Question 2:
Diagram below shows a semicircle PTQ, with centre O and quadrant of a circle RST, with centre R.
[Use π = 3.142]
Calculate
(a) the value of θ, in radians,
(b) the perimeter, in cm, of the whole diagram,
(c) the area, in cm2, of the shaded region.
Solution:
(a)sin∠ROT=2.55 ∠ROT=30oθ=180o−30o=150o =150×π180 =2.618 rad
(b)Length of arc PT=rθ =5×2.618 =13.09 cmLength of arc ST=π2×2.5 =3.9275 cmOR2+2.52=52 OR2=52−2.52OR=4.330Perimeter=13.09+3.9275+2.5+4.330+5 =28.8475 cm
(c)Area of shaded region=Area of quadrant RST−Area of quadrant RQTArea of quadrant RQT=Area of OQT−Area of OTR=12(5)2×(30×π180)−12(4.33)(2.5)=1.1333 cm2Area of shaded region=Area of quadrant RST−Area of quadrant RQT=12(2.5)2×(90×π180)−1.1333=3.7661 cm2
Diagram below shows a semicircle PTQ, with centre O and quadrant of a circle RST, with centre R.

Calculate
(a) the value of θ, in radians,
(b) the perimeter, in cm, of the whole diagram,
(c) the area, in cm2, of the shaded region.
Solution:
(a)sin∠ROT=2.55 ∠ROT=30oθ=180o−30o=150o =150×π180 =2.618 rad
(b)Length of arc PT=rθ =5×2.618 =13.09 cmLength of arc ST=π2×2.5 =3.9275 cmOR2+2.52=52 OR2=52−2.52OR=4.330Perimeter=13.09+3.9275+2.5+4.330+5 =28.8475 cm
(c)Area of shaded region=Area of quadrant RST−Area of quadrant RQTArea of quadrant RQT=Area of OQT−Area of OTR=12(5)2×(30×π180)−12(4.33)(2.5)=1.1333 cm2Area of shaded region=Area of quadrant RST−Area of quadrant RQT=12(2.5)2×(90×π180)−1.1333=3.7661 cm2