Question

Please solve ques 20 and 21 correctly. Iwill Mark you brainiest

Answer 1

20)

for cuboidal room;

length, l = 9 m

breadth, b = 8 m

height, h = 6.5 m

for white washing the walls, we need to consider the lateral surface area of the room

= 2(bh + lh) = 2h(l + b) = 2 x 6.5 (9 + 8) = 13 x 17

= 221 m²

LSA(room) = 221 m² ... (i)

for rectangular doors;

length, l = 2 m

breadth, b = 1.5 m

area(door) = l x b = 2 x 1.5 = 3 m²

area(door) = 3 m² ... (ii)

for rectangular windows;

length, l = 1.5 m

breadth, b = 1 m

area(1 window) = 1.5 x 1 = 1.5 m²

area(3 windows) = 1.5 x 3 = 4.5 m²

area(3 windows) = 4.5 m² ... (iii)

Now, total non-required area for white washing

= ar(door) + ar(3 windows)

= 3 + 4.5 m² ... (from i and ii)

= 7.5 m² ... (iv)

Hence, area required for white washing

= LSA(room) - (non-required area)

= 221 - 7.5 m² ... (from i and iv)

= 213.5 m²

since the rate of white washing is 3.80 Rs. of 1 m²

hence, cost of white washing 213.5 m²

= rate of white washing 1 m² x 213.5 Rs.

= 3.80 x 213.5 Rs

= 811.30 Rs.

Hence, cost of white washing the walls of the room is 811.30 Rs.

____________________________________________

21)

For a cuboidal hall;

length, l = 36 m

breadth, b = 24 m

height = h

area required for papering would be the lateral surface area of the hall

= 2(bh + lh) = 2h(l + b) = 2h(36 + 24) = 2h(60)

= 120h m²

LSA(hall) = 120h m² ... (i)

net area of windows and doors

= 80 m² ... (ii)

hence, area that will be papered

= LSA(hall) - net area of windows and doors

= 120 h - 80 m² ... (from i and ii)

area(walls) = 120h - 80 m² ... (iii)

now, we know that

cost of papering = rate of papering 1 m² x area(walls)

hence,

9408 = 8.4 (120h - 80) ... (from iii and given)

$\frac{9408}{8.4} = 120h - 80 \\ \frac{94080}{84} = 120h - 80$

(multiplying both numerator and denominator in LHS by 10)

1120 = 120h - 80

120h = 1120 + 80

120h = 1200

$h = \frac{1200}{120}$

h = 10 m

Hence, height of the hall is 10 m.

Answer 2

Answer:

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