Sunday, 22 January 2017

CBSE Class 9 Maths - SURFACE AREAS AND VOLUMES (NCERT EXERCISE 13.1) (#cbsenotes) (#ncertsolutions)

SURFACE AREAS AND VOLUMES

Class 9 Maths 

NCERT EXERCISE 13.1

CBSE Class 9 Maths - SURFACE AREAS AND VOLUMES (NCERT EXERCISE 13.1) (#cbsenotes) (#ncertsolutions)


Q1: 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.

Answer: Given,
l = 1.5 m,
b = 1.25 m,
h = 65 cm = 0.65 m

(i) ∵ the box is open at the top, it has only five faces.

Surface Area (SA) for five faces = lb + 2(bh + hl)

= 1.5 × 1.25 m² + 2 (1.25 × 0.65 + 0.65 × 1.5) m²
= 1.875 + 2 (1.7875) m² = 1.875 + 3.575
= 5.45 m²

(ii) Cost of 1 m² of the sheet = ₹ 20
∴ cost of 5.45 m² of the sheet = 20 × 5.45 m² = ₹ 109 (Answer)


Q2: 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.



Answer: Given, l = 5 m, b = 4 m, h = 3 m
SA for 4 walls and ceiling = 2hl + 2 hb + lb 
    = 2h (l + b) + lb
    = [2 × 3 (5 + 4) + 5 × 4] m²
    = 54 + 20 = 74 m²

Cost of white washing = ₹ 7.50 per m²
∴ total cost of white washing the walls and the ceiling = 7.50 ×  74 = ₹ 555


Q3: 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 15000, find the height of the hall.
[Hint : Area of the four walls = Lateral surface area.]

Answer: Let length = l
breadth = b
height = h
Perimeter of the floor of the hall = 2 (l + b) = 250 m.
Area of the four walls of the hall (LSA) = 2h (l + b)         ... (i)

Also area of four walls = ₹15000 / ₹10 = 1500m²               ... (ii)

From ⅰ and ⅱ, we have
2h (l + b) = 1500
⇒   h × 2 (l + b ) = 1500
⇒   h × 250 = 1500
h = 1500 / 250 = 6m


Q4: The paint in a certain container is sufficient to paint an area equal to 9.375 m². How many bricks of dimensions 22.5 cm × 10 cm × 7.5 cm can be painted out of this container?

Answer: Given,
l = 22.5 cm
b = 10 cm
h = 7.5 cm
Total surface area (TSA) of 1 brick = 2 (lb + bh + hl)
= 2(22.5 × 10 + 10 × 7.5 + 7.5 × 22.5) cm²
= 2(225 + 75 + 168.75) cm² = 937.5 cm² = 937.5 / 10000 m² = 0.09375m²

∴ required number of bricks = 9.375/0.09375 = 100 (Answer) 


Q5: A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm 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?


Answer: Given,  a = 10 cm, l = 12.5 cm, b = 10 cm, h = 8 cm

(i) LSA of cube = 4a²
= 4 × 10 × 10 = 400 cm²

LSA of the cuboidal box = 2h (l + b)
= 2 × 8 (12.5 + 10)
= 16 × 22.5
= 360 cm²
Difference in LSAs = 400 - 360 = 40cm²
Hence, the cubical box has greater lateral surface area by 40cm² (Answer)

(ii) TSA of a cube = 6a²
= 6 × 10 × 10 = 600 cm²

TSA of a cuboid box = 2(lb + bh + hl)
= 2 (12.5 × 10 + 10 × 8 + 8 × 12.5)
= 2 × 305
=  610 cm²
Difference of TSAs = 610 - 600 = 10 cm²
Hence, the cubical box has smaller total surface area by 10 cm² (Answer)

CBSE Class 9 Maths - SURFACE AREAS AND VOLUMES (NCERT EXERCISE 13.1) (#cbsenotes) (#ncertsolutions)

Q6: 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?

Answer: Given,
l = 30 cm,
b = 25 cm,
h = 25 cm.

(i) TSA of the herbarium (cuboid) = 2(lb + bh + hl)
= 2(30 × 25 + 25 × 25 + 25 × 30)
= 2(750 + 625 + 750)
= 2 × 2125 cm2 = 4250
Thus, area of the glass = 4250 cm² (Answer)

(ii) A cuboid has 12 edges. It consist of 4 lengths, 4 breadths and 4 heights.
∴ length of the tape required = 4l + 4b + 4h
= (4 × 30 + 4 × 25 + 4 × 25)
= (120 + 100 + 100) cm = 320 cm (Answer)


Q7: 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 dimesnsions 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.

Answer:
For Bigger Boxes:
l = 25 cm, b = 20 cm, h = 5 cm
TSA  = 2(lb + bh + hl)
= 2(25 × 20 + 20 × 5 + 5 × 25)
=  (500 + 100 + 125) cm2 = 1450 cm²

Area of cardboard required for overlaps = 5% of TSA
= (5/100) × 1450
= 72.5 cm²

Total area of cardboard need of 1 bigger box = 1450 + 72.5 = 1522.5 cm²
Total area of cardboard needed for 250 bigger boxes = 1522.5 × 250 = 380625 cm²

For smaller boxes:
l = 15 cm, b = 12 cm, h = 5 cm
TSA = 2(lb + bh + hl)
= 2(15 × 12 + 12 × 5 + 5 × 15)
= 2 (180 + 60 + 75)  = 630 cm²

Area required for overlaps = 5% of 630 = (5/100)× 630 = 31.5 cm²
Total area of cardboard needed for 1 smaller box = 630 + 31.5 = 661.5 cm²
Total area of cardboard needed for 250 smaller boxes = 661.5 ×  250 = 165375cm²


Total cardboard required for 500 boxes (big and small) = 380625 + 165375 = 546000cm²

Cost of 1000 cm² of cardboard = ₹ 4
∴ cost of 546000 cm² cardboard = (4/1000) × 546000 = ₹2184 


Q8: 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?

Answer:
Given, l = 4 m, b = 3 m, h = 2.5 m
Tarpaulin needed to cover 5 faces of cuboid (Surface Area) = lb + 2 (bh + hl)

= 4 × 3 + 2(3 × 2.5 + 2.5 × 4)
= 12  + 2(7.5 + 10)
= (12 + 35)  = 47 m²

Thus, 47 m² of tarpaulin is required to make the shelter 

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