please solve this given question
Attachments:
Answers
Answered by
1
PM is the radius which is 10cm
According to the property of circles
=>>>>>A radius done perpendicular to the chod bisects the chord equally
Hence =>> AM = MB
Join AP and PB
AP=PB = 6cm (AB is equidistant from the point p or the centre of the circle)
Now in right angled AMP
AP^2 = PM^2 + AM^2
6^2 = 10^2 + AM^2
36 = 100+AM^2
AM => √64 => 8cm
AM = MB proved above
So, AM = MB =>8cm
length of AB => AM + MB => 8+8= 16cm
Hope it will help you
According to the property of circles
=>>>>>A radius done perpendicular to the chod bisects the chord equally
Hence =>> AM = MB
Join AP and PB
AP=PB = 6cm (AB is equidistant from the point p or the centre of the circle)
Now in right angled AMP
AP^2 = PM^2 + AM^2
6^2 = 10^2 + AM^2
36 = 100+AM^2
AM => √64 => 8cm
AM = MB proved above
So, AM = MB =>8cm
length of AB => AM + MB => 8+8= 16cm
Hope it will help you
Answered by
2
Answer:
Step-by-step explanation:
given:AB is a chord
AB⊥PM
radius=10 cm and PM=6cm
construction:Join AP and BP
solution:In ΔAPM and ΔBPM
AP=BP=10cm (radii of the same circle)
PM=PM=6cm (common)
∠PMA=∠PMB=90° (PM⊥AB)
by RHS congruency
ΔAPM≅ΔBPM
by CPCT,
AM=BM
In ΔPMB
(PM)²+(MB)²=(PB)² (pythagoras theorem)
36+(MB)²=100
(MB)²=64
MB=8cm
AB=2MB=2×8=16cm
Similar questions