Question 66
In $$\triangle$$ABC, D and E are the points on sides AB and BC respectively such that DEĀ $$\parallel$$ AC. If AD : DB = 5 : 3, then what isĀ the ratio of the area of $$\triangle$$BDE to that of the trapezium ACED?
Solution
$$\triangle ABC $$~$$ \triangle DEB$$
($$\because$$ DE is parallel to AC)
$$\angle$$ B is a common angle.
So, ratio area of theĀ $$ \triangle BDE :Ā \triangle ABC $$
(3)^2 : (3 +Ā 5)^2 = 9 : 64
Area ofĀ trapezium ACED =Ā area of theĀ $$\triangle ABC - \triangle BDE$$ = 64 - 9 = 55
Ratio of the area of $$\triangle$$BDE to that of the trapezium ACED = 9 : 55
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