AndyB wrote:
In triangle ABC shown above, DE || BC and DE/BC = 1/4. If area of triangle ADE is 10, find the area of the trapezium BCED and the area of triangle CED.1
Plug in values that satisfy the given conditions:
In the figure above, DE =1.
Since DE/BC = 1/4, BC = 4.
Since the area of ∆ADE is 10:
(1/2)(DE)(AF) = 10
(1/2)(1)(AF) = 10
AF = 20.
Since DE||BC, ∆ADE is similar to ∆ABC.
Since ∆ADE is similar to ∆ABC, and DE/BC = 1/4, all corresponding lengths must yield a ratio of 1:4.
Thus, AF/AG = 1/4:
20/AG = 1/4
AG = 80.
Since AG = 80 and AF = 20, FG = 80-20 = 60. This is the height both of trapezoid BCED and of ∆CED.
Trapezoid BCED:
The area of a trapezoid = (b1 + b2)(h)/2.
Thus, the area of trapezoid BCED = (1+4)(60)/2 = 150.
∆CED:
If we consider DE the base, then the area = (1/2)bh = (1/2)(1)(60) = 30.
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