Question

A horse is tied to a peg at vertex A of a equilateral triangular grass field of side 30 m by
means of a 10 m long rope . Find
(i) the area of the part of the field in which the horse can graze.

Answer 1

Step-by-step explanation:

Given :-

A horse is tied to a peg at vertex A of an equilateral triangular grass field of side 30 m by means of a 10 m long rope .

To find :-

Find the area of the part of the field in which the horse can graze.

Solution :-

Given that

Length of the side of an equilateral triangle = 30 m

Let a = 30 m

We know that

Area of an equilateral triangle whose side is a units is (√3/4)a² sq.units

Area of the equilateral triangle whose side is 30 m

=> (√3/4)×(30)² m²

=> (√3/4)×(30×30) m²

=> (√3/4)×900 m²

=> (900×√3)/4 m²

=> 225√3 m²

Area of the equilateral triangle =

225√3 m² -----(1)

Length of the rope which is tied to a peg with horse = 10 m

The angle in an equilateral triangle = 60°

If the horse tied at the vertex then it can graze the part is the area of the sector.

So, We have to find the area of the sector

The area of the field that the horse can graze = Area of a sector

Area of a sector = (X°/360°)×πr² sq.units

We have ,

X° = 60°

r = 10 m

On Substituting these values in the above formula

=> (60°/360°)×(22/7)×10² m²

=> (1/6)×(22/7)×100

=> (1×22×100)/(6×7)

=> 2200/42

=> 1100/21

=> 52.38 m²

Area of the sector = 52.38 m²

Answer:-

The area of the part of the field in which the horse can graze is 52.38 m²

Used formulae:-

• Area of a sector = (X°/360°)×πr² sq.units
• r = radius of the sector
• X° = Angle subtended by the arc at the centre
• Area of an equilateral triangle is (√3/4)a² sq.units
• a = Side of the equilateral triangle.