9th Maths 13.1
Chapter 13
Surface Areas and Volumes
NCERT Class 9th solution of Exercise 13.2
NCERT Class 9th solution of Exercise 13.3
NCERT Class 9th solution of Exercise 13.4
NCERT Class 9th solution of Exercise 13.5
NCERT Class 9th solution of Exercise 13.6
NCERT Class 9th solution of Exercise 13.7
NCERT Class 9th solution of Exercise 13.8
Exercise 13.1
Q1. A plastic box 1.5m long, 1.25m wide and 65cm deep is to be made. It is opened 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 1m2 costs ₹20.
Sol. :
Given:
L=1.5m, B=1.25m, H=65cm=0.65m
To Find:
Area of the sheet.
Solve:
i) Surface Area of box open at the top
SA=2(L+B)H+LB
SA=2(1.5+1.25)0.65+1.5×1.25
SA=2×2.75×0.65+1.875
SA=3.575+1.875
SA=5.45m2
Answer:
Surface Area of the sheet required to making the box is 5.45m2.
ii) Cost=Area×Rate
Cost=₹5.45×20
Cost=₹109
Answer:
Cost of the sheet ₹109.
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 ₹7.50 per m2.
Sol. :
Given:
L=5m, B=4m, H=3m and Rate=₹7.50 per m2.
To Find:
Cost of white washing.
Solve:
Surface Area of room and ceiling
SA=2(L+B)H+L×B
SA=2(5+4)3+5×4
SA=2×9×3+20
SA=54+20
SA=74m2
Cost=Area×Rate
Cost=₹74×7.50
Cost=₹555.
Answer:
Cost of white washing of the room and ceiling ₹555.
Q3. The floor of a rectangular hall has a perimeter 250m. If the cost of painting the four walls at the rate of ₹10 per m2 is ₹15000, find the height of the hall.
[Hint: Area of the four walls=lateral surface area.]
Sol. :
Given:
Perimeter=250m, Rate=₹10, Total cost=₹15000
To Find:
The Height of the hall.
Solve:
Area of four walls=TotalcostRate
Area of four walls=1500010
Area of four walls=1500m2
Area of four walls = lateral surface area
Area of four walls=2(L+B)H
1500=250×H
H=1500250
H=6m
Answer:
The Height of the hall is 6m.
Q4. The paint in a certain container is sufficient to paint an area equal to 9.375m2. How many bricks of dimensions 22.5cmtimes10cmtimes7.5cm` can be painted out of this container?
Sol. :
Given:
Painted Area=9.375m2 and
Dimensions of Bricks L=22.5cm=22.5100m,
B=10cm=10100m, H=7.5cm=7.5100m
To Find:
The number of bricks can be painted.
Solve:
SA of Brick=2(LB+BH+HL)
SA=2(22.5100×10100+10100×7.5100+7.5100×22.5100)
SA=0.09375m2
No. of Bricks =(Painted Area)/(Area of a Brick)
No. of Bricks=9.3750.09375
No. of Bricks=9375×1000009375×1000
No. of Bricks=100
Answer:
100 bricks can be painted.
Q5. A cubical box has each edge 10cm and another cuboidal box is 12.5cm long, 10cm wide and 8cm 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?
Sol. :
Given:
Edge of cubical box=10cm, and Cuboidal box of L=12.5,B=10cm, and H=8cm.
To Find:
Lateral Surface Area.
Solve:
i) Lateral SA of Cubical box=4a2
LSA of Cubical box=4(10)2=4×10×10=4×100
=400cm2
Lateral Surface Area of Cuboidal box=2(L+B)H
LSA of Cuboidal box=2(12.5+10)8
LSA of Cuboidal box=2(22.5)8=360.0cm2
Difference of Lateral Surface Area=400-360=40cm2
Answer:
The Cubical box has greater Surface Area by 40cm2.
ii) Total Surface Area of Cubical box=6a2
Total Surface Area of Cubical box=6(10)2=600cm2.
Total Surface Area of Cuboidal box=2(LB+BH+HL)
Total Surface Area of Cuboidal box=2(12.5×10+10×8+8×12.5)
Total Surface Area of Cuboidal box=2(125+80+100)
Total Surface Area of Cuboidal box=2×305=610cm2
Difference of Total Surface Area=610-600=10cm2
Answer:
The Cubical box has smaller Surface Area by 10cm2.
Q6. A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30cm long, 25cm wide and 25cm high.
i) What is the area of the glass?
ii) How much of tape is needed for all the 12 edges?
Sol. :
Given:
L=30cm, B=25cm. H=25cm
To Find:
i) Area of the glass.
ii) How much tape is needed.
Solve:
i)
Area of glass=2(LB+BH+HL)
Area of glass=2(30×25+25×25+25×30)
Area of glass=2(750+625+750)
Area of glass=2×2125
Area of glass=4250cm2
Answer:
Area of glass is 4250cm2.
ii)
The length of Tape=4(30+25+25)
The length of Tape=4×80=320cm.
Answer:
The length of Tape 320cm.
Q7. Shanti Sweets Stall was placing an order for making cardboard boxes for packing 25cm×20cm×5cm and the smaller of dimensions 15cm×12cm×5cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is ₹4 for 1000cm2, find the cost of cardboard required for supplying 250 boxes of each kind.
Sol. :
Given:
Big Cardboard box of L=25cm, B=20cm, H=5cm
Small Cardboard box of L=15cm, B=12cm, H=5cm
To Find:
Cost of cardboard
Solve:
Total Surface Area of a Big box=2(LB+BH+HL)
Total Surface Area of a Big box=2(25×20+20×5+5×25)
Total Surface Area of a Big box=2(500+100+125)
Total Surface Area of a Big box=2×725=1450cm2
Total Surface Area of a small box=2(LB+BH+HL)
Total Surface Area of a small box=2(15×12+12×5+5×15)
Total Surface Area of a small box=2(180+60+75)=2×315
=630cm2
Total cardboard required for 250 boxes=250(1450+630)105100
Total cardboard required for 250 boxes=250×2080×105100=546000cm2
Cost of cardboard=Area×Rate
Cost of cardboard=546000×41000=₹2184
Answer:
Cost of cardboard ₹2184.
Q8. Praveen 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 2.5m, with base dimensions 4m×3m?
Sol. :
Given:
L=4m, B=3m, H=2.5m
To Find:
Area of Tarpaulin.
Solve:
The Area of four sides and top of structure=2(L+B)H+LB
The Area of four sides and top of structure=2(4+3)2.5+4×3
The Area of four sides and top of structure=35+12=47m2
Answer:
Area of Tarpaulin is 47m2.
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