Question

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➥An aquarium is in the form of a cuboid whose external measures are
80 tm x 30 cm x 40 cm. The base, side faces and back face are to be covered with a
coloured paper. Find the area of the paper needed?

➥The internal measures of a cuboidal room are 12 mx8mx 4 m. Find the
total cost of whitewashing all four walls of a room, if the cost of white washing is Rs 5 per
m2. What will be the cost of white washing if the ceiling of the room is also whitewashed.

➛ Need quality answers.
➛ No irrelevant answers plz.
➛ Best answers will be market brainliest!
➛ Answer both questions.​

Answer 1

Answer and Step-by-step explanation:

1)

$\diamond$ Length of aquarium = l = 80 cm

$\diamond$ Breadth  of aquarium = b = 30 cm

$\diamond$ Height of aquarium = h = 40 cm

$\implies$ Area of paper needed = Total surface area of aquarium

$\implies$ Area of paper needed = 2(lb + bh + lh)

$\implies$ Area of paper needed = 2[ (80*30) + (30*40) + (80*40) ]

$\implies$ Area of paper needed = 2(2400 + 1200 + 3200)

$\implies$ Area of paper needed = 2(7800)

$\implies$ Area of paper needed = 15,600 cm²

2)

$\diamond$ Length of room = l = 12 m

$\diamond$ Breadth  of room = b = 8 m

$\diamond$ Height of room = h = 4 m

$\implies$ Total area to be painted = LSA of room

$\implies$ Total area to be painted = 2h(l + b)

$\implies$ Total area to be painted = 2*4 (12 + 8)

$\implies$ Total area to be painted = 8 * 20

$\implies$ Total area to be painted = 160 m²

Hence,

Cost to whitewash 1 m² = Rs 5

Cost to whitewash 160m² = Rs 5 * 160 = Rs 800

$\implies$ Area to be painted (if ceiling is alsow whitewashed) = LSA + Area of ceiling

$\implies$ Area to be painted = 2h(l + b) + lb

$\implies$ Area to be painted = 160 + 12*8

$\implies$ Area to be painted = 160 + 96

$\implies$ Area to be painted = 256 m²

Hence,

Cost to whitewash 1 m² = Rs 5

Cost to whitewash 256 m² = Rs 5 * 256 = Rs 1280