Question

The dimension of Mr Narula's bedroom is 8 m x 7m x 5 m. It has two windows, each measuring 2m x 1m and one door measuring 3 m x 1m. Find the cost of white washing the room at 50 per square metre.​

Answer 1

Answer:

Given that :-

• The dimension of Mr. Narula's bedroom is 8m×7m×5m.
• It has two windows, each measuring 2m×1m and one door measuring 3m×1m.

To find :-

• The cost of white washing the room at 50 per sq m?

Solution :-

Area of room = Area of cuboid

»» 2(lb+bh+hl)

=> 2(8×7 + 7×5 + 5×8)m

=> 2(56 + 35 + 40)

=> 2(131)

=> 262m²

Therefore, Area of bedroom is 262m².

Area of 2 windows = Area of rectangle

»» l × b

=> (2×1)m

=> 2m²

One window = 2m².

2 windows = 2×2 = 4m²

Therefore, Area of two windows are 4m².

Then,

Area of door = Area of rectangle

»» l×b

=> (3×1)m

=> 3m²

Therefore, Area of the door is 3m².

So, Area of white washing of the room

=> Area of the room - (Area of two windows + Area of the door)

=> 262 - (4+3)

=> 262 - 7

=> 255m²

»» Cost of white washing of the room at Rs.50 per square metre :-

=> 255 × 50

=> Rs.12,750

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

The dimension of Mr Narula's bedroom is 8 m x 7m x 5 m.

It means,.

Length of room, l = 8 m

Breadth of room, b = 7 m

Height of room, h = 5m

Further given that,

Room has 2 windows, each measuring 2 m × 1 m

One door, measuring 3m × 1m

Now, Toral area to be white washed is evaluated as

$\bf\: Area_{(to \: be \: white \: washed)} \: \:$

$\rm \: = Area_{(4\:walls)} + Area_{(ceiling)} - Area_{(2\:windows)} - Area_{(1\:door)}$

$\: = 2(l + b) \times h + l \times b - 2(2 \times 1) - (3 \times 1)$

$\: = 2(8+ 7) \times 5 + 8\times 7 - 4 - 3$

$\: = 150 + 56 - 4 - 3$

$\rm \: = \: 199 \: {m}^{2}$

$\rm\implies \:\bf\: Area_{(to \: be \: white \: washed)} = 199 \: {m}^{2} \: \:$

Now, Given that

$\rm \: Cost \: of \: white \: washing \: {1 \: m}^{2} = 50$

So,

$\rm \: Cost \: of \: white \: washing \: {199 \: m}^{2} = 199 \times 50 = 9950$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$