#### Class nine Maths NCERT Sol Surface Areas and Vol

## Chapter 13: Surface Areas and Volumes

#### TEXTBOOK’S EXERCISE – 13.1

**5. 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? **

**Sol.
**Here, a = 10 cm, l = 12.5 cm, b = 10 cm, h = 8
cm

(i) Lateral surface area of the cubical box

= 4a^{2} = 4 × (10)^{2}
cm^{2} = 400 cm^{2}

Lateral surface area of the cuboidal box

= 2h (l + b)

= 2 × 8 (12.5 + 10)
cm^{2}

= 16 × 22.5 cm^{2}
= 360 cm^{2}

Difference in the
lateral surface areas of the two boxes = (400 – 360) cm2 = 40 cm^{2}.

Hence, the cubical box has greater lateral surface area by 40 cm2

(ii) Total surface area of the cubical box

= 6a^{2} = 6 × (10)^{2}
cm^{2} = 600 cm^{2}

Total surface area of the cuboidal box

= 2(lb + bh + hl)

= 2(12.5 × 10 + 10 × 8 + 8 × 12.5) cm^{2}
= 2(125 + 80 + 100) cm^{2}

= 2 × 305 cm^{2}
= 610 cm^{2}

Difference in the
total surface areas of the two boxes = (610 – 600) cm^{2} = 10 cm^{2}

Hence, the cubical box has smaller total surface
area by 10 cm^{2}

**6. 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?

**Sol.
**Here, l = 30 cm, b = 25 cm, h = 25 cm.

(i) Total surface area of the herbarium

= 2(lb + bh + hl)

=
2(30 × 25 + 25 × 25 + 25 × 30) cm^{2}

=
2(750 + 625 + 750) cm^{2}

= 2
× 2125 cm^{2} = 4250 cm^{2}

Hence,
area of the glass = 4250 cm^{2}

(ii) A cuboid has 12 edges. These consist of 4 lengths, 4 breadths and 4 heights.

∴ length of the tape required

= 4l + 4b + 4h

= (4 × 30 + 4 × 25 + 4 × 25) cm

= (120 + 100 + 100) cm = 320 cm

## Chapter 13: Surface Areas and Volumes

**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 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 cm**^{2}**, find the cost of cardboard required for supplying 250 boxes of each kind. **

**Sol.
For bigger boxes : **

l = 25 cm, b = 20 cm, h = 5 cm

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

= 2(25 × 20 + 20 × 5
+ 5 × 25) cm^{2}

= 2 (500 + 100 +
125) cm2 = 1450 cm^{2}

Area of cardboard required for overlaps

= 5% of 1450 cm^{2} = 1450 5/100× cm^{2} = 72.5 cm^{2}

Total area of cardboard needed for

1 bigger box = (1450
+ 72.5) cm^{2}

= 1522.5 cm^{2}

Total area of
cardboard needed for 250 bigger boxes = 1522.5 × 250 cm2 = 380625 cm^{2}

**For
smaller boxes : **

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

Total surface area of 1 smaller box

= 2 (lb + bh + hl)

= 2(15 × 12 + 12 × 5
+ 5 × 15) cm^{2}

= 2 (180 + 60 + 75)
cm2 = 630 cm^{2}

Area of cardboard required for overlaps

= 5% of 630 cm^{2}

= 630× 5/ 100
cm^{2} = 31.5 cm^{2}

Total area of cardboard needed for

1 smaller box = (630
+ 31.5) cm^{2} = 661.5 cm^{2}

Total area of
cardboard needed for 250 smaller boxes = 661.5 × 250 cm^{2} = 165375 cm^{2}

Now, total area of cardboard needed for 500 boxes (250 bigger and 250 smaller boxes)

= (380625 + 165375)
cm^{2} = 546000 cm^{2 }

Cost of 1000 cm^{2}
of cardboard = Rs 4

∴ Cost of 546000 cm2 of cardboard = Rs 4/ 1000 × 546000 **= **Rs 2184.

**8. Parveen wanted to make a temporary shelter for her car, by making a box-like structure with a 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 many tarpaulins would be required to make the shelter of height 2.5 m, with base dimensions 4 m × 3 m? **

**Sol.
**Here, l = 4 m, b = 3 m, h = 2.5 m

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

The surface area of the shelter

= lb + 2 (bh + hl)

= 4 × 3 m^{2}
+ 2(3 × 2.5 + 2.5 × 4) m^{2}

= 12 m^{2} +
2(7.5 + 10) m^{2}

= (12 + 35) m^{2}
= 47 m^{2}
Hence, 47 m^{2} of tarpaulin is required to make the shelter.