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10th-class
mathematics
Mathematics - 2026
Variations
EXERCISE 3.6
Topic 3.10
4 mins read
hard Level
2024 Edition
EXERCISE 3.6
Textbook Content
EXERCISE 3.6
If
a
:
b
=
c
:
d
a:b = c:d
a
:
b
=
c
:
d
,
(
a
,
b
,
c
,
d
≠
0
)
(a,b,c,d\neq 0)
(
a
,
b
,
c
,
d
=
0
)
, then show that
4
a
−
9
b
4
a
+
9
b
=
4
c
−
9
d
4
c
+
9
d
4
a
−
9
b
4
a
+
9
b
=
4
c
−
9
d
4
c
+
9
d
{\frac{4a-9b}{4a+9b}}={\frac{4c-9d}{4c+9d}}
4
a
+
9
b
4
a
−
9
b
=
4
c
+
9
d
4
c
−
9
d
4
a
+
9
b
4
a
−
9
b
=
4
c
+
9
d
4
c
−
9
d
a
b
=
a
2
+
c
2
b
2
+
d
2
a
b
=
a
2
+
c
2
b
2
+
d
2
{\frac{a}{b}}=\sqrt{{\frac{a^{2}+c^{2}}{b^{2}+d^{2}}}}
b
a
=
b
2
+
d
2
a
2
+
c
2
b
a
=
b
2
+
d
2
a
2
+
c
2
{\frac{p(a+b)+qb:p(c+d)+qd=a:c}}{a^{2}+b^{2}:{\frac{a^{3}}{a+b}}=c^{2}+d^{2}:{\frac{c^{3}}{c+d}}}
a
a
−
b
:
a
+
b
b
=
c
c
−
d
:
c
+
d
d
a
a
−
b
:
a
+
b
b
=
c
c
−
d
:
c
+
d
d
{\frac{a}{a-b}}:{\frac{a+b}{b}}={\frac{c}{c-d}}:{\frac{c+d}{d}}
a
−
b
a
:
b
a
+
b
=
c
−
d
c
:
d
c
+
d
a
−
b
a
:
b
a
+
b
=
c
−
d
c
:
d
c
+
d
Topic Tags:
#exercise 3.6
#K-method
#proportions
3.4 K-Method (Continued)
3.4(ii) Real life problems based on variation