In a ∆ABC, D and E are two points on the side BC such that they trisect the line BC. M and N are the two point on line AB and AC. ME || AC and ND || AB. The line ME and ND Intersect at O. Find (Area of ∆DOE + Area of □AMON)/(Area of ∆ABC)
Answers
Solution :-
Construction :- Draw a perpendicular from vertices A on BC through point O such that, it cuts MN At Q and BC at P .
Let us assume that,
- AQ = a unit .
- QO = b unit .
- OP = c unit .
- AP = (a + b + c) unit .
we know that,
- Area of ∆ = (1/2) * Base * Height .
so,
→ Area (∆AMN) = (1/2) * MN * AQ = (1/2) * MN * a
→ Area (∆MON) = (1/2) * MN * QO = (1/2) * MN * b
then,
→ Area (□AMON) = Area (∆AMN) + Area (∆MON) = (1/2) * MN * a + (1/2) * MN * b = (1/2)MN(a + b)
and,
→ Area (∆DOE) = (1/2) * DE * PO = (1/2) * DE * c
- put DE = (1/3)BC { given that D and E Trisect the line BC .}
→ Area (∆DOE) = (1/2) * (1/3)BC * c = (1/6) * BC * c
also,
→ Area (∆ABC) = (1/2) * BC * AP = (1/2) * BC * (a + b + c)
then,
→ (Area ∆DOE + Area □AMON)/(Area ∆ABC) = [{(1/6) * BC * c} + (1/2)MN(a + b)] / [ (1/2) * BC * (a + b + c) ]
now, given that,
- ME || AC .
- ND || AB .
- so, MN ll BC and MN = (1/2) BC
therefore,
→ (Area ∆DOE + Area □AMON)/(Area ∆ABC) = [{(1/6) * BC * c} + (1/2) * (1/2) * BC * (a + b)] / [ (1/2) * BC * (a + b + c) ]
→ (Area ∆DOE + Area □AMON)/(Area ∆ABC) = (1/2)BC[(c/3) + (a + b)/2] / (1/2)BC (a + b + c)
→ (Area ∆DOE + Area □AMON)/(Area ∆ABC) = (2c + 3a + 3b)/(6a + 6b + 6c) (Ans.)
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