# CLASS-9CIRCUMFERENCE & AREA OF CIRCLE - EQUILATERAL TRIANGLE

EQUILATERAL TRIANGLE -

In a equilateral triangle of side a units, we have

3a

(i)     Height of the triangle, h = ------- units

2

3a²

(ii)     Area of the Triangle = (------) sq.units

4

(iii)     Radius of Incircle, = h/3 = (a/23) units

(iv)     Radius of Circumcircle, = 2h/3 = (a/3) units

Thus, r = a/23 and R = a/3

Example.1) In an equilateral triangle of side 24 cm, a circle is inscribed, touching its sides. Find the area of the remaining portion of the triangle. Take √3 = 1.73 and π = 3.14

Ans.) Let ABC be the given equilateral triangle in which a circle is inscribed.

Side of the Triangle, a = 24 cm

√3               √3

Height of the triangle, h = (------ X a) = (------ X 24) = 12√3 cm

2                2

1            1

Radius of the incircle, r = ------ h = ----- X 12√3 cm = 4√3 cm

3            3

Required area = Area of the shaded region

=  (Area of the ABC) – (Area of incircle)

√3

= [(------ a²) – (πr²)

4

√3

= [{(------ X 24 X 24) – {3.14 X (4√3)²}] cm²

4

= [(√3 X 6 X 24) – (3.14 X 48)] cm²

= (144√3 – 150.72) cm²

= [(1.73 X 144) – 150.72] cm²

= (249.12 – 150.72) cm²

=  98.4 cm²

so, area of the remaining portion of the triangle is 98.4 cm²  (Ans.)

Example.2) The figure shows a running track surrounding a grass enclosure PQRSTU. The enclose consist of a rectangle PQST with a semi circular region at each end. Given, PQ = 200m and PT = 70m

(i) Calculate the area of grassed enclosure in m²

(ii) Given that the track is of constant width 7m, calculate the outer perimeter ABCDEF of the track.

Ans.)

Ans.) Diameter of each semicircular region of grassed enclosure = PT = 70 m

Radius of each one of them = 35 m.

Area of grassed enclosure –

= (Area of Rect. PQST) + 2 (Area of semi-circular region with radius 35 m)

1

=  (PQ X PT) + 2 X ------ πr²

2

22

=  [(200 X 70) + ------ X 35 X 35] m²

7

=   17850 m² …………………………………………………(i)      (Ans.)

Diameter of each outer semi-circle of the track

= AE = (PT + 7 + 7) m = 84 m

So, radius of each one of them = 42 m

Outer perimeter ABCDEF = (AB + DE + Semi-circle BCD + Semi-circle EFN)

= (2PQ + 2 X circumference of the semi-circle with radius 42 m)

= {(2 X 200) + (2 X π X 42)} m

22

= {400 + (2 X ------- X 42)} m

7

= {400 + (12 X 22)} m = (400 + 264) m

= 664 m  ……………………………(ii)      (Ans.)