Question 1:
Solution:
(a)
For triangle OPQsin30∘=8OPOP=8sin30∘=16 cmRadius of sector SPT=16+8=24 cm∠SPT=60×3.142180=1.047 radianLength of arc ST=24×1.047=25.14 cm
(b)
For triangle OPQ:tan30∘=8QPPQ=8tan30∘=13.86 cm∠QOR=2(60∘)=120∘Reflex angle ∠QOR=360∘−120∘=240∘240∘=3.142180∘×240∘=4.189 radianArea of shaded region=( Area of sector SPT)−(Area of major sector OQR)−(Area of triangle OPQ and OPR)=12(24)2(1.047)−12(8)2(4.189)−2(12×8×13.86)=301.54−134.05−110.88=56.61 cm2
Diagram shows a circle, centre O and radius 8 cm inscribed in a sector SPT of a circle at centre P. The straight lines, SP and TP, are tangents to the circle at point Q and point R, respectively.
[Use p= 3.142]
Calculate
(a) the length, in cm, of the arc ST,
(b) the area, in cm2, of the shaded region.
Solution:
(a)
For triangle OPQsin30∘=8OPOP=8sin30∘=16 cmRadius of sector SPT=16+8=24 cm∠SPT=60×3.142180=1.047 radianLength of arc ST=24×1.047=25.14 cm
(b)
For triangle OPQ:tan30∘=8QPPQ=8tan30∘=13.86 cm∠QOR=2(60∘)=120∘Reflex angle ∠QOR=360∘−120∘=240∘240∘=3.142180∘×240∘=4.189 radianArea of shaded region=( Area of sector SPT)−(Area of major sector OQR)−(Area of triangle OPQ and OPR)=12(24)2(1.047)−12(8)2(4.189)−2(12×8×13.86)=301.54−134.05−110.88=56.61 cm2