Question
If two medians BE and CF of a triangle ABC, intersect each other at G and if BG = CG, ∠BGC = 60°, BC = 8 cm, then area of the triangle ABC is:
Answer: Option B
According to question,
⇒ ∵ ∠BGC = 60° (given)
⇒ ∠GBC = ∠GCB = x°
⇒ x° + x° + 60° = 180°
⇒ x = 60°
⇒ So ΔBGC is an equilateral triangle with side 8 cm each
Then,
Area pf triangle ΔBGC
= $$\frac{{\sqrt 3 }}{4}$$ a2
= $$\frac{{\sqrt 3 }}{4}$$ 82
= 16$${\sqrt 3 }$$ cm2
⇒ Area of ΔABC
= Area (ΔBGC + ΔAGC + ΔAGB)
⇒ Area of ΔABC = 3 × 16$${\sqrt 3 }$$
⇒ Area of ΔABC = 48$${\sqrt 3 }$$ cm2 {∵ ΔBGC = ΔAGC = ΔAGB}
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According to question,
⇒ ∵ ∠BGC = 60° (given)
⇒ ∠GBC = ∠GCB = x°
⇒ x° + x° + 60° = 180°
⇒ x = 60°
⇒ So ΔBGC is an equilateral triangle with side 8 cm each
Then,
Area pf triangle ΔBGC
= $$\frac{{\sqrt 3 }}{4}$$ a2
= $$\frac{{\sqrt 3 }}{4}$$ 82
= 16$${\sqrt 3 }$$ cm2
⇒ Area of ΔABC
= Area (ΔBGC + ΔAGC + ΔAGB)
⇒ Area of ΔABC = 3 × 16$${\sqrt 3 }$$
⇒ Area of ΔABC = 48$${\sqrt 3 }$$ cm2 {∵ ΔBGC = ΔAGC = ΔAGB}
Was this answer helpful ?
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