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Standard VII
Mathematics
Question
If in a triangle
A
B
C
, right angle at
B
,
s
−
a
=
3
and
s
−
c
=
2
, then
a
=
2
;
c
=
3
a
=
3
;
c
=
4
a
=
4
;
c
=
3
a
=
6
;
c
=
8
A
a
=
3
;
c
=
4
B
a
=
6
;
c
=
8
C
a
=
2
;
c
=
3
D
a
=
4
;
c
=
3
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Solution
Verified by Toppr
(
s
−
a
)
=
3
;
(
s
−
c
)
=
2
3
+
a
=
2
+
c
a
+
1
=
c
−
−
−
e
q
n
.1
s
=
a
+
b
+
c
2
s
−
a
=
b
+
c
−
a
2
2
(
3
)
=
b
+
1
⇒
6
=
b
+
1
⇒
b
=
5
s
−
c
=
b
+
a
−
c
2
⇒
4
=
b
−
1
⇒
b
=
5
a
2
+
c
2
=
b
2
(Pythagoras theorem)
⇒
a
2
+
(
a
+
1
)
2
=
25
⇒
a
2
+
a
2
+
2
a
+
1
=
25
⇒
2
a
2
+
2
a
−
24
=
0
⇒
a
=
3
⇒
c
=
a
+
1
=
4
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