Area of a Triangle

If ∆ be the area of a triangle ABC, Proved that, ∆ = ½ bc sin A = ½ ca sin B = ½ ab sin C

That is,

(i) ∆ = ½ bc sin A

(ii) ∆ = ½ ca sin B

(iii) ∆ = ½ ab sin C

Proof:

(i) ∆ = ½ bc sin A

Let ABC is a triangle. Then the following three cases arise:

Case I: When the triangle ABC is acute-angled:

Now form the above diagram we have,

sin C = AD/AC

sin C = AD/b, [Since, AC = b]

 AD = b sin C ……………………….. (1)

 Therefore, ∆ = area of triangle ABC

= 1/2 base × altitude

Area of Acute-angled Triangle

= ½ ∙ BC ∙ AD  

= ½ ∙ a ∙ b sin C, [From (1)]

= ½ ab sin C


Case II: When the triangle ABC is obtuse-angled:

Now form the above diagram we have,

sin (180° - C) = AD/AC

sin C = AD/AC, [Since, sin (π - θ) = sin θ]

sin C = AD/b, [Since, AC = b]

AD = b sin C ……………………….. (2)

Therefore, ∆ = area of the triangle ABC

Area of Obtuse-angled Triangle

= ½ base x altitude

= ½ ∙ BC ∙ AD

= ½ ∙ a ∙ b sin C, [From (1)]  

= ½ ab sin C


Case III: When the triangle ABC is right-angled

Now form the above diagram we have,

∆ = area of triangle ABC

= ½ base x altitude

= ½ ∙ BC ∙ AD  

= ½ ∙ BC ∙ AC

= ½ ∙ a ∙ b

Area of Right-angled Triangle

= ½ ∙ a ∙ b ∙ 1, [Since, ∠C = 90°. Therefore, sin C = sin 90° = 1]

= ½ ab sin C

Therefore, in all three cases, we have ∆ = ½ ab sin C

In a similar manner we can prove the other results, (ii) ∆ = ½ ca sin B and (iii) ∆ = ½ ab sin C.






11 and 12 Grade Math

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