\r\n Class 9 Mathematics\r\n Nepal CDC Aligned\r\n
\r\nArea of Four Walls, Ceiling & Floor
\r\n\r\n \r\n Essential Formulas Reference\r\n
\r\n\r\n \r\n 10 Comprehensive Solved Questions\r\n
\r\nClick on any question to toggle the step-by-step left-aligned mathematical solution.
\r\nSolution Step-by-Step
\r\n \r\nPart (a): Finding Dimensions
\r\nLet breadth of room ($b$) = $x\\text{ m}$
\r\nThen length ($l$) = $2x\\text{ m}$
\r\nGiven height ($h$) = $4\\text{ m}$
\r\nRate of coloring ($R$) = $\\text{Rs } 15/\\text{m}^2$
\r\nTotal Cost = $\\text{Rs } 2,160$
\r\n$$\\text{Area of 4 walls } (A_w) = \\frac{\\text{Total Cost}}{\\text{Rate}}$$
\r\n$$\\text{or, } A_w = \\frac{2160}{15} = 144\\text{ m}^2$$
\r\nUsing formula for 4 walls:
\r\n$$A_w = 2h(l + b)$$
\r\n$$\\text{or, } 144 = 2 \\times 4 \\times (2x + x)$$
\r\n$$\\text{or, } 144 = 8 \\times 3x$$
\r\n$$\\text{or, } 144 = 24x$$
\r\n$$\\text{or, } x = \\frac{144}{24}$$
\r\n$$\\therefore x = 6\\text{ m}$$
\r\n\r\n• Breadth ($b$) = $6\\text{ m}$
\r\n• Length ($l$) = $2 \\times 6 = 12\\text{ m}$
\r\n• Height ($h$) = $4\\text{ m}$
\r\nPart (b): Length of Carpet Required
\r\n$$\\text{Area of floor} = l \\times b$$
\r\n$$= 12 \\times 6 = 72\\text{ m}^2$$
\r\n$$\\text{Length of Carpet } (L) = \\frac{\\text{Area of floor}}{\\text{Width of carpet}}$$
\r\n$$= \\frac{72}{3} = \\mathbf{24\\text{ m}}$$
\r\n\r\nPart (c): Cost of Plastering Ceiling
\r\n$$\\text{Area of ceiling} = l \\times b = 72\\text{ m}^2$$
\r\n$$\\text{Total Cost} = \\text{Area} \\times \\text{Rate}$$
\r\n$$= 72 \\times 50 = \\mathbf{\\text{Rs } 3,600}$$
\r\nSolution Step-by-Step
\r\n\r\nStep 1: Express dimensions in terms of variable $x$
\r\nGiven $l = 2b = 3h$. Let length ($l$) = $6x$.
\r\nThen Breadth ($b$) = $\\frac{6x}{2} = 3x$
\r\nHeight ($h$) = $\\frac{6x}{3} = 2x$
\r\n\r\nStep 2: Find floor area and value of $x$
\r\n$$\\text{Area of floor} = \\frac{\\text{Total Carpeting Cost}}{\\text{Rate}}$$
\r\n$$\\text{or, } \\text{Area of floor} = \\frac{19440}{120} = 162\\text{ m}^2$$
\r\nNow, $\\text{Area of floor} = l \\times b$
\r\n$$\\text{or, } (6x) \\times (3x) = 162$$
\r\n$$\\text{or, } 18x^2 = 162$$
\r\n$$\\text{or, } x^2 = \\frac{162}{18}$$
\r\n$$\\text{or, } x^2 = 9$$
\r\n$$\\therefore x = 3\\text{ m}$$
\r\n\r\n• Length ($l$) = $6 \\times 3 = 18\\text{ m}$
\r\n• Breadth ($b$) = $3 \\times 3 = 9\\text{ m}$
\r\n• Height ($h$) = $2 \\times 3 = 6\\text{ m}$
\r\nStep 3: Calculate area of 4 walls + ceiling and cost
\r\n$$\\text{Area of 4 walls } (A_w) = 2h(l + b)$$
\r\n$$= 2 \\times 6 \\times (18 + 9)$$
\r\n$$= 12 \\times 27 = 324\\text{ m}^2$$
\r\n$$\\text{Area of ceiling } (A_c) = l \\times b$$
\r\n$$= 18 \\times 9 = 162\\text{ m}^2$$
\r\n$$\\text{Total Area to Plaster} = 324 + 162 = 486\\text{ m}^2$$
\r\n$$\\text{Total Cost} = 486 \\times 80 = \\mathbf{\\text{Rs } 38,880}$$
\r\nSolution Step-by-Step
\r\n\r\nHeight ($h$) = $5\\text{ m}$
\r\nBreadth ($b$) = $l - 3 \\implies l - b = 3\\text{ m}$ ........ (Equation 1)
\r\n$$\\text{Area of 4 walls } (A_w) = \\frac{\\text{Total Cost}}{\\text{Rate}}$$
\r\n$$\\text{or, } A_w = \\frac{2114}{7} = 302\\text{ m}^2$$
\r\nUsing formula $2h(l + b) = 302$:
\r\n$$\\text{or, } 2 \\times 5 \\times (l + b) = 302$$
\r\n$$\\text{or, } 10(l + b) = 302$$
\r\n$$\\text{or, } l + b = 30.2\\text{ m}$$ ........ (Equation 2)
\r\n\r\nSolving Equations (1) and (2):
\r\nAdding both equations:
\r\n$$(l + b) + (l - b) = 30.2 + 3$$
\r\n$$\\text{or, } 2l = 33.2$$
\r\n$$\\therefore l = 16.6\\text{ m}$$
\r\nBreadth ($b$) = $16.6 - 3 = 13.6\\text{ m}$
\r\n\r\nArea and Cost Calculations:
\r\n$$\\text{Area of floor} = l \\times b$$
\r\n$$= 16.6 \\times 13.6 = 225.76\\text{ m}^2$$
\r\n$$\\text{Cost of Carpeting} = 225.76 \\times 200$$
\r\n$$= \\mathbf{\\text{Rs } 45,152}$$
\r\nSolution Step-by-Step
\r\n\r\n$$\\text{Gross Area of 4 walls} = 2h(l + b)$$
\r\n$$= 2 \\times 4 \\times (10 + 8)$$
\r\n$$= 8 \\times 18 = 144\\text{ m}^2$$
\r\n$$\\text{Area of 1 door} = 2 \\times 1.5 = 3\\text{ m}^2$$
\r\n$$\\text{Area of 2 windows} = 2 \\times (1.5 \\times 1) = 3\\text{ m}^2$$
\r\n$$\\text{Total Area of Openings} = 3 + 3 = 6\\text{ m}^2$$
\r\n$$\\text{Net Painted Area} = 144 - 6 = 138\\text{ m}^2$$
\r\n$$\\text{Total Painting Cost} = 138 \\times 45 = \\mathbf{\\text{Rs } 6,210}$$
\r\nSolution Step-by-Step
\r\n\r\n(a) Edge length ($a$):
\r\n$$\\text{TSA of cubic room} = 6a^2 = 150\\text{ m}^2$$
\r\n$$\\text{or, } a^2 = \\frac{150}{6}$$
\r\n$$\\text{or, } a^2 = 25$$
\r\n$$\\therefore a = \\mathbf{5\\text{ m}}$$
\r\n\r\n(b) Area of 4 walls ($A_w$):
\r\n$$A_w = 4a^2$$
\r\n$$= 4 \\times 25 = \\mathbf{100\\text{ m}^2}$$
\r\n\r\n(c) Cost of carpeting floor:
\r\n$$\\text{Floor area} = a^2 = 25\\text{ m}^2$$
\r\n$$\\text{Cost} = 25 \\times 150 = \\mathbf{\\text{Rs } 3,750}$$
\r\nSolution Step-by-Step
\r\n\r\n$$\\text{Area of 4 walls} = 2h(l + b)$$
\r\n$$= 2 \\times 3 \\times (6 + 4.5)$$
\r\n$$= 6 \\times 10.5 = 63\\text{ m}^2$$
\r\nWallpaper width = $75\\text{ cm} = 0.75\\text{ m}$
\r\n$$\\text{Length of wallpaper } (L) = \\frac{\\text{Area of 4 walls}}{\\text{Width of wallpaper}}$$
\r\n$$\\text{or, } L = \\frac{63}{0.75} = \\mathbf{84\\text{ m}}$$
\r\n$$\\text{Total Cost} = 84 \\text{ m} \\times \\text{Rs } 40/\\text{m} = \\mathbf{\\text{Rs } 3,360}$$
\r\nSolution Step-by-Step
\r\n\r\nPerimeter of floor ($P$) = $2(l + b) = 48\\text{ m}$
\r\nHeight ($h$) = $4\\text{ m}$
\r\n$$\\text{Area of 4 walls } (A_w) = 2h(l + b) = h \\times P$$
\r\n$$= 4 \\times 48 = \\mathbf{192\\text{ m}^2}$$
\r\n$$\\text{Cost of papering} = 192 \\times 35 = \\mathbf{\\text{Rs } 6,720}$$
\r\nSolution Step-by-Step
\r\n\r\nPart (a): Finding Dimensions
\r\nSince the floor is square, let length = breadth = $x\\text{ m}$ ($l = x$, $b = x$).
\r\nGiven height ($h$) = $3.5\\text{ m}$
\r\nArea of 4 walls ($A_w$) = $112\\text{ m}^2$
\r\nUsing standard formula for 4 walls:
\r\n$$A_w = 2h(l + b)$$
\r\n$$\\text{or, } 112 = 2 \\times 3.5 \\times (x + x)$$
\r\n$$\\text{or, } 112 = 7 \\times (2x)$$
\r\n$$\\text{or, } 112 = 14x$$
\r\n$$\\text{or, } x = \\frac{112}{14}$$
\r\n$$\\therefore x = 8\\text{ m}$$
\r\n• Length ($l$) = $8\\text{ m}$
\r\n• Breadth ($b$) = $8\\text{ m}$
\r\nPart (b): Number of Floor Tiles & Total Cost
\r\n$$\\text{Area of floor} = l \\times b = 8 \\times 8 = 64\\text{ m}^2$$
\r\nSide of 1 tile = $50\\text{ cm} = 0.5\\text{ m}$
\r\n$$\\text{Area of 1 tile} = 0.5 \\times 0.5 = 0.25\\text{ m}^2$$
\r\n$$\\text{Number of tiles } (N) = \\frac{\\text{Floor Area}}{\\text{Area of 1 tile}}$$
\r\n$$\\text{or, } N = \\frac{64}{0.25} = \\mathbf{256\\text{ tiles}}$$
\r\n$$\\text{Total Cost of Tiles} = 256 \\times 40 = \\mathbf{\\text{Rs } 10,240}$$
\r\nSolution Step-by-Step
\r\n\r\n$$\\text{Area of 4 walls} = 2h(l + b)$$
\r\n$$= 2 \\times 3 \\times (8 + 5) = 78\\text{ m}^2$$
\r\n$$\\text{Area of ceiling} = l \\times b = 8 \\times 5 = 40\\text{ m}^2$$
\r\n$$\\text{Total painting area} = 78 + 40 = 118\\text{ m}^2$$
\r\n$$\\text{Cost of painting} = 118 \\times 60 = \\mathbf{\\text{Rs } 7,080}$$
\r\n$$\\text{Cost of carpeting floor} = 40 \\times 250 = \\mathbf{\\text{Rs } 10,000}$$
\r\nComparison: Carpeting the floor is more expensive than painting the walls and ceiling by $\\text{Rs } 2,920$ ($\\text{Rs } 10,000 - \\text{Rs } 7,080$).
\r\nSolution Step-by-Step
\r\n\r\n$$\\text{Gross Area of 4 walls} = 2h(l + b)$$
\r\n$$= 2 \\times 4 \\times (12 + 9) = 168\\text{ m}^2$$
\r\n$$\\text{Area of ceiling} = l \\times b = 12 \\times 9 = 108\\text{ m}^2$$
\r\n$$\\text{Area of 2 doors} = 2 \\times (2 \\times 1.2) = 4.8\\text{ m}^2$$
\r\n$$\\text{Area of 4 windows} = 4 \\times (1.5 \\times 1.2) = 7.2\\text{ m}^2$$
\r\n$$\\text{Total Openings Area} = 4.8 + 7.2 = 12\\text{ m}^2$$
\r\n$$\\text{Net Area to Plaster} = (168 - 12) + 108 = 264\\text{ m}^2$$
\r\n$$\\text{Total Plastering Cost} = 264 \\times 95 = \\mathbf{\\text{Rs } 25,080}$$
\r\n\r\n \r\n Quick Concept Check Quiz\r\n
\r\n\r\n What is the area of four walls of a cubic room with edge length $a = 4\\text{ m}$?\r\n
\r\n\r\n A room has floor area $50\\text{ m}^2$. How many meters of carpet of width $2\\text{ m}$ are required to cover it?\r\n
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