Module-1 Algebra Quadratic Equations Worksheet With Answers Page 9
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14MODULE -
1
Quadratic Equations
Algebra
Hence, perimeter of bigger square = 4x
and perimeter of smaller square = 4y
Therefore,
4x – 4y = 24
Notes
or
x – y = 6
or
x = y + 6
....................(1)
2
Also, since sum of areas of two squares is 468 m
2
2
Therefore,
x
+ y
= 468 ....................(2)
Substituting value of x from (1) into (2), we get
2
2
(y + 6)
+ y
= 468
2
2
or
y
+ 12y + 36 + y
= 468
2
or
2y
+ 12y – 432 = 0
2
or
y
+ 6y – 216 = 0
−
±
+
−
±
6
36
864
6
900
=
Therefore
y =
2
2
−
6 ±
30
or
y =
2
−
6 +
−
6 −
30
30
Therefore,
y =
or
2
2
or
y = 12 or – 18
Since, side of square can not be negative, so y = 12
Therefore,
x = y + 6 = 12 + 6 = 18
Hence, sides of squares are 18 m and 12 m.
Example 6.11:
The product of digits of a two digit number is 12. When 9 is added to the
number, the digits interchange their places. Determine the number.
Solution:
Let the digit at ten's place be x
and digit at unit's place be y
Therefore, number = 10 x + y
When digits are interchanged, the number becomes 10y + x
Therefore
10x + y + 9 = 10y + x
178
Mathematics Secondary Course
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