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Byju's Answer
Standard VI
Mathematics
Area of Square
Sum of area o...
Question
Sum of area of two squares is
468
m
2
. If the difference of their permeter is 24 m, and the sides of the two squares.
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Solution
Sum of the areas of two squares
=
468
m
2
Let
a
and
b
be the sides of the two squares.
⇒
a
2
+
b
2
=
468
.....
(
1
)
Also given that,
the difference of their perimeters
=
24
m
⇒
4
a
−
4
b
=
24
⇒
a
−
b
=
6
⇒
a
=
b
+
6
.......
(
2
)
We need to find the sides of the two squares.
Substituting the value of
a
from equation
(
2
)
in equation
(
1
)
, we have,
(
b
+
6
)
2
+
b
2
=
468
⇒
b
2
+
12
b
+
36
+
b
2
=
468
⇒
2
b
2
+
12
b
+
36
−
468
=
0
⇒
2
b
2
+
12
b
−
432
=
0
⇒
b
2
+
6
b
−
216
=
0
⇒
b
2
+
18
b
−
12
b
−
216
=
0
⇒
b
(
b
+
18
)
−
12
(
b
+
18
)
=
0
⇒
(
b
+
18
)
(
b
−
12
)
=
0
⇒
b
=
−
18
or
12
Since side cannot be negative.Hence
b
=
12
m
∴
a
=
b
+
6
=
12
+
6
=
18
m
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. If the difference of their perimeters is 24 m, find the sides of the two squares.
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