass?
(ii) How much of tape is needed for all the 12 edges?
Solution: The green house is a cuboid in shape in which ℓ = 30 cm, b = 25 cm and h = 25 cm
(i) Area of the glass
= Surface area of cubical green house.
= 2(ℓ × b + b × h + h × ℓ)
= 2(30 × 25 + 25 × 25 + 25 × 30) cm2
= 2(750 + 625 + 750)cm2 = 4250 cm2
(ii) Length of the tape
= Length of the 12 edges of the cubical greenhouse = 4(ℓ + b + h)
= 4(30 + 25 + 25)cm = 320 cm
Example 7: Shanti sweets stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is Rs. 4 for 1000 cm2, find the cost of cardboard required for supplying 250 boxes of each kind.
Solution: For each bigger box :
ℓ = 25 cm, b = 20 cm and h = 5 cm
∴ Surface area
= 2(ℓ × b + b × h + h × ℓ)
= 2(25 × 20 + 20 × 5 + 5 × 25)cm2
= 2(500 + 100 + 125)cm2 = 1450 cm2
For each smaller box :
ℓ = 15 cm, b = 12 cm and h = 5 cm
∴ Surface area
= 2(ℓ × b + b × h + h × ℓ)
= 2(15 × 12 + 12 × 5 + 5 × 15)cm2
= 2(180 + 60 + 75) cm2 = 630 cm2
Total surface area of 250 boxes of each kind
= 250 × 1450 cm2 + 250 × 630 cm2
= 362500 cm2 + 157500 cm2
= 520000 cm2
Cardboard required for overlaps
= 5% of 520000 cm2
= \(\frac{5}{{100}} \times 52000c{m^2}\) = 26000 cm2
∴ Total area of the cardboard used
= 520000 cm2 + 26000 cm2
= 546000 cm2
Given, cost of 1000 cm2 of cardboard = Rs.4
⇒ cost of 1 cm2 of cardboard = Rs. \(\frac{4}{{1000}}\)
⇒ cost of 546000 cm2 of cardboard
= Rs. \(\frac{4}{{1000}}\) × 546000 = Rs. 2184
Example 8: The volume of a cubical solid is 3240 cm3, find, its
(i) height, if length =18 cm and breadth = 15cm
(ii) breadth, if length =24 cm and height = 10cm
(iii) length, if breadth = 9cm and height = 20cm
Solution: Volume = length × breadth × height
⇒ (i) height = \(\frac{{Volume}}{{length \times breadth}}\)
(ii) breadth = \(\frac{{Volume}}{{length \times height}}\)
(iii) Length = \(\frac{{Volume}}{{breadth \times height}}\)
(i) Height = \(\frac{{Volume}}{{length \times breadth}}\)
= \(\frac{{3240\,\,c{m^3}}}{{18\,\,cm \times 15\,\,cm}}\) = 12 cm
(ii) Breadth = \(\frac{{Volume}}{{length \times height}}\)
= \(\frac{{3240\,\,c{m^3}}}{{24\,\,cm \times 10\,\,cm}}\) = 13.5 cm
(iii) Length = \(\frac{{Volume}}{{breadth \times height}}\)
= \(\frac{{3240\,\,c{m^3}}}{{9\,\,cm \times 20\,\,cm}}\) = 18 cm
Example 9: A matchbox measures 6 cm × 4 cm × 2.5 cm. What will be the volume of a packet containing 24 such boxes ?
Solution: The shape of a matchbox is a cuboid
∴ Volume of 1 matchboxes
= its length × breadth × height
= 6 × 4 × 2.5 cm2 = 60 cm3
⇒ Volume of a packet containing 24 such boxes
= Volume of 24 matchboxes
= 24 × 60 cm3 = 1440 cm3
Example 10: A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many liters of water can it hold? (1 m2 = 1000 ℓ)
Solution: Volume of water which tank can hold
= Volume of the tank
= Its length × width × height
= 6 m × 5 m × 4.5 m = 135 m3
= 135 × 1000 ℓ = 135000 ℓ
Example 11: Find the cost of digging a cuboidal pit 10 m long, 7.5 m broad and 80 cm deep at the rate of Rs. 16 per m3.
Solution: Given :
Length of the pit = 10 m, breadth = 7.5 m and depth = 80 cm = 0.8 m
∴ Volume of cuboidal pit
= Its length × breadth × depth (or height)
= 10 m × 7.5 m × 0.8 m = 60 m3
Rate of digging the pit = Rs. 16 per m3
∴ Cost of digging 60 × Rs. 16 = Rs. 960
Example 12: Wooden crates each measuring 1.5 m × 1.25 m × 0.5 m are to be stores in a godown. Find the largest number of wooden crates which can be stores in a godown measuring :
(i) 45 m × 25 m × 10m.
(ii) 40 m × 25 m × 10m
Solution: (i) Since, the length of godown is 45 m and the length of a crate is 1.5 m
∴ The largest no. of crates that can stores along the length of the godown
= \(\frac{{45m}}{{1.5m}} = 30\)
Similarly, the largest no. of crates stores along the width of the godown
= \(\frac{{25m}}{{1.25m}} = 20\)
And, the largest no. of crates stored along the length of the godown
= \(\frac{{10m}}{{0.5m}} = 20\)
∴ The largest no. of crates which can be stored in the godown
= 30 × 20 × 20
= 12000
(ii) No. of crates along the length
= \(\frac{{40m}}{{1.5m}} = 26\frac{2}{3}\) = 26
[NO. of crates can not be in fraction]
No. of crates along the width
= \(\frac{{25m}}{{1.25m}} = 20\)
No. of crates along the height
= \(\frac{{10m}}{{0.5m}} = 20\)
∴ The largest no. of crates which can be stored in the godown
= 26 × 20 × 20
= 10400
Example 13: The capacity of a cuboidal tank is 50000 liters of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.
Solution: The capacity of cuboidal tank is 50000 liters of water.
⇒ Volume of the tank = 50000 liters
= \(\frac{{50000}}{{1000}}{m^3}\) [∵ 1 m3 = 1000 liters]
= 50 m3
⇒ Length of the tank × its breadth × its height = 50 m3
⇒ 2.5 m × breadth × 10 m = 50 m3
⇒ Breadth = \(\frac{{50}}{{2.5 \times \,\,10}}m = 2m\)