Question

Given a line segment AB, P is a point on the perpendicular bisector of line AB, such that AP = 10cm also AB = 12cm .What is the distance of point P from line segment AB

Answer 1

AP=10cm
AB=12cm
AC=1/2AB=1/2×12=6cm
As Triangle PCB is a right angled triangle we apply pythagoras theorem:
AP^2=PC^2+AC^2
=10^2=PC^2+6^2
=>PC^2=100-36=64=8^2
=>PC=8cm
Therefore the distance of point P from line segment AB=8cm.
Hope this helps :)

Answer 2

Answer:

LetthelengthofAB=xcm

Pisontheperpedicularbisector,

AP=x+7

BP=x+7(sincePisonthe⊥bisector;AP=BP)

x+x+7+x+7=38

3x=24

x=8

x+7=15cm

Sothethreesidesare8,15,15cm