Question

In triangle ABC, DE parallel BC and AD:DB = 3:5 than
AE:EC is equal to.​

Answer 1

BASIC PROPORTIONALITY THEOREM(BPT) :

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points than the other two sides are divided in the same ratio.

SOLUTION:

In ∆ABC , DE||BC

AD/DB= AE/EC

[ By basic PROPORTIONALITY theorem]

AD/DB= AE/(AC-AE)

3/5 = AE / (4.8 - AE)

3(4.8 - AE) = 5 AE

14.4 - 3AE = 5AE

14.4 = 5AE+3AE

14.4 = 8AE

AE = 14.4/8

AE= 1.8

Hence, AE is = 1.8 cm

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Answer 2

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(BPT) :

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points than the other two sides are divided in the same ratio.

SOLUTION:

In ∆ABC , DE||BC

AD/DB= AE/EC

[ By basic PROPORTIONALITY theorem]

AD/DB= AE/(AC-AE)

3/5 = AE / (4.8 - AE)

3(4.8 - AE) = 5 AE

14.4 - 3AE = 5AE

14.4 = 5AE+3AE

14.4 = 8AE

AE = 14.4/8

AE= 1.8

Hence, AE is = 1.8 cm