Long Questions (Question 2)


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)sinROT=2.55 ROT=30oθ=180o30o=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=522.52OR=4.330Perimeter=13.09+3.9275+2.5+4.330+5   =28.8475 cm


(c)Area of shaded region=Area of quadrant RSTArea of quadrant RQTArea of quadrant RQT=Area of OQTArea of OTR=12(5)2×(30×π180)12(4.33)(2.5)=1.1333 cm2Area of shaded region=Area of quadrant RSTArea of quadrant RQT=12(2.5)2×(90×π180)1.1333=3.7661 cm2