Question

1) In adjoining figure AE perpendicular seg BC, seg DF perpendicular line BC,
AE = 4, DF = 6, than find
A(AABC)/A(ADBC)​

Answer 1

Step-by-step explanation:

AE=4

DF=6

Then,

=A(triangleABC)/A(triangleDBC)

=AE/DF

=4/6

=2/3

therefore,A(ABC)/A(DBC)is 2/3.

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Answer 2

Given : seg AE perpendicular side BC,

seg DF perpendicular line BC  

AE =4, DF= 6

To find : Ar (Δ ABC )  / Ar (Δ DBC )

​

Solution:

Area of triangle = (1/2) * Base * Height  

Ar (Δ ABC ) = (1/2) * BC  * AE

Ar (Δ DBC ) = (1/2) * BC  * DF

Ar (Δ ABC )  / Ar (Δ DBC )   =    (1/2) * BC  * AE /  (1/2) * BC  * DF

=> Ar (Δ ABC )  / Ar (Δ DBC )   =     AE /   DF

=> Ar (Δ ABC )  / Ar (Δ DBC )   =     4  /    6

=> Ar (Δ ABC )  / Ar (Δ DBC )   =   2  / 3

Ar (Δ ABC )  / Ar (Δ DBC )   =   2  /  3

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