circle of PQ ana PR are two chouds of radius & such PR = po If a and b are . dutances of Pig and Pre from the centure Ice PR that 46² = Q2 + 3x2 2 2 fecare that
Answers
Answer:
AB and AC are two chords of a circle with center O. Such that AB=2AC
p and q are ⊥ distances of AB and AC from center O i.e., OM=p and ON=q
r is the radius of the circle
To prove that:
4q
2
=p
2
+3r
2
Proof:
Join OA.
OM and ON are ⊥ distances of AB and AC from center O.
Here,
AN=
2
AC
(perpendicular from center to chord intersect at mid-point of the chord)
AM=
2
AB
(perpendicular from center to chord intersect at mid-point of the chord)
In right angled ΔOMA,
OM
2
+AM
2
=OA
2
p
2
+AM
2
=r
2
AM
2
=r
2
−p
2
…… (1)
In right angled ΔONA,
ON
2
+AN
2
=OA
2
q
2
+AN
2
=r
2
AN
2
=r
2
−q
2
…… (2)
Since,
AM=
2
AB
=
2
2AC
=AC=2AN
From equations (1) and (2), we have
r
2
−p
2
=AM
2
r
2
−p
2
=4AN
2
r
2
−p
2
=4[r
2
−q
2
]
4q
2
=p
2
+3r
2
LHS=RHSAB and AC are two chords of a circle with center O. Such that AB=2AC
p and q are ⊥ distances of AB and AC from center O i.e., OM=p and ON=q
r is the radius of the circle
To prove that:
4q
2
=p
2
+3r
2
Proof:
Join OA.
OM and ON are ⊥ distances of AB and AC from center O.
Here,
AN=
2
AC
(perpendicular from center to chord intersect at mid-point of the chord)
AM=
2
AB
(perpendicular from center to chord intersect at mid-point of the chord)
In right angled ΔOMA,
OM
2
+AM
2
=OA
2
p
2
+AM
2
=r
2
AM
2
=r
2
−p
2
…… (1)
In right angled ΔONA,
ON
2
+AN
2
=OA
2
q
2
+AN
2
=r
2
AN
2
=r
2
−q
2
…… (2)
Since,
AM=
2
AB
=
2
2AC
=AC=2AN
From equations (1) and (2), we have
r
2
−p
2
=AM
2
r
2
−p
2
=4AN
2
r
2
−p
2
=4[r
2
−q
2
]
4q
2
=p
2
+3r
2
LHS=RHSAB and AC are two chords of a circle with center O. Such that AB=2AC
p and q are ⊥ distances of AB and AC from center O i.e., OM=p and ON=q
r is the radius of the circle
To prove that:
4q
2
=p
2
+3r
2
Proof:
Join OA.
OM and ON are ⊥ distances of AB and AC from center O.
Here,
AN=
2
AC
(perpendicular from center to chord intersect at mid-point of the chord)
AM=
2
AB
(perpendicular from center to chord intersect at mid-point of the chord)
In right angled ΔOMA,
OM
2
+AM
2
=OA
2
p
2
+AM
2
=r
2
AM
2
=r
2
−p
2
…… (1)
In right angled ΔONA,
ON
2
+AN
2
=OA
2
q
2
+AN
2
=r
2
AN
2
=r
2
−q
2
…… (2)
Since,
AM=
2
AB
=
2
2AC
=AC=2AN
From equations (1) and (2), we have
r
2
−p
2
=AM
2
r
2
−p
2
=4AN
2
r
2
−p
2
=4[r
2
−q
2
]
4q
2
=p
2
+3r
2
LHS=RHSAB and AC are two chords of a circle with center O. Such that AB=2AC
p and q are ⊥ distances of AB and AC from center O i.e., OM=p and ON=q
r is the radius of the circle
To prove that:
4q
2
=p
2
+3r
2
Proof:
Join OA.
OM and ON are ⊥ distances of AB and AC from center O.
Here,
AN=
2
AC
(perpendicular from center to chord intersect at mid-point of the chord)
AM=
2
AB
(perpendicular from center to chord intersect at mid-point of the chord)
In right angled ΔOMA,
OM
2
+AM
2
=OA
2
p
2
+AM
2
=r
2
AM
2
=r
2
−p
2
…… (1)
In right angled ΔONA,
ON
2
+AN
2
=OA
2
q
2
+AN
2
=r
2
AN
2
=r
2
−q
2
…… (2)
Since,
AM=
2
AB
=
2
2AC
=AC=2AN
From equations (1) and (2), we have
r
2
−p
2
=AM
2
r
2
−p
2
=4AN
2
r
2
−p
2
=4[r
2
−q
2
]
4q
2
=p
2
+3r
2
LHS=RHS