NCERT Class 9 Solutions: Surface Areas and Volumes (Chapter 13) Exercise 13.1 – Part 1

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Perimeter and area

Perimeter and Area

Perimeter and area

Q-1 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:

  1. The area of the sheet required for making the box.

  2. The cost of sheet, if a sheet measuring 1m2 costs Rs20 .


Quadrilateral poly1 Quadrilateral poly1: Polygon A, B, C, D Quadrilateral poly2 Quadrilateral poly2: Polygon A, E, F, B Quadrilateral poly3 Quadrilateral poly3: Polygon F, G, C, B Segment a Segment a: Segment [A, B] of Quadrilateral poly1 Segment b Segment b: Segment [B, C] of Quadrilateral poly1 Segment c Segment c: Segment [C, D] of Quadrilateral poly1 Segment d Segment d: Segment [D, A] of Quadrilateral poly1 Segment a_1 Segment a_1: Segment [A, E] of Quadrilateral poly2 Segment e Segment e: Segment [E, F] of Quadrilateral poly2 Segment f Segment f: Segment [F, B] of Quadrilateral poly2 Segment b_1 Segment b_1: Segment [B, A] of Quadrilateral poly2 Segment f_1 Segment f_1: Segment [F, G] of Quadrilateral poly3 Segment g Segment g: Segment [G, C] of Quadrilateral poly3 Segment c_1 Segment c_1: Segment [C, B] of Quadrilateral poly3 Segment b_2 Segment b_2: Segment [B, F] of Quadrilateral poly3 Segment h Segment h: Segment [H, I] Segment i Segment i: Segment [I, J] Segment j Segment j: Segment [I, K] 0.65 m text1 = "0.65 m" 1.5 m text2 = "1.5 m" 1.25 m text3 = "1.25 m"

Plastic Box

Plastic box length1.5m,width=1.25m and depth=65cm

  • Plastic box length is (l)=1.5m

  • Plastic box width is (b)=1.25m

  • Plastic box depths is (h)=65cm=0.65m

Solution (i) The area of sheet required to make the box is equal to the surface area of the box excluding the top.

  • Surfaceareaofthebox=Lateralsurfacearea+Areaofthebase

  • 2(l+b)×h+(l×b)

  • 2(1.5+1.25)×0.65+(1.5×1.25)

  • (3.575+1.875)m2

  • 5.45m2

The sheet required required to make the box is 5.45m2

Solution (ii) Cost of 1m2 of sheet =Rs20

∴ Cost of 5.45m2 of sheet =Rs(20×5.45)=Rs109

Q-2 The length, breadth and height of a room are 5m,4mand3m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs7.50perm2


  • The room length is (l)=5m

  • The room breadth is (b)=4m

  • The room height is (h)=3m

Area of four walls including the ceiling = Area of walls ( 2×l×h+2×l×b) + area of ceiling ( l×b )

  • 2(l+b)×h+(l×b)

  • 2(5+4)×3+(5×4)m2

  • (54+20)m2

  • 74m2

Cost of white washing =Rs7.50perm2

Total cost =Rs.(74×7.50)=Rs.555

Q-3 The floor of a rectangular hall has a perimeter 250m . If the cost of painting the four walls at the rate of Rs.10perm2 is Rs15000 , find the height of the hall.


  • Perimeter of rectangular hall =2(l+b)=250m

  • Total cost of painting =Rs15000

  • Rate per m2=Rs.10

Area of four walls = 2×l×h+2×b×h=(2×l+2×b)×h

  • Now (2×l+2×b) is the perimeter of the floor which is 250 m. Therefore, area of four walls is (250×h)m2

  • Total cost of painting= Area of four walls hall × rate per square m =(250×h)×10=Rs15000

  • Therefore, 2500×h=Rs15000 or h=150002500m=6m

So, the height of the hall is 6m.

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