Question

in triangle abc,DE is parallel to BC and AD/DB=3/5.if AC=4.8 cm,find AE

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

HOPE THIS WILL HELP YOU..

Answer 2

Answer:In ∆ABC , DE||BC

AD/DB= AE/EC

[ By B.P.T]

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