mensuration class 9

Exercise 1.

1 A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is opened at the top. Ignoring the thickness of the plastic sheet, determine a) The area of the sheet required for making the box. b) The cost of sheet for it, if a sheet measuring 1 m² cost Rs 20.

Solution : Given length=l=1.5 m, width=1.25 m and depth=h=65 cm= 0.65m

Given it is to be open at the top.

a) Surface are of the box= Total surface area- Area of the top.

Area of the box=2(lb+bh+hl)-lb

=2[(1.5 ×1.25)+(1.25 ×0.65)+(0.65 ×1.5)]-(1.5 ×1.25)

=2(1.875+0.8125+0.975)-1.875

=2(3.6625)-1.875

=7.325-1.875=5.45 m².

b) Given Cost of 1m² of the sheet= Rs 20

∴ Cost of 5.45 m² of the sheet=Rs 20 ×5.45 m²= Rs 109.

2 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 length=l=5 m, breadth=b=4m and height=h=3 m.

Lateral surface area of cuboid=2h(l+b)

Surface area of the walls of the room+Ceiling area=2h(l+b)+lb

=2×3(5+4)+(5 ×4) =(6 ×9)+20=74 m²

Given cost of white washing =7.50 per m²

∴ Total cost of white washing the walls and the ceiling of the room= 74 × 7.50= Rs 555

3 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 Rs 15,000, find the height of the hall.

Solution: Lett length, breadth and height of the hall be ‘l’,’b’ and ‘h’ respectively.

Given parimeter of the hall=2(l+b)=250 m.—(1)

Area of the four walls=(Total cost of painting the four walls)÷(cost of painting per m²)

=15000÷10=1500 m²

∴ 2h(l+b)=1500

250h=1500 {∵ from (1)}

h=1500÷250=6

height of the wall =6m

4 The paint in a certain container is sufficient to point 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?

Solution : Given paint in a certain container is sufficient to paint an area =9.375 m²=9.375×100×100 cm² {∵1m²=10000cm²}

=93750 cm².

Given l=22.5 cm, b=10 cm, h=7.5 cm.

Total surface area 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.

∴ By using the Container number of bricks to be painted=

$\frac{93750}{937.5}=100$

bricks.

5 A cubical box has each edge 10cm and another cuboidal box is 12.5 cm long, 10cm wide and 8cm high. a) Which box has the greater lateral surface area and by how much? b) Which box has the smaller total surface area and by how much?

Solution: Given length of edge of cubical box=a=10 cm,

Cuboidal box dimensions are length=l=12.5 cm, breadth=b=10cm and height=h=8cm.

a) Lateral surface area of the cubical box=4a²=4 ×(10)²=400 cm²

Lateral surface area of cuboidal box=2h(l+b)

=2 (8)(12.5+10)

=16(22.5)=360 cm².

Differences in the lateral surface areas of the two boxes=(400-360) cm²=40 cm²

The cubical box has greater lateral surface area by 40 cm².

b) Total surface area of cubical box=6a²

=6(10)²=600 cm²

Total surface area of cuboidal box=2(lb+bh+hl)

=2[(12.5 ×10)+(10 ×8)+(8 ×12.5)]

=2(125+80+100) cm²

=2 ×305 cm²=610 cm².

Difference in the total surface areas of the two boxes=(610-600) cm²=10 cm².

The cubical box has smaller total surface area by 10cm².

6 A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30cm long, 25 cm wide and 25 cm high. a) What is the area of the glass? b) How much of tape is needed for all the 12 edges?

Solution: Given dimensions are length=l=30cm, breadth=b=25cm and h=25cm.

a) Total surface area of the herbarium=2(lb+bh+hl)

=2[(30 ×25)+(25 ×25)+(25 ×30)]

=2(750+625+750)

=2 ×2125 cm²=4250 cm²

∴ area of the glass=4250 cm²

b) A cuboid as 12 edges. These consist of 4 lengths 4 breadths and 4 heights.

∴ Length of the tape required=4l+4b+4h

=4(l+b+h)

=4(30+25+25)

=4(80)=320 cm.

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 thousand cm², find the cost of cardboard required for supplying 250 boxes of each kind.

Solution: Given dimensions of bigger box are l=25cm, b=20cm and h=5cm.

Total surface area of 1 bigger box=2(lb+bh+hl)

=2[(25 ×20)+(20 ×5)+(5 ×25)]

=2[500+100+125]

=2(725)

=1450 cm²

Area of cardboard required for overlaps=5% of 1450 cm²=$\frac{5\times 1450}{100}=72.5cm\xb2$

Total area of cardboard needed for 1 bigger box=(1450+72.5)cm²=1522.5 cm².

Total area of cardboard needed for 250 bigger boxes=1522.5 ×250 =380625 cm²

Given dimensions of smaller box are

l=15 cm, b=12cm and h=5 cm.

Total surface area of 1 smaller box=2(lb+bh+lh)

=2[(15 ×12)+(12 ×5)+(5 ×15)]

=2[180+60+75] cm²=630 cm²

Area of cardboard required for overlaps=5% of 630 cm²=

$\frac{5\times 630}{100}=31.5cm\xb2$

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 cm²=165375 cm²

Total area of cardboard needed for 500 boxes (250 bigger boxes and 250 smaller boxes)=

(380625+165375) cm²=546000 cm²

Cost of 1000 cm² of cardboard= Rs 4.

∴ Cost of 546000 cm² of cardboard=

$4\times \frac{546000}{1000}=Rs2184$

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 dimensions are,

l=4 m, b=3 m and h=2.5 m.

The tarpaulin is needed to cover 5 faces only excluding the floor.

Surface area of the shelter=2(lb+bh+lh)-lb

=lb+2(bh+lh)

=(4 ×3)+2[(3 ×2.5)+(2.5 ×4)]

=12+2(7.5+10)

=12+35

=47 m².

∴ 47 m² of tarpaulin is required to make the shelter.