Question 66
In $$\triangle$$ABC, D and E are the points on sides AB and BC respectively such that DE $$\parallel$$ AC. If AD : DB = 5 : 3, then what is 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 the $$ \triangle BDE : \triangle ABC $$
(3)^2 : (3 + 5)^2 = 9 : 64
Area of trapezium ACED = area of the $$\triangle ABC - \triangle BDE$$ = 64 - 9 = 55
Ratio of the area of $$\triangle$$BDE to that of the trapezium ACED = 9 : 55
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