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If in a triangle ABC, D and E are on the sides AB and AC, such that, DE is parallel to BC and $$\frac{{AD}}{{BD}}$$ = $$\frac{3}{5}$$. If AC = 4 cm, then AE is
Answer & Solution
Correct Answer:
Option
A
According to question,
Given :
AD = 3
BD = 5
AB = 8
AC = 4
AE = ?
By applying B. P. T
$$\eqalign{ & \frac{{AD}}{{AB}} = \frac{{AE}}{{AC}} = \frac{{DE}}{{BC}} \cr & \frac{3}{8} = \frac{{AE}}{4} \cr & AE = \frac{3}{2} = 1.5\,{\text{cm}} \cr} $$
Given :
AD = 3
BD = 5
AB = 8
AC = 4
AE = ?
By applying B. P. T
$$\eqalign{ & \frac{{AD}}{{AB}} = \frac{{AE}}{{AC}} = \frac{{DE}}{{BC}} \cr & \frac{3}{8} = \frac{{AE}}{4} \cr & AE = \frac{3}{2} = 1.5\,{\text{cm}} \cr} $$
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