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 Equation long, Equation wide and Equation 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 Equation costs Equation .

Solution:

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 Equation

  • Plastic box width is Equation

  • Plastic box depths is Equation

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

  • Equation

  • Equation

  • Equation

  • Equation

  • Equation

The sheet required required to make the box is Equation

Solution (ii) Cost of Equation of sheet Equation

∴ Cost of Equation of sheet Equation

Q-2 The length, breadth and height of a room are Equation respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Equation

Solution:

  • The room length is Equation

  • The room breadth is Equation

  • The room height is Equation

Area of four walls including the ceiling = Area of walls ( Equation + area of ceiling ( Equation )

  • Equation

  • Equation

  • Equation

  • Equation

Cost of white washing Equation

Total cost Equation

Q-3 The floor of a rectangular hall has a perimeter Equation . If the cost of painting the four walls at the rate of Equation is Equation , find the height of the hall.

Solution:

  • Perimeter of rectangular hall Equation

  • Total cost of painting Equation

  • Rate per Equation

Area of four walls = Equation

  • Now Equation is the perimeter of the floor which is 250 m. Therefore, area of four walls is Equation

  • Total cost of painting= Area of four walls hall × rate per square m Equation

  • Therefore, Equation or Equation

So, the height of the hall is Equation

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