Get live Maths 1-on-1 Classs - Class 6 to 12
Example 1 - Chapter 11 Class 7 Perimeter and Area
Last updated at March 23, 2023 by Teachoo
Examples
Example 1
You are here
Example 2
Example 3 Important
Example 4 Important
Example 5
Example 6
Example 7 Important
Example 8 Important
Example 9 (i)
Example 9 (ii) Important
Example 10
Example 11 Important
Example 12 Important
Example 13
Example 14
Example 15 Important
Example 16 Important
Example 17 Important
Example 18
Example 19 Important
Example 20
Example 21 Important
Example 22 Important
Transcript
Example 1 A door-frame of dimensions 3 m × 2 m is fixed on the wall of dimension 10 m × 10 m. Find the total labour charges for painting the wall if the labour charges for painting 1 m2 of the wall is Rs 2.50. Area to be painted = Area of square wall − Area of rectangular door frame Wall Side of wall = a = 10 m Area of wall = a × a = 10 × 10 = 100 m2 Door Frame Length = l = 3 m Breadth = b = 2 m Area of door frame = l × b = 3 × 2 = 6 m2 Therefore, Area to be painted = 100 − 6 = 94 m2 Now, Labour charge for painting 1m2 of wall = Rs 2.50 Labour charge for painting 94m2 of wall = Rs 2.50 × 94 = Rs 25/10 × 94 = Rs 5/2 × 94 = Rs 5 × 47 = Rs 235 ∴ Labour will charge Rs 235
