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# The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m² is ₹ 4

**Solution:**

Let ABDC be a rhombus as shown below with AD and BC as the diagonals.

Let BC = 30 cm, AD = 45 cm

Area of rhombus ABDC = Area of ΔABC + Area of ΔDCB

= 1/2 × (BC × AO) + 1/2 × (BC × OD)

= 1/2 × BC × (AO + OD)

= 1/2 × BC × AD

= 1/2 × 30 cm × 45 cm

= 675 cm²

Area of each tile = 675 cm²

Area covered by 3000 tiles = (675 × 3000) cm² = 2025000 cm² = 202.5 m²

Given that the cost of polishing is Rs. 4 per m²

Cost of polishing for 202.5 m² area = Rs. 4 × 202.5 = Rs. 810

**☛ Check: **NCERT Solutions for Class 8 Maths Chapter 11

**Video Solution:**

## The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m² is ₹ 4

Class 8 Maths NCERT Solutions Chapter 11 Exercise 11.2 Question 7

**Summary:**

The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. If the cost per m² is ₹ 4, the total cost of polishing the floor is ₹ 810.

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