# The sides of a triangular field are 41 m, 40 m and 9 m. Find the number of rose beds that can be prepared in the field, if each rose bed, on an average needs 900 cm² space.

**Solution:**

Given, the sides of a triangular field are 41 m, 40 m and 9 m.

Each rose bed needs an average of 900 cm² space

We have to find the number of rose beds that can be prepared in the field.

By __Heron’s formula__,

Area of triangle = √s(s - a)(s - b)(s - c)

Where s= semiperimeter

s = (a + b + c)/2

So, s = (41 + 40 + 9)/2

= 90/2

s = 45 m

Now, area = √[45(45 - 41)(45 - 40)(45 - 9)]

= √[45 × 4 × 5 × 36]

= √[9 × 5 × 5 × 4 × 4 × 9]

= 9 × 5 × 4

= 180 m²

We know, 1 m = 100 cm

1 cm = 1/100 m

i.e., 1 cm = 0.01 m

Similarly, 900 cm² = 0.09 m²

Area needed for one rose bed = 0.09 m²

Number of rose beds = entire area of triangular field / area needed for one rose bed

= 180/0.09

= 2000

Therefore, the number of rose beds is 2000.

**✦ Try This:** The sides of a right triangular field are in the ratio 5:3:4 and its perimeter is 180 m. Find the cost of levelling the field at the rate of Rs.10 per square metre.

**☛ Also Check: **NCERT Solutions for Class 9 Maths Chapter 12

**NCERT Exemplar Class 9 Maths Exercise 12.3 Sample Problem 1**

## The sides of a triangular field are 41 m, 40 m and 9 m. Find the number of rose beds that can be prepared in the field, if each rose bed, on an average needs 900 cm² space.

**Summary:**

The sides of a triangular field are 41 m, 40 m and 9 m. The number of rose beds that can be prepared in the field, if each rose bed, on an average needs 900 cm² space is 2000

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