Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 3 Algebra Ex 3.3
Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 3 Algebra Ex 3.3
Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 3 Algebra Ex 3.3
Miscellaneous Practice problems
Question 1.
Using identity, find the value of
(i) (4.9)2
(ii) (100.1)2
(iii) (1.9) × (2.1)
Solution:
(i) (4.9)2
(4.9)2 = (5 – 0.1)2
Substituting a = 5 and b = 0.1 in
(a – b)2 = a2 – 2ab + b2, we have
(5 – 0.1)2 = 52 – 2(5) (0.1) + (0.1)2
(4.9)2 = 25 – 1 + 0.01 = 24 + 0.01
(4.9)2 = 24.01
(ii) (100.1)2
(100.1)2 = (100 + 0.1)2
Substituting a = 100 and b = 0.1 in
(a + b)2 = a2 + 2ab + b2, we have
(100 + 0.1)2 = (100)2 + 2(100) (0.1) + (0.1)2
(100.1)2 = 10000 + 20 + 0.01
(100.1)2 = 10020.01
(iii) (1.9) × (2.1)
(1.9) × (2.1) = (2 – 0.1) × (2 + 0.1)
Substituting a = 100 and b = 0.1 in
(a – b) (a + b) = a2 – b2 we have
(2 – 0.1) (2 + 0.1) = 22 – (0.1)2
(1.9) × (2.1) = 4 – 0.01
(9.9) (2.1) = 3.99
Question 2.
Factorise: 4x2 – 9y2
Solution:
4x2 – 9y2 = 22 x2 – 32 y2 = (2x)2 – (3y)2
Substituting a = 2x and b = 3y in
(a2 – b2) = (a + b) (a – b), we have
(2x)2 – (3y)2 = (2x + 3y) (2x – 3y)
∴ Factors of 4x2 – 9y2 are (2x + 3y) and (2x – 3y)
Question 3.
Simplify using identities
(i) (3p + q) (3p + r)
(ii) (3p + q) (3p – q)
Solution:
(i) (3p + q) (3p + r)
Substitute x = 3p,a = q and b = r in
(x + a) (x + b) = x2 + x(a + b) + ab
(3p + q)(3p + r) = (3p)2 + 3p (q + r) + (q × r)
= 32 p2 + 3p (q + r) + qr
(3p + q)(3p + r) = 9p2 + 3p(q + r) + qr
(ii) (3p + q) (3p – q)
Substitute a = 3p and b = q in
(a + b) (a – b) = a2 – b2, we have
(3p + q) (3p – q) = (3p)2 – q2 = 32 p2 – q2
(3P + q) (3p – q) = 9p2 – q2
Question 4.
Show that (x + 2y)2 – (x – 2y)2 = 8xy.
Solution:
LHS = (x + 2y)2 – (x – 2y)2
= x2 + (2 × x × 2y) + (2y)2 – [x2 – (2 × x × 2y) + (2y)2]
= x2 + 4xy + 4y2 – [x2 – 4xy + 22y2]
= x2 + 4xy + 4y2 – x2 + 4xy – 4y2
= x2 – x2 + 4xy + 4xy + 4y2 – 4y2
= x2 (1 – 1) + xy (4 + 4) + y2 (4 – 4)
= 0x2 + 8xy + 0y2 = 8xy = RHS
∴ (x + 2y)2 – (x – 2y)2 = 8xy
[∵ (a + b)2 = a2 + 2ab + b2 (a – b)2 = a2 – 2ab + b2]
Question 5
The pathway of a square paddy field has 5 m width and length of its side is 40 m. Find the total area of its pathway. (Note: Use suitable identity)
Solution:
Given side of the square = 40 m
Also width of the pathway = 5 m
∴ Side of the larger square = 40m + 2(5)m = 40m + 10m = 50m
Area of the path way = area of large square – area of smaller square
= 502 – 402
Substituting a = 50 and b = 40 in
a2 – b2 = (a + b) (a – b) we have
502 – 402 = (50 + 40) (50 – 40)
Area of pathway = 90 × 10
Area of the pathway = 900 m2
Challenge Problems
Question 1.
If X = a2 – 1 and Y = 1 – b2, then find X + Y and factorize the same.
Solution:
Given X = a2 – 1
Y = I – b2
X + Y = (a2 – 1) + (1 – b2)
= a2 – 1 + 1 – b2
We know the identity that a2 – b2 = (a + b) (a – b)
∴ X + Y = (a + b) (a – b)
Question 2.
Find the value of (x – y) (x + y) (x2 + y2).
Solution:
We know that (a – b) (a + b) = a2 – b2
Put a = x and b = y in the identity (1) then
(x – y) (x + y) = x2 – y2
Now (x – y) (x + y)(x2 + y2) = (x2 – y2) (x2 + y2)
Again put a = x2 and b = y2 in (1)
We have (x2 – y2) (x2 + y2) = (x2)2 – (y2)2 = x4 – y4
So (x – y) (x + y) (x2 + y2) = x4 – y4
Question 3.
Simplify (5x – 3y)2 – (5x + 3y)2.
Solution:
We have the identities
(a + b)2 = a2 + 2ab + b2
(a – b)2 = a2 – 2ab + b2
So (5x – 3y)2 – (5x + 3y)2 = (5x)2 – (2 × 5x × 3y) + (3y)2
= 52x2 – 30xy + 32 y2 – [52x2 – 30xy + 32 y2]
= 25x2 – 30xy + 9y2 – [25x2 + 30xy + 9y2]
= 25x2 – 30xy + 9y2 – 25x2 – 30xy – 9y2
= x2 (25 – 25) – xy (30 + 30) + y2 (9 – 9)
= 0x2 – 60xy + 0y2 = – 60 xy
∴ (5x – 3y)2 – (5x + 3y)2 = -60xy
Question 4.
Simplify : (i) (a + b)2 – (a – b)2
(ii) (a + b)2 + (a – b)2
Solution:
Applying the identities
(a + b)2 = a2 + 2ab + b2
(a – b)2 = a2 – 2ab + b2
(i) (a + b)2 – (a – b)2 = a2 + 2ab + b2 – [a2 – 2ab + b2]
= a2 + 2ab + b2 – a2 + 2ab – b2
= a2 (1 – 1) + ab (2 + 2) + b2 (1 – 1)
= 0a2 + 4 ab + 0b2 = 4ab
(a + b)2 – (a – b)2 = 4ab
(ii) (a + b)2 + (a – b)2 = a2 + 2ab + b2 + (a2 – 2ab + b2)
= a2 + 2ab + b2 + a2 – 2ab + b2
= a2 (1 + 1) + ab (2 – 2) + b2 (1 + 1)
= 2a2 + 0 ab + 2b2 = 2a2 + 2b2 = 2 (a2 + b2)
∴ (a + b)2 – (a – b)2 = 2 (a2 + b2)
Question 5.
A square lawn has a 2 m wide path surrounding it. If the area of the path is 136 m2, find the area of lawn.
Solution:
Let the side of the lawn = a m
then side Of big square = (a + 2(2)) m
= (a + 4)m
Area of the path – Area Of large square – Area of smaller square
136 = (a + 4)2 – a2
136 = a2 + (2 × a × 4) + 42 – a2
136 = a2 + 8a + 16 – a2
136 = 8a + 16
136 = 8 (a + 2)
Dividing by 8
17 = a + 2
Subtracting 2 on both sides
17 – 3 = a + 2 – 2
15 = a
∴ side of small square = 15 m
Area of square = (side × side) Sq. units
∴ Area of the lawn = (15 × 15)m2 = 225 m2
∴ Area of the lawn = 225 m2
Question 6.
Solve the following inequalities.
(i) 4n + 7 > 3n + 10, n is an integer
(ii) 6(x + 6) > 5 (x – 3), x is a whole number.
(iii) -13 < 5x + 2 < 32, x is an integer.
Solution:
(i) 4n + 7 > 3n + 10, n is an integer.
4n + 7 – 3n > 3n + 10 – 3n
n(4 – 3) + 7 > 3n + 10 – 3n
n (4 – 3) + 7 > n (3 – 3) + 10
n + 7 > 10
Subtracting 7 on both sides
n + 7 – 7 > 10 – 7
n > 3
Since the solution is an integer and is greater than or equal to 3, the solution will be 3,
4, 5, 6, 7, …..
n = 3, 4, 5, 6,7, ….
(ii) 6 (x + 6) > 5 (x – 3), x is a whole number.
6x + 36 > 5x – 15
Subtracting 5x on both sides
6x + 36 – 5x > 5x – 15 – 5x
x (6 – 5) + 36 > x(5 – 5) – 15
x + 36 > -15
Subtracting 36 on both sides
x + 36 – 36 > -15 -36
x > -51
The solution is a whole number and which is greater than or equal to -51
∴ The solution is 0, 1, 2, 3, 4,…
x = 0,1,2, 3,4,…
(iii) -13 < 5x + 2 < 32, x is an integer.
Subtracting throughout by 2
-13 – 2 < 5x + 2 – 2 < 32 – 2
-15 < 5x < 30
Dividing throughout by 5
−15/5 < 5x/5 < 30/5
– 3 < x < 6
∴ Since the solution is an integer between -3 and 6 both inclusive, we have the solution
as -3, -2, -1,0, 1,2, 3, 4, 5, 6.
i.e. x = -3, -2, 0, 1, 2, 3,4, 5 and 6.
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