NCERT Class 9 Solutions: Surface Areas and Volumes (Chapter 13) Exercise 13.9 – Part 1

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Surface area of cuboid=2(lb+bh+lh)

Surface Area of Cuboid

Surface area of cuboid=2(lb+bh+lh)

Q-1 A wooden bookshelf has external dimensions as follows: Height Equation , depth Equation , breadth Equation (see figure). The thickness of the plank is Equation everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is Equation paise per Equation and the rate of painting is Equation paise per Equation find the total expenses required for polishing and painting the surface of the bookshelf.

Quadrilateral poly2 Quadrilateral poly2: Polygon A, E, F, B Quadrilateral poly3 Quadrilateral poly3: Polygon F, G, C, B Quadrilateral poly4 Quadrilateral poly4: Polygon L, M, N, O Quadrilateral poly4_1 Quadrilateral poly4_1: Polygon L_1, M_1, N_1, O_1 Quadrilateral poly4_2 Quadrilateral poly4_2: Polygon L_2, M_2, N_2, O_2 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 l Segment l: Segment [L, M] of Quadrilateral poly4 Segment m Segment m: Segment [M, N] of Quadrilateral poly4 Segment n Segment n: Segment [N, O] of Quadrilateral poly4 Segment o Segment o: Segment [O, L] of Quadrilateral poly4 Segment l_1 Segment l_1: Segment [L_1, M_1] of Quadrilateral poly4_1 Segment m_1 Segment m_1: Segment [M_1, N_1] of Quadrilateral poly4_1 Segment n_1 Segment n_1: Segment [N_1, O_1] of Quadrilateral poly4_1 Segment o_1 Segment o_1: Segment [O_1, L_1] of Quadrilateral poly4_1 Segment l_2 Segment l_2: Segment [L_2, M_2] of Quadrilateral poly4_2 Segment m_2 Segment m_2: Segment [M_2, N_2] of Quadrilateral poly4_2 Segment n_2 Segment n_2: Segment [N_2, O_2] of Quadrilateral poly4_2 Segment o_2 Segment o_2: Segment [O_2, L_2] of Quadrilateral poly4_2 Segment k Segment k: Segment [A_1, B_1] Segment p Segment p: Segment [B_1, C_1] Segment q Segment q: Segment [B_1, D_1] Segment p_1 Segment p_1: Segment [B_2, C_2] Segment p_2 Segment p_2: Segment [B_3, C_3] Segment r Segment r: Segment [B_2, A_2] Segment s Segment s: Segment [B_2, E_1] Segment t Segment t: Segment [B_3, A_3] Segment g_1 Segment g_1: Segment [B_3, F_1] Vector u Vector u: Vector[P, Q] Vector u Vector u: Vector[P, Q] Vector v Vector v: Vector[P, R] Vector v Vector v: Vector[P, R] Vector w Vector w: Vector[S, T] Vector w Vector w: Vector[S, T] Vector h Vector h: Vector[S, U] Vector h Vector h: Vector[S, U] Vector i Vector i: Vector[V, W] Vector i Vector i: Vector[V, W] Vector j Vector j: Vector[V, Z] Vector j Vector j: Vector[V, Z] 85 cm text1 = "85 cm" 110 cm text2 = "110 cm" 25 cm text3 = "25 cm"

Wooden Bookshelf

Wooden bookshelf with external dimensions of height, length and breadth being 110cm,85cm,25cm respectively

Solution:

Given

  • Height Equation

  • Depth Equation

  • Breadth Equation

  • Plank Thickness Equation

  • Rate of polishing is Equation

  • Rate of painting is Equation

First lets calculate the outer area to be polished.

Now the shelf is open from the front, therefore there is only one face with sides of height and length.

Hence, external surface area of the shelf = Equation

  • Equation

  • Equation

  • Equation

Now front face has some area, due to the thickness of the planks.

Bookshelf with front face having some area due to plant thinkness

Front Face of a Bookshelf

Bookshelf with front face having some area due to plant thinkness

The area inside the outer edge of front face is Equation , the area inside the inner edge of front face = Equation . Therefore the area of front face occupied due to think planks = Outside area – Inside area

  • Equation

  • Equation

  • Equation

Therefore, total area to be polished = Outer area + Wooden area on the front face

  • Equation

  • Equation

Since cost of polishing Equation , therefore, cost of polishing Equation area Equation

Now let’s calculate the inner area to be painted.

Internal depth, breadth and height are calculated by subtracting the length so planks.

  • Inner breadth = Outer breadth – width of right plank – width of left plank = Equation .

  • Similarly, External height = Outer height – width of top plank – width of bottom plank = Equation .

  • Since the bookshelf is open from the front, therefore the inner depth = outer depth – width of plank on the back Equation

Therefore the inner surface are (since there is no front face) Equation

  • Equation

  • Equation

  • Equation

Now, cost of painting Equation

Therefore, cost of painting Equation

Thus, total expense required for polishing and painting Equation

So, it would take Equation for polishing and painting the surface of the bookshelf.

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