1) In the figure is the centre of the circle.lf ZBAC = 60°,BC = 10 then A 60 8 10 С a) Draw the diametre from B which meet the circle at P b) Draw triangle BPC, write the measure of ZBPC? c) What is the diametre of the circle? What is its radius? d) Calculate the area of triangle BPC?
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Answer:
In triangle ABC, applying Pythagoras theorem,
AB
2
=AC
2
+BC
2
AB
2
=24
2
+10
2
AB
2
=576+100
AB
2
=676
AB=26
Area of the triangle ABC=
2
1
×BC×AC
=
2
1
×24×10
=120Area of the semi circle =
2
πr
2
here r=
2
AB
=
2
26
=13
So,
Area of semi circle =
2
π(13)
2
=
7
22
×169×
2
1
=265.57
Area of shaded part = Area of semicircle - Area of triangle ABC
=265.57−120
=145.57
Thus area of shaded part is 145.57units
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