10.3 Area of Triangle


10.3 Area of Triangle





Example:



Calculate the area of the triangle above.

Solution:
A r e a = 1 2 a b sin C A r e a = 1 2 ( 7 ) ( 4 ) sin 40 o A r e a = 9  cm 2


Example:


Diagram above shows a triangle ABC, where AC = 8 cm and ∠ C = 32o . Point D lies on straight line BC where BD = 10 cm and ∠ ADB = 70o . Calculate
(a) the length, in cm, of CD,
(b) the area, in cm2 of ∆ ADC,
(c) the area, in cm2 of ∆ ABC,
(d) the length, in cm, of AB.

Solution:
(a)
ADC+ 70 o = 180 o ADC= 110 o CAD= 180 o 110 o 32 o CAD= 38 o Using Sine Rule, 8 sin 110 o = CD sin 38 o CDsin 110 o =8sin 38 o CD( 0.940 )=8( 0.616 ) CD=5.243

(b)
Area of ADC = 1 2 ( 8 )( 5.243 )sin 32 o =20.972( 0.530 ) =11.12  cm 2

(c)
Area of ABC = 1 2 ( 8 )( 10+5.243 )sin 32 o =60.972( 0.530 ) =32.315  cm 2

(d)
Use Cosine Rule for ABC, A B 2 = 15.243 2 + 8 2 2( 15.243 )( 8 )cos 32 o A B 2 =232.35+64206.83 A B 2 =89.52 AB=9.462 cm

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