A circle has radius 2 cm, (in that circle there is a traingle AOB)?

Question

O is the centre of the circle.

the chord AB has length 3 cm. .
BO is 2cm
AO is 2cm

(i) Show that ∠AOB is 1.696 radians (to three decimal places).

(ii) Hence find the area of the (smaller) sector AOB.

(iii) Find the area of the triangle AOB.

(iv) Hence find the (smaller) area enclosed between the chord AB and the circle

(v) Express both the area of the sector and the area from part (a)(iv) as percentages of the area of the circle.

Answer 1

i) Angle AOB = 2arsin(1.5/2) =97.1808 degrees
Change to radians:
97.1808/180 = x/(pi)
x = 97.1808(pi)/180 = 1.696 radians
ii) Area of smaller sector AOB:
(1.696/(2pi))((pi)r^2) = (1.696/2(2^2) = (1.696)(2) = 3.392cm^2
iii) Height of triangle AOB = 2cos(97.1808/2) = 1.32288cm
Area of triangle AOB = (1/2)bh = (1/2)(3)(1.32288) = 1.98433cm^2
iv) Smaller area enclosed between chord AB and the circle:
3.392 - 1.32288 = 2.06912cm^2
v) Area percentages of the whole circle:
Sector percentage:
3.392/((pi)r^2) = 3.392/((pi)2^2) = 26.9927 percent
Smaller area percentage:
2.06912/((pi)2^2) = 16.4655 percent