Surface Areas and Volumes: 9 Math
NCERT Exercise 13.1 part 2: 9th math
Surface Areas And Volumes Class nine Math NCERT Exercise 13.1 Question(5) A cubical box has each edge 10 cm and another cuboidal box is 12.5cm long, 10 cm wide and 8 cm high.
(i) Which box has the greater lateral surface area and by how much?
(ii) Which box has the smaller total surface area and by how much?
Solution
Given, Length of cubical box = 10 cm = Edge of cubical box
Length of cuboidal box = 12.5 cm
Breadth of cuboidal box = 10 cm
Height of cuboidal box = 8 cm
(i) Which box has the greater lateral surface area and by how much?
Calculation of Lateral Surface Area of Given Cube
We know that, Lateral surface area of a cube `= 4 l^2`
= 4 × (10 cm)2
= 4 × 100 cm2
= 400 cm2
Calculation of Lateral Surface Area of Given Cuboidal
We know that, Lateral surface area of a cuboidal `= 2 h (l +b)`
Therefore, lateral surface area of given cuboidal
= 2 × 8 cm (10 cm + 12.5 cm)
= 16 cm × 22.5 cm
= 360 cm2
Here it is clear that the lateral surface area of cube is greater than that of cuboidal box.
Therefore, Lateral surface area cube is greater by = 400 cm2 – 360 cm2
= 40 cm2
Thus, cube has greater surface area by 40 cm2 Answer
(ii) Which box has the smaller total surface area and by how much?
Calculation of total surface area of given cube
We know that, Total surface area of a cube = `6l^2`
Therefore, total surface area of given cubical box
= 6 × (10 cm)2
= 6 × 100 cm2
= 600 cm2
Thus, total surface area of given cubical box = 600 cm2
Calculation of total surface area of given cuboidal box
We know that, total surface area of a cuboid `= 2(lb + bh + hl)`
Thus, total surface area of given cuboidal box
= 2 [(12.5 cm × 10 cm) + (12.5 cm × 8 cm) + (10 cm × 8 cm)]
= 2 (125 cm2 + 100 cm2 + 80 cm2)
= 2 × 305 cm2
= 610 cm2
Thus, total surface area of given cuboidal box = 610 cm2
Here clearly the total surface area of cuboidal box is greater than that of cubical box.
Therefore, Total surface area of cube is smaller by
= 610 cm2 – 600 cm2
= 10 cm2
Thus, cubical box has smaller total surface area by 10 cm2 Answer
Surface Areas And Volumes Class nine Math NCERT Exercise 13.1 Question(6) A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. 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
Given,
Length of the green house = 30 cm
Breadth of the green house = 25 cm
Height of the green house = 25 cm
Thus, (i) What is the area of the glass?
Here total surface area is equal to the area of the glass used in herbarium.
Now, we know that,
Total surface area of a cuboid `= (lb + bh + hl)`
Thus, total surface area of the given cuboid
= 2 [(30 cm × 25 cm) + (25 cm × 25 cm) + (30 cm × 25 cm)]
= 2 (750 cm2 + 625 cm2 + 750 cm2)
= 2 × 2125 cm2
= 4250 cm2
Therefore, total surface area of the glass of the given herbarium = total area of the glass = 4250 cm2
Thus, total surface are of the glass = 4250 cm2 Answer
(ii) How much of tape is needed for all the 12 edges?
Now, the length of total 12 edges = length of tape needed for all the 12 edges
Now, length of 12 edges of the herbarium `= 4h + 4b + 4l`
`= 4 (h + b + l)`
= 4 (30 cm + 25 cm + 25 cm)
= 4 × 80 cm
= 320 cm
Thus, length of tape needed for all the 12 edges = 320 cm Answer
Surface Areas And Volumes Class nine Math NCERT Exercise 13.1 Question(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 × 12cm ×5cm. 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
Calculation for bigger box
Given, Dimension of bigger box
Length = 25 cm
Breadth = 20 cm
And, Height = 5 cm
And dimension of
Now, we know that Total surface area of a cuboid `= 2 ( bl +bh + hl)`
Thus, total surface area of given big box
= 2 [(25 cm × 20 cm) + (20 cm × 5 cm) + (25 cm × 5 cm)]
= 2 (500 cm2 + 100 cm2 + 125 cm2)
= 2 × 725 cm2
= 1450 cm2
Thus, total surface area of one given bigger box = 1450 cm2
Now, 5% of extra cardboard is required due to overlap
Thus total cardboard needed to make one bigger box
= Total surface area of one bigger + 5% of total surface area of one bigger box
Thus, 5% of the total surface area of one cardboard =
`=1450 cm^2 + (1450 cm^2 xx5)/100`
= 1450 cm2 + 5 × 14.5 cm2
= 1450 cm2 + 72.5 cm2
= 1522.5 cm2
That is total cardboard needed to make one bigger box = 1522.5 cm2
Therefore, area for making of 250 bigger boxes
= 250 × 1522.5 cm2
= 380625 cm2
Thus, total cardboard needed to make 250 bigger boxes = 380625 cm2
Calculation for Smaller box
Given,
Length of smaller box = 15 cm
Breadth of smaller box = 12 cm
Height of smaller box = 5 cm
Now, we know that, Total surface area of a cuboid `= 2 ( bl + bh + hl)`
Thus, total surface area of given smaller box
= 2 (15 cm × 12 cm + 12 cm × 5 cm + 5 cm × 15 cm)
= 2 (180 cm2 + 60 cm2 + 75 cm2)
= 2 × 315 cm2
= 630 cm2
Thus, total surface area of one smaller box = 630 cm2
Now, as given in question, due to cardboard overlaps 5% is required extra
Thus, cardboard needed to make one smaller box
= total surface area of one smaller box +5% of total surface area of one smaller box
`= 630 cm^2 + 5/100 xx 630cm^2`
= 630 cm2 + 5 × 6.3 cm2
= 630 cm2 + 31.5 cm2
= 661.5 cm2
Thus, total cardboard need to make one smaller box = 661.5 cm2
Now, cardboard needed to make 250 smaller boxes
= 250 × 661.5 cm2
= 165375 cm2
Thus, total cardboard needed to make 250 smaller boxes = 165375 cm2
Now, total area of cardboard needed to make all the bigger and smaller boxes
= Cardboard needed to make 250 bigger boxes + cardboard needed to make 250 smaller boxes
= 380625 cm2 + 165375 cm2
= 546000 cm2
Thus, total cardboard needed to make all the boxes = 546000 cm2
Calculation of cost of cardboard
As given in question, the cost is Rs 4 for per 1000 cm2 of cardboard.
Cost of cardboard required for supplying for 250 boxes of each kind @ 4/1000 cm2
`= 546000 cm^2xx (4/(1000cm^2))`
= Rs 2184
Thus, cost of cardboard for supplying the given number of boxes = Rs 2184 Answer
Surface Areas And Volumes Class nine Math NCERT Exercise 13.1 Question(8) Parveen wanted to make a temporary shelter for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m with base dimensions 4 m × 3 m?
Solution:
Given,
Length of the shelter = 4 cm
Breadth of the shelter = 3 m
Height of the shelter = 2.5 m
Thus, area of tarpaulin required to make the given shelter = ?
The area of tarpaulin required to make the shelter = total surface area of the shelter
Now, we know that, total surface area of a cuboidal shaped box without base. Since the base of the shelter is not to be covered by tarpaulin.
`= lb + 2lh + 2bh`
= (4 m × 3 m) + (2 × 4 m × 2.5 m) + (2 × 3 m × 2.5 m)
= 12 m2 + 20 m2 + 15 m2
= 47 m2
Thus, tarpaulin needed to make the shelter = 47 m2 Answer
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