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 `1m^2` 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+1.5times1.25`
`SA =2times2.75times0.65+1.875`
`SA =3.575+1.875`
`SA =5.45m^2`
Answer:
Surface Area of the sheet required to making the box is `5.45m^2.`
ii) `Cost = AreatimesRate`
`Cost =₹5.45times20`
`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 `m^2`.
Sol. :
Given:
`L=5m,` `B=4m,` `H=3m` and Rate`=₹7.50` per `m^2`.
To Find:
Cost of white washing.
Solve:
Surface Area of room and ceiling
`SA=2(L+B)H+LtimesB`
`SA=2(5+4)3+5times4`
`SA=2times9times3+20`
`SA=54+20`
`SA=74m^2`
`Cost=AreatimesRate`
`Cost=₹74times7.50`
`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 `m^2` 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`=(Total cost)/(Rate)`
Area of four walls`=15000/10`
Area of four walls`=1500m^2`
Area of four walls = lateral surface area
Area of four walls`=2(L+B)H`
`1500=250timesH`
`H=1500/250`
`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 `9.375m^2`. How many bricks of dimensions 22.5cmtimes10cmtimes7.5cm` can be painted out of this container?
Sol. :
Given:
Painted Area`=9.375m^2` and
Dimensions of Bricks `L=22.5cm=(22.5)/(100)m,`
`B=10cm=(10)/(100)m,` `H=7.5cm=(7.5)/(100)m`
To Find:
The number of bricks can be painted.
Solve:
`SA` of Brick`=2(LB+BH+HL)`
`SA=2((22.5)/(100)times(10)/(100)+(10)/(100)times(7.5)/(100)+(7.5)/(100)times(22.5)/(100))`
`SA=0.09375m^2`
No. of Bricks =(Painted Area)/(Area of a Brick)
No. of Bricks`=(9.375)/(0.09375)`
No. of Bricks`=(9375times100000)/(9375times1000)`
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`=4a^2`
`LSA` of Cubical box`=4(10)^2=4times10times10=4times100`
`=400cm^2`
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=360.0cm^2`
Difference of Lateral Surface Area`=400-360=40cm^2`
Answer:
The Cubical box has greater Surface Area by `40cm^2`.
ii) Total Surface Area of Cubical box`=6a^2`
Total Surface Area of Cubical box`=6(10)^2=600cm^2.`
Total Surface Area of Cuboidal box`=2(LB+BH+HL)`
Total Surface Area of Cuboidal box`=2(12.5times10+10times8+8times12.5)`
Total Surface Area of Cuboidal box`=2(125+80+100)`
Total Surface Area of Cuboidal box`=2times305=610cm^2`
Difference of Total Surface Area`=610-600=10cm^2`
Answer:
The Cubical box has smaller Surface Area by `10cm^2`.
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(30times25+25times25+25times30)`
Area of glass`=2(750+625+750)`
Area of glass`=2times2125`
Area of glass`=4250cm^2`
Answer:
Area of glass is `4250cm^2`.
ii)
The length of Tape`=4(30+25+25)`
The length of Tape`=4times80=320cm.`
Answer:
The length of Tape `320cm`.
Q7. Shanti Sweets Stall was placing an order for making cardboard boxes for packing `25cmtimes20cmtimes5cm` and the smaller of dimensions `15cmtimes12cmtimes5cm`. For all the overlaps, `5%` of the total surface area is required extra. If the cost of the cardboard is `₹4` for `1000cm^2`, 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(25times20+20times5+5times25)`
Total Surface Area of a Big box`=2(500+100+125)`
Total Surface Area of a Big box`=2times725=1450cm^2`
Total Surface Area of a small box`=2(LB+BH+HL)`
Total Surface Area of a small box`=2(15times12+12times5+5times15)`
Total Surface Area of a small box`=2(180+60+75)=2times315`
`=630cm^2`
Total cardboard required for `250` boxes`=250(1450+630)(105)/(100)`
Total cardboard required for `250` boxes`=250times2080times(105)/(100)=546000cm^2`
Cost of cardboard`=AreatimesRate`
Cost of cardboard`=546000times4/(1000)=₹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 `4mtimes3m`?
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+4times3`
The Area of four sides and top of structure`=35+12=47m^2`
Answer:
Area of Tarpaulin is `47m^2`.
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