Question

A black blackboard of side 5m50cm and4m40cm is to be painted find the cost at the rate of rupees 10 per square metre

Answer 1

Given :

• Length of the blackboard = 5 m 50 cm
• Breadth of the blackboard = 4 m 40 cm
• Rate of painting the balckboard = Rs. 10

To find :

• Cost of painting the blackboard

Concept :

To find the cost of painting the balckboard :-

• Find the area of the blackboard.
• For that firstly we will convert the length and breadth of the blackboard from 5 m 50 cm and 4 m 40 cm into m.
• To find the cost multiply the rate of painting by area of the blackboard

→ Formula to calculate area of rectangle :-

$\boxed{ \sf \pmb{Area \: of \: rectangle = l \times b}}$

→ Unit conversion :-

• 1 cm = 0.01 m

To convert the value from cm into m divide the value by 100.

Solution :

• Length = 5 m 50 cm

→ 50 cm = 50 ÷ 100 = 0.5 m

→ Length = 5 + 0.5 = 5.5 m

• Length of the Blackboard = 5.5 m

• Breadth = 4 m 40 cm

→ 40 cm = 40 ÷ 100 = 0.4 m

→ Breadth = 4 + 0.4 = 4.4 m

• Breadth of the Blackboard = 4.4 m

Area of the blackboard :-

$\\ \dashrightarrow \sf \quad Area \: of \: blackboard = l \times b$

$\\ \dashrightarrow \sf \quad Area = 5.5 \times 4.4$

$\\ \dashrightarrow \sf \quad Area = 24.2$

• Area of blackboard = 24.2 m²

Cost of painting the blackboard :-

$\\ \dashrightarrow \sf \quad Cost = Area_{(Blackboard)} \times Rate_{(Painting)}$

$\\ \dashrightarrow \sf \quad Cost = 24.2 \times 10$

$\\ \dashrightarrow \sf \quad Cost = 242$

• Cost of painting the blackboard = Rs. 242

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Know More :

Some related formulae :-

• Perimeter of reactangle = 2(l + b)
• Diagonal of rectangle = √l² + √b²
• Area of square = side × side
• Perimeter of square = 4 × side
• Diagonal of square = √2a
• Total surface Area of cuboid = 2(lb + bh + hl)
• Volume of cuboid = length × breadth × height
• Perimeter of cuboid = 4 (length + breadth + height)
• Total surface area of cube = 6 × side²
• Volume of cube = side³
• Perimeter of cube = 12 × side