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TRINA DAS

· started a discussion

· 1 Months ago

Not Clear (Question:451133.0)

Please explain 2nd line of solution in detail.

Question:
In the figure given below, AD bisects ∠BAC. If the area of ∆ABD = 40cm2 and AC = 3AB, then the area of ∆ABC: 

Options:
A) 150cm2
B) 100cm2
C) 120cm2
D) 160cm2
Solution:

\(\cfrac{Area of ∆ABD}{Area of ∆ADC}\)    = \(\cfrac{\cfrac{1}{2} × (BD) × (OA)}{\cfrac{1}{2} × (DC) × (OA)}\)

= \(\cfrac{BD}{DC}\) = \(\cfrac{AB}{AC}\) = \(\cfrac{1}{3}\) (∵ AD is the angle bisector of ∠BAC)

⇒ Area of ∆ADC = 3 × 40 = 120cm2

 

∴ Area of ∆ABC = 120 + 40 = 160cm2

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