Question

a box is 6.5 cm long and 5.4 cm wide and 1.4 cm high then find the cost of painting the four sides faces and top at the rate of ₹ 6.50per cm2 ​

Answer 1

Solution

Length of the box = 6.5cm

Breadth of the box = 5.4cm

Height of the box = 1.4cm

Calculating Surface Area of the Box

Surface area of cuboid = 2(lb + bh + hl)

= 2(6.5×5.4 + 5.4×1.4 + 1.4×6.5)

= 2(35.1 + 7.56 + 9.1)

= 2(51.76)

= 2 × 51.76

= 103.52cm².

Calculating Cost of Painting

Cost of painting 1cm² = ₹6.50

Cost of painting 103.52cm²

= ₹6.50 × 103.52

= ₹672.88.

Therefore, cost of painting is ₹672.88.

More formulas to know

SA of cube = 6a²

Edge of cube = √⅙s

CSA of cylinder = 2 πrh

TSA of cylinder = 2 πr (r + h).

Answer 2

To Find :

The cost of painting the outer surface of box

Given :

• Length = 6.5 cm
• Width = 5.4 cm
• Hieght = 1.4 cm

We know that :

To paint the box , means it is covering the total surface area of the box .

• TSA of cuboid = 2(lb+bh+lh)

__________________________

Solution :

Given, length = 6.5 cm

Width = 5.4 cm

hieght = 1.4 cm

so, TSA = 2[(5.4×6.5)+(5.4×1.4)+(1.4×6.5)]

$\tt{ \implies \: tsa = 2(35.1 + 7.56 + 9.1)}$

$\tt{ \implies tsa = 2 \times 51.76}$

$\implies \tt{tsa = 103.52 \: {cm}^{2} }$

So, Cost of 1cm² = 6.5

so, total cost = 103.52 × 6.5

or, total cost = ₹672.88

Hence, Cost of painting = ₹672.88

__________________________

Additional information:

• TSA of cuboid = 2(lb+bh+lh)
• LSA of cuboid = 2h(l+b)
• Volume of cuboid = l×b×h

• TSA of cube = 6a²
• LSA of cube = 4a²
• Volume of cube = a³

• TSA of cylinder = 2pi r(h+r)
• CSA of cylinder = 2pi rh
• Volume of cylinder = pi r²h