Math, asked by kartik2070, 1 year ago

An open box is made of wood 3 cm thick. Its external dimensions are 1.46 m, 1.16 m and 8.3 dm . Find the cost of painting the inner surface of the box at 75 paise per 100 cm²​

Answers

Answered by Utkarsheenikairvi
1

Given thickness of wood = 3 cm Length of open box = 1.48 m = 148 cm Breadth of open box = 1.16 m = 116 cm Height of open box = 8.3 dm = 83 cm Inner length, l = outer length – twice width = 148 – 6 = 142 cm Outer breath = 116 – 6 = 110 cm Outer height = 83 – 3 = 80 cm [since it is an open box] Area of inner surface = 2(lb + bh + hl) – lb = 2[142 x 110 + 110 x 80 + 80 x 142] – [142 x 110] = 55940 sq cm Cost of painting inner surface = Rs (50/100) x (55940/100) = Rs 279.90

or for inner surface area you can also apply 2(l+b)h×lb formula


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Answered by paryuljain23
1

Given: Thickness of wood = 3 cm

Length of open box = 1.46 m = 146cm

Breadth of open box = 1.16 m = 116 cm

Height of open box = 8.3 dm = 83 cm Inner length,

l = outer length – twice width = 146 – 6 = 140 cm

Outer breath = 116 – 6 = 110 cm Outer height = 83 – 3 = 80 cm

[since it is an open box]

Area of inner surface = 2(lb + bh + hl) – lb = 2[140 x 110 + 110 x 80 + 80 x 140] – [140 x 110] = 20000 sq cm

Cost of painting inner surface = Rs (75/100) x (20000/100) = Rs 150

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