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Byju's Answer
Standard XII
Mathematics
Pythagoras Theorem
In PQR, PQ=...
Question
In
△
P
Q
R
,
P
Q
=
24
c
m
,
Q
R
=
7
c
m
and
∠
P
Q
R
=
90
∘
. Find the radius (in
c
m
) of the inscribed circle.
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Solution
By Pythagoras' theorem,
P
R
2
=
P
Q
2
+
Q
R
2
P
R
2
=
24
2
+
7
2
=
576
+
49
=
625
∴
P
R
=
25
c
m
Let the inradius of
△
P
Q
R
be
x
c
m
.
□
O
A
Q
C
is a square. Hence
Q
A
=
x
c
m
and
A
R
=
(
7
−
x
)
c
m
R
A
and
R
B
act as tangents to the incircle from point
R
, hence their lengths are equal.
∴
R
B
=
A
R
=
(
7
−
x
)
c
m
.
Similarly,
P
B
=
P
C
=
(
24
−
x
)
c
m
P
R
=
P
B
+
R
B
⇒
25
=
(
24
−
x
)
+
(
7
−
x
)
⇒
25
=
31
−
2
x
⇒
x
=
3
c
m
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