Surface Areas & Volumes | Exercise 13.1 | Q4

Question 4: The paint in a certain container is sufficient to paint an area equal to 9.375 m². How many bricks of dimensions 22.5 cm x 10 cm x 7.5 cm can be painted out of this container?

Explanatory Answer | Exercise 13.1 Question 4

Given Data

Area that can be painted with available paint = 9.375 m²
Size of each brick = 22.5 cm × 10 cm × 7.5 cm
Note: Dimensions of a brick is given in cm and area that can be painted is given in square metre.
Let us convert the area that can be painted to square centimetres.
Area that can be painted = 9.375 m² = 9.375 × 100 × 100 cm²

Step 1: Compute Total Surface Area of Brick

Area of each brick covered by paint = Total surface area of the brick
2(lb + lh + bh) = 2 [(22.5 × 10) + (10 × 7.5) + (22.5 × 7.5)] cm²
2(lb + lh + bh) = 2 × (225 + 75 + 168.75) = 937.5 cm²

Step 2: Compute Number of Bricks

Number of bricks that can be painted = \\frac{\text{Total area that can be painted}}{\text{Total surface area of each brick}})
= \\frac{9.375 × 100 × 100}{937.5})
= \\frac{93750}{937.5})
= 100 bricks

Exercise 13.1 | Surface Area of Cubes & Cuboids

1. Class 9 Exercise 13.1 Question 3 | Height of hall ▶
2. Class 9 Exercise 13.1 Question 6 | TSA of Herbarium ▶
3. Class 9 Exercise 13.1 Question 7 | Shanti Sweets ▶

Exercise 13.2 | Surface Area of Cylinders

1. Class 9 Exercise 13.2 Question 3 | CSA TSA of Metal Pipe ▶
2. Class 9 Exercise 13.2 Question 4 | Roller & Playground ▶
3. Class 9 Exercise 13.2 Question 9 | Cylindrical Petrol Tank ▶
4. Class 9 Exercise 13.2 Question 10 | CSA of Decorative Lampshade ▶

Exercise 13.3 | Surface Area of Cones

1. Class 9 Exercise 13.3 Question 5 | Length of Tarpaulin 3 m wide ▶
2. Class 9 Exercise 13.3 Question 8 | Cost of painting conical barricades ▶

Exercise 13.4 | Surface Area of Spheres & Hemispheres

1. Class 9 Exercise 13.4 Question 8 | Outer CSA of hemispherical bowl ▶
2. Class 9 Exercise 13.4 Question 9 | Surface area of sphere & Cylinder ▶