Special Formula For Finding Area

Geometry Level 2

A triangle has sides of 12 m, 16 m, and 22 m. Find the area of the triangle, giving your answer to three significant figures.


The answer is 93.7.

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1 solution

Solution 1: Heron's Formula

Let the sides of the triangle be a = 12 , b = 16 a=12, b=16 and c = 22 c=22 . Then the semi-perimeter is

s = a + b + c 2 = 12 + 16 + 22 2 = 25 s=\dfrac{a+b+c}{2}=\dfrac{12+16+22}{2}=25

So the area is

A = s ( s a ) ( s b ) ( s c ) = 25 ( 25 12 ) ( 25 16 ) ( 25 22 ) 93.7 A=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{25(25-12)(25-16)(25-22)}\approx \boxed{93.7}

Solution 2: Cosine Rule

Let the sides of the triangle be a = 12 , b = 16 a=12, b=16 and c = 22 c=22 and let the opposite angle of side a a be A A , of side b b be B B and of side c c be C C . By cosine rule, we have

c 2 = a 2 + b 2 2 a b cos C c^2=a^2+b^2-2ab \cos C

2 2 2 = 1 2 2 + 1 6 2 2 ( 12 ) ( 16 ) cos C 22^2=12^2+16^2-2(12)(16) \cos C

C 102.6 4 C \approx 102.64 ^\circ

So the area is

A = 1 2 a b sin C = 1 2 ( 12 ) ( 16 ) ( sin 102.64 ) 93.7 A=\dfrac{1}{2}ab \sin C = \dfrac{1}{2}(12)(16)(\sin 102.64) \approx \boxed{93.7}

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