CLASS-9
CIRCUMFERENCE & AREA OF CIRCLE - EQUILATERAL TRIANGLE

                                      √3a

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

                                       2

                                  √3a²

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

                                   4  

(iii)     Radius of Incircle, r = h/3 = (a/2√3) units

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

Thus, r = a/2√3 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.)