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Question 1. 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 a sheet for it, if a sheet measuring 1m² costs Rs. 20.

Answer.

(i) Box length (l) = 1.5m 

The box's breadth (b) is 1.25 m. 

Depth of the box (h) = 0.65m

I Box is to be open at top 

Required sheet area. 

= 2lh+2bh+lb 

= [2 \( \times \) 1.5 \( \times \) 0.65+2 \( \times \)1.25×0.65+1.5 \( \times \)1.25]m² 

5.45 m² = (1.95+1.625+1.875) m² 

(ii) Sheet cost per square metre = Rs.20

The cost of a 5.45 m² sheet is Rs. (5.4520). 

= Rs. 109

Question 2. The length, breadth, and height of a room are 5 m, 4 m, and 3 m respectively. Find the cost of whitewashing the walls of the room and ceiling at the rate of Rs 7.50 per m².

Answer. Room Length (l) = 5m 

The room's breadth (b)= 4 m

Room height (h) = 3m

It can be seen that the room's four walls and ceiling will be whitewashed. 

The total area to be whitewashed equals the area of the room's walls plus the area of the room's ceiling. 

= 2lh+2bh+lb 

= [2×5×3+2×4×3+5×4] 

= (30+24+20) 

= 74 

Area = 74 m² 

Furthermore, 

Whitewash costs Rs.7.50 per m² area (given).

Whitewashing a 74 m² area costs Rs. (747.50). 

= Rs. 555

Question 3. The floor of a rectangular hall has a perimeter of 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 the rectangular hall's length, width, and height be l, b, and h, respectively. 

The area of four walls = 2lh+2bh 

= 2 (l+b) h 

Perimeter of the hall's floor = 2 (l + b) 

= 250 metres 

2(l+b) h = 250h m² = four-wall area

Painting costs per square metre are Rs. 10. 

Painting costs Rs. 2500h (250h 10) per square metre area.

The expense of painting the walls, however, is estimated to be Rs. 15,000. 

15000 hours = 2500 hours

Alternatively, h = 6 

As a result, the hall's height is 6 metres.

Question 4. 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 \( \times \) 10 cm \( \times \) 7.5 cm can be painted out of this container?

Answer. The total surface area of a brick is equal to 2 (lb + bh + lb). 

= [2 (22.5 \( \times \) 10+ 10 \( \times \) 7.5 +22.5 \( \times \) 7.5)] cm² 

= 2(225 + 75 + 168.75) cm² 

= (2 \( \times \) 468.75) cm² 

= 937.5 cm² 

The container's paint can be used to cover n bricks. 

Area of n bricks = (n \( \times \) 937.5) cm² = 937.5n cm² 

Area that can be painted using the container's paint = 9.375 m2 = 93750 cm² according to the recommendations 

As a result, we have 93750 = 937.5n.

100 = n

As a result, the container's paint can be used to paint out 100 bricks.

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?

Answer. From the question statement, we have the following information: 

cube's edge = 10 cm 

length (l) = 12.5 cm

 breadth (b) = 10cm

 height (h) = 8 cm

(i) For both figures, calculate the lateral surface area. 

Cubical box lateral surface area = 4 (edge) 

= 4(10)² 

= 400 m²…… (1) 

cuboidal box's lateral surface area = 2 [lh + bh] 

= [2 (12.5 \( \times \) 8 + 10 × 8)] 

= (2 \( \times \) 180) 

= 360 

As a result, the cuboidal box's lateral surface area is 360 cm². ..... (2)

As seen in (1) and (2), the lateral surface area of the cubical box is greater than the lateral surface area of the cuboidal box. There is a 40 cm2 difference between the two lateral surfaces. 

(Cubical box lateral surface area - cuboidal box lateral surface area = 400 cm² – 360 cm² = 40 cm²) 

(ii) Calculate the total surface area of both figures.

The cubical box's entire surface area is 6 (edge) 

2 = 6 (10 cm) 

2 = 600 cm²…(3) 

The cuboidal box's total surface area is 

= 2 [lh+bh+lb] 

= [2 (12.5 \( \times \) 8 + 10 \( \times \) 8+ 12.5 \( \times \) 100)] 

= 610 …..(4) 

This means that the cuboidal box's total surface area is 610 cm²

As seen in (3) and (4), the cubical box has a smaller total surface area than the cuboidal box. And the difference between them is 10cm². 

As a result, the cubical box's total surface area is 10 cm² smaller than the cuboidal box's.

Question 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.

(i)What is the area of the glass?

(ii)How much of tape is needed for all the 12 edges?

Answer.

Greenhouse length, let l = 30 cm 

Greenhouse breadth, let b = 25 cm 

Greenhouse height, let h = 25 cm

(i) Total greenhouse surface area = 

Area of the glass = 2[lb + lh + bh] 

= [2(30 × 25 + 30 × 25 + 25 × 25)] 

= [2 (750 + 750 + 625)] 

= (2 × 2125) 

= 4250 

Glass has a total surface area of 4250 cm². 

(ii) Tape is required on sides AB, BC, CD, DA, EF, FG, GH, HE AH, BE, DG, and CF, according to the diagram.

total tape length= 4 (l + b + h)

= [4 (30 + 25 + 25)] (after substituting the sum) 

= 320 

As a result, all 12 edges will require 320 cm of tape.

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\( \times \)20cm\( \times \)5cm and the smaller of dimension 15cm\( \times \)12cm\( \times \)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.

Answer. Let l, b, and h be the box's length, breadth, and height, respectively. 

The bigger box: 

l = 25cm 

b = 20 cm 

h = 5 cm 

total surface area of bigger box= 2 (lb + lh + bh) 

= [2 (2520 + 255 + 205)] 

= [2 (500 + 125 + 100)] 

= 1450 cm² 

Extra area required for overlapping 1450 \( \times \) 5/100 cm² 

= 72.5 cm² 

When calculating the total surface area of a larger box, consider all the overlapps. 

= (1450 + 72.5) cm² 

= 1522.5 cm² 

For 250 such larger boxes, the area of a cardboard sheet is required. 

= (1522.5 × 250) cm² 

= 380625 cm² 

The Smaller Box: 

Similarly, [2(1512 + 155 + 125)] is the entire surface area of the smaller box. cm² 

= [2(180 + 75 + 60)] 

= (2315) cm² 

= 630  cm² 

As a result, the additional space needed for overlapping= 630 \( \times \) 5/100 cm² 

= 31.5 cm². 

When all overlaps are taken into account, the total surface area of one smaller box = (630 + 31.5) cm² 

= 661.5 cm²

Area of cardboard sheet required for 250 smaller boxes = (250 \( \times \) 661.5) cm² 

= 165375 cm²

Box	Dimensions (in cm)	Total surface area (in cm2 )	Extra area required for overlapping (in cm2)	Total surface area for all overlaps (in cm2)	Area for 250 such boxes (in cm2)
Bigger Box	L= 25 B= 20 H= 5	1450	1450 \( \times \) 5/100 = 72.5	1450 + 72.5 = 1522.5	1522.5 \( \times \) 250 = 380625
Smaller Box	L= 15 B= 12 H= 5	630	630 \( \times \) 5/100 = 31.5	630 + 31.5 = 661.5	661.5 \( \times \) 250 = 165375

 

Now, the total amount of cardboard required is (380625+165375) cm². 

= 54,6000 cm²

Assume that a 1000 cm² cardboard sheet costs Rs. 4.

As a result, the cost of a 546000 cm² cardboard sheet is equal to Rs. (546000 \( \times \) 4)/1000 = Rs. 2184. 

As a result, the cardboard cost for supplying 250 boxes of each type will be Rs. 2184.

Question 8.  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 \( \times \) 3m?

Answer. Let l, b, and h be the shelter's length, breadth, and height, respectively. 

Given: 

l = 4m 

b = 3m 

h = 2.5m 

The shelter's top and four wall sides will need to be covered with tarpaulin. 

Using the formula, the needed tarpaulin area is 2 (lh + bh) +lb. 

When the variables of l, b, and h are combined, we get 

= [2(42.5 + 32.5 + 43)] m² 

= [2(10+7.5) + 12] m² 

= 47 m² 

As a result, a 47 m² tarpaulin would be needed.