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Class 8 Maths Chapter 6 We Distribute Yet Things Multiply Solutions
Ganita Prakash Class 8 Chapter 6 Solutions
Class 8 Maths Ganita Prakash Chapter 6 Solutions We Distribute Yet Things Multiply
Intext (NCERT Page 138)
Question 1.
What would we get if we had expanded (a +1) (b +1) by first taking (b + 1) as a single term? Try it?
Solution:
We have, (a + 1) (b + 1)
Taking (b + 1) first term, we get
(b + 1) (a + 1) = (b + 1) a + (b + 1)1 [by distributive property]
= ba + a + b + 1
= ab + a + b + 1 [∵ ab = ba]
Question 2.
Will the product always increase? Find 3 examples where the product decreases.
Solution:
No, the product does not always increases.
Here are three examples where the product decreases.
(i) Let a = 5 and b = 3
Original product = 5 × 3 = 15
When one number is increased by 1 and other is decreased by 1.
∴ New product = (5 + 1) × (3 – 1) = 6 × 2 = 12
Hence, the product decreases.
(ii) Let a = 10 and b = 8
Original product = 10 × 8 = 80
When one number is increased by 1 and other is decreased by 1.
∴ New product = (10 + 1) × (8 – 1) = 11 × 7 = 77
Hence, the product decreases.
(iii) Let a = 2 and b = 0
Original product = 2 × 0 = 0
When one number is increased by 1 and other is decreased by 1.
∴ New product = (2 + 1) × (0 – 1) = 3 × (- 1) = – 3
Hence, the product decreases.
Intext (NCERT Page 140)
Question 1.
Use Identity 1 to find how the product changes when
(i) one number is decreased by 2 and the other increased by 3.
Solution:
We know that,
(a + m) (b + n) = ab + mb + an + mn …(i)
Here, m = – 2 and n = 3
∴ (a – 2) (b + 3) = ab – 2b + 3a – 6 [from Eq. (i)]
= ab + (3a – 2b – 6)
Therefore, the product changes by (3a – 2b – 6).
(ii) both numbers are decreased, one by 3 and the other by 4.
Solution:
Here, m= -3 andn = -4
(a – 3) (b – 4) = ab – 3b – 4a + 12
= ab + (-3b – 4a + 12)
Therefore, the product changes by (-3b – 4a + 12).
Figure it Out (NCERT Page 142-143)
Question 1.
Observe the multiplication grid below. Each number inside the grid is formed by multiplying two numbers. If the middle number of a 3 × 3 frame is given by the expression pq, as shown in the figure, write the expressions for the other numbers in the grid.
Solution:
Since, the middle number of the 3 × 3 frame is pq.
Row number = p and column number = q
Now, row number above p = p – 1,
row number below p = p + 1,
column number to left of q = q – 1
and column number to the right of q = q + 1
The expressions for other numbers in the 3 × 3 frame are
Top left = (p – 1) (q – 1)
Top middle = (p – 1)q
Top right = (p – 1)(q + 1)
Middle left = p(q – 1)
Middle right = p(q + 1)
Bottom left = (p + 1)(q – 1)
Bottom middle = (p + 1)q
Bottom right = (p + 1) (q + 1)
Question 2.
Expand the following products.
(i) (3 + u)(v – 3)
Solution:
We have, (3 + u) (v – 3) = (3 + u) v – (3 + u)3 [by distributive property]
= 3v + uv – 9 – 3u
= 3v + uv – 3u – 9
(ii) \(\frac{2}{3}\) (15 + 6a)
Solution:
We have, \(\frac{2}{3}\) (15 + 6a) = \(\frac{2}{3}\) × 15 + \(\frac{2}{3}\) × 6a
[by distributive property] = 2 × 5 + 2 × 2a
= 10 + 4a
(iii) (10a + b) (10c + d)
Solution:
We have, (10a + b) (10c + d)= (10a + b) 10c + (10a + b)d
[by distributive property]
= 100ac + 10bc + 10ad + bd
(iv) (3 – x) (x – 6)
Solution:
We have, (3 – x) (x – 6) = (3 – x) x + (3 – x) (-6)
[by distributive property]
= 3x – x2 – 18 + 6x
= – x2 + 9x – 18
(v) (-5a + b)(c + d)
Solution:
We have, (-5a + b) (c + d) = (-5a + b) c + (-5a + b)d
[by distributive property]
= – 5ac + bc – 5ad + bd
(vi)(5 + z)(y+ 9)
Solution:
We have, (5 + z)(y + 9) = (5 + z)y + (5 + z) 9
[by distributive property]
= by + yz + 45 + 9z
Question 3.
Find 3 examples where the product of two numbers remains unchanged when one of them is increased by 2 and the other is decreased by 4.
Solution:
Let the two numbers be x and y.
Product = xy
If one number is increased by 2 and the other is decreased by 4.
Then the new numbers become (x + 2) and (y – 4).
∴ Product = (x + 2) (y – 4)
According to the question,
xy = (x + 2) (y – 4)
⇒ xy = x (y – 4) + 2 (y – 4)
⇒ xy = xy – 4x + 2y – 8
⇒ 0 = – 4x + 2y – 8
⇒ 4x – 2y + 8 = 0
⇒ 2x – y + 4 = 0
⇒ y = 2x + 4
Example 1
Let x = 1
Then, y = 2 × 1 + 4 = 2 + 4 = 6
∴ Original product = 1 × 6 = 6
New numbers are 1 + 2 and 6 – 4 i.e. 3 and 2.
New product = 3 × 2 = 6
∴ The product remains unchanged.
Example 2 Let x = 2
Then, y = 2 × 2 + 4 = 4 + 4 = 8
∴Original product = 2 × 8 = 16
New numbers are 2 + 2 and 8 – 4 i.e. 4 and 4.
∴New product = 4 × 4 = 16
∴The product remains unchanged.
Example 3 Let x = 3
Then, y = 2 × 3 + 4 = 6 + 4 = 10
∴Original product = 3 × 10 = 30
New numbers are 3 + 2 and 10 – 4 i.e. 5 and 6.
New product = 5 × 6 = 30
∴ The product remains unchanged.
Question 4.
Expand
(i) (a + ab – 3b2)(4 + b)
Solution:
We have, (a + ab – 3b2)(4 + b)
= (a + ab – 3b2) 4 + (a + ab – 3b2)b
= 4a + 4ab – 12b2 + ab + ab2 – 3b3
= -3b3 – 12b2 + ab2 + 5ab + 4a
(ii) (4y + 7) (y + 11z – 3)
Solution:
We have, (4y + 7) (y + 11z – 3)
= (4y + 7)y + (4y + 7)11z + (4y + 7) (-3)
= 4y2 +7y + 44yz + 77z – 12y – 21
= 4y2 + 44yz – 5y + 77z – 21
Question 5.
Expand
(i) (a – b) (a + b)
(ii) (a – b) (a2 + ab + b2)
(iii) (a – b)(a3 + a2b + ab2 + b3).
Do you see a pattern? What would be the next identity in the pattern that you see? Can you check it by expanding?
Solution:
(i) We have, (a – b) (a + b) = (a – b) a + (a – b)b [by distributive property]
= a2 – ba + ab – b2
= a2 – b2 [vab = ba]
(ii) We have, (a – b) (a2 + ab + b2)
= (a – b)a2 + (a – b)ab + (a – b)b2
= a2 – ba2 + a2b – ab2 + ab2 – b3
= a3 – b3
(iii) We have, (a – b) (a3 + a2b + ab2 + b3)
= (a – b)a3 + (a – b)a2b + (a – b)ab2 + (a – b)b3
= a4 – ba3 + a3b – a2b2 + a2b2 – ab3 + ab3 – b4
= a4 + (a3b – a3b) + (a2b2 – a2b2) + (ab3 – ab3) – b4
= a4 – b4
The pattern observed is that the product of (a – b) and a sum of power of a and b (where the powers of a decreases from n to 0 and the powers of b increases from 0 to n results in a(n + 1) – b(n + 1) .
Generally, for an integer n ≥ 2
(a – b)(a4 + an-1 + …+ abn-1 + bn ) = a(n + 1) – b(n + 1)
Following this pattern, the next identify would be for n = 5
(a – b) (a4 + a3 b + a2 b2 + ab3 + b4 ) = a5 – b5
Now, checking by expanding
LHS = (a – b)(a4 + a3 b + a2 b2 + ab3 + b4 )
= (a – b) a4 + (a – b)a3 b + (a – b)a2 b2 + (a – b)ab3 + (a – b)b4
= a3 – ba4 + a4 b – a3 b2 + a3 b2 – a3b3 + a2 b3 – ab4 + ab4 – b3
= a5 + (a4b – ba4) + (a3b2 – a3b2) + (a2b3 – a2b3) + (ab4 – ab4) – b3
= a3 – b3
= RHS
Intext (NCERT Page 144)
Question 1.
Describe a general rule to multiply a number (of any number of digits) by 11 and write the product in one line.
Evaluate
(i) 94 × 11
Solution:
Let a be any number, then a × 11 = a × 10 + a
We have, 94 × 11
∵ a × 11 = a × 10 + a
∴ 94 × 11 = 94 × 10 + 94 [∵ a = 94]
= 940 + 94
= 1034
(ii) 495 × 11
Solution:
We have, 495 × 11 = 495 × 10 + 495 [∵ a = 495]
= 4950 + 495
= 5445
(iii) 3279 × 11
Solution:
We have, 3279 × 11 = 3279 × 10 + 3279 [∵ a = 3279]
= 32790 + 3279
= 36069
(iv) 4791256 × 11
Solution:
We have, 4791256 × 11 = 4791256 × 10 + 4791256 [∵ a = 4791256]
= 47912560 + 4791256
= 52703816
Question 2.
Use this to multiply 3874 × 101 in one line.
Solution:
We have, 3874 × 101
We know that
dcba × 101 = dcba × 100 + dcba
∴ 3874 × 101 = 391274
Question 3.
What could be a general rule to multiply a number by 101 and write the product in one line? Extend this rule for multiplication by 1001,10001
Solution:
General rule to multiply a number by 101.
If a be any number then a × 101 = a × 101 + 1 + a.
General rule for multiplication by 1001, 10001,….
If a be any number multiply by 1001, 10001,….
Then the result is a × 10k + 1 + a, where k is number of zeros between two Is.
Intext (NCERT Page 145-146)
Question 1.
What if we write 652 as (30 + 35)2 or (52 + 13)2?
Draw the figures and check the area that you get.
Solution:
We have, (30 + 35)2 = (30 + 35) (30 + 35)
= 30 × 30 + 30 × 35 + 35 × 30 + 35 × 35
= 302 + 2 × 30 × 35 + 352
= 900 + 2100 + 1225
= 4225 sq units
Now, (52 + 13)2 = (52 + 13) (52 + 13)
= 52 × 52 + 52 × 13 + 13 × 52 + 13 × 13
= 522 + 2 × 52 × 13+ 132
= 2704 + 1352 + 169
= 4225 sq units
Question 2.
If a and b are any two integers, is (a + b)2 always greater than a2 + b2? If not, when is it greater?
Solution:
No, (a + b)2 is not always greater than a2 + b2.
For example, take a = 3, to = – 5.
Then, (a + b)2 = (3 + (- 5))2
= (3 – 5)2
= (- 2)2
= 4
and a2 + b2 = (3)2 + (- 5)2
= 9 + 25
= 34
Here, (a + b)2 = 4 < 34 = a2 + b2.
Now, let a and to are any two integers.
Let us suppose (a + b)2 > a2 + b2
⇒ a2 + 2ab + b2 > a2 + b2
⇒ 2ab > 0
⇒ ab > 0
a and b must either both positive or both negative.
∴ (a + b)2 is greater than a2 + b2 only if a and to have the same sign.
Question 3.
Use Identity 1A to find the values of 1042,372.
(Hint Decompose 104 and 37 into sums or differences of numbers whose squares are easy to compute.)
Solution:
We have, 1042 = (100 + 4)2
= (100)2 + 2 × 100 × 4 + (4)2 [v(a + b)2 = a2 + 2ab + b2]
= 10000 + 800 + 16
= 10816
Also, we have, 372 = (30 + 7)2
= (30)2 + 2 × 30 × 7 + (7)2
[∵ (a + b)2 = a2 + 2ab + b2]
= 900 + 420 + 49
= 1369
Question 4.
Use Identity 1A to write the expressions for the following.
(i) (m + 3)2
Solution:
We have, (m + 3)2 = m2 + 2 × m × 3 + (3)2
[∵ (a + b)2 = a2 + 2ab + b2]
= m2 + 6m + 9
(ii) (6 + p)2
Solution:
We have, (6 + p)2 = (62) + 2 × 6 × p + (p)2
= 36 + 12p + p2
Question 5.
Expand (3j + 2k)2 using both the identity and by applying the distributive property.
Solution:
We have, (3; + 2k)2 = (3j)2 + 2 × 3j × 2k + (2k)2
[∵(a + b)2 = a2 + 2ab + b2]
= 9j2 + 12 jk + 4k2
Also, (3j + 2k)2 = (3j + 2k) (3j + 2k)
= (3j + 2k)3j + (3j + 2k)2k
[by distributive property]
= (3j × 3j) + (2k × 3j) + (3j × 2k) + (2k × 2k)
= (3j)2 + 2(2k × 3f) + (2k)2
= 9j2 + 12 jk + 4k2
Intext (NCERT Page 147)
Question 1.
Find the general expansion of (a – b)2 using geometry, as we did for 552.
Solution:
Let ABCD be a square with length of side be a and take a smaller square in this square at one corner with length of side be to.
Now, area of square with side (a – b)
= area of larger square ABCD – area of two rectangles ROQC and APOS – area of smaller square PBQO
= a2 – 2b(a – b) – b2
= a2 – 2ab + 2b2 – b2
⇒ (a – b)2 = a2 – 2ab + b2
Question 2.
Use the identity (a – b)2 to find the values of
(i) 992
Solution:
We have, 992 = (100 – 1)2
= (100)2 – 2 × 100 × 1 + (1)2
[∵ (a – b)2 = a2 – 2ab + b2]
= 10000 – 200 + 1
= 10001 – 200
= 9801
(ii) 582
Solution:
We have, 582 = (60 – 2)2
= (60)2 – 2 × 60 × 2 + (2)2
= 3600 – 240 + 4
= 3604 – 240
= 3364
Question 3.
Expand the following using both Identity IB and by applying the distributive property.
(i) (b – 6)2
Solution:
We have, (b – 6)2
Using identity (x – y)2 = x2 – 2xy + y2,
we get (b – 6)2 = b2 – 2 × b × 6 + 62
= b2 – 12b + 36
Using distributive property, we get
(b – 6)2 = (b – 6) (b – 6)
= (b – 6)b + (b – 6) (—6)
= b2 – 6b – 6b + 36
= b2 – 12b + 36
(ii) (-2a + 3)2
Solution:
We have, (- 2a + 3)2 = (3 – 2a)2
Using identity (x – y)2 = x2 – 2xy + y2,
we get (3 – 2a)2 = (3)2 – 2 × 3 (2a) + (2a)2
= 9 – 12a + 4a2
Using distributive property, we get
(3 – 2a)2 = (3 – 2a) (3 – 2a)
= (3 – 2a) 3 + (3 – 2a) (-2a)
= 9 – 6a – 6a + 4a2
(iii) (7y – \(\frac{3}{4z}\))2
Solution:
We have, (7y – \(\frac{3}{4z}\))2
Using identity (a – b)2 = a2 – 2ab + b2, we get
(7y – \(\frac{3}{4z}\))2 = (7y)2 – 2 × 7y × \(\frac{3}{4 z}+\left(\frac{3}{4 z}\right)^2\)
= 49y2 – \(\frac{21 y}{2 z}+\frac{9}{16 z^2}\)
Using distributive property, we get
Intext (NCERT Page 148)
Question 1.
Take a pair of natural numbers. Calculate the sum of their squares. Can you write twice this sum as a sum of two squares?
Solution:
Let the two numbers be 6 and 7.
Then, 62 + 72 = 36 + 49 = 85
Yes, we can write twice of this sum as a sum of two squares.
We know that, 2(a2 + b2) = (a + b)2 + (a – b)2
2(72 + 62) = (7 + 6)2 + (7 – 6)2
= 132 + 12
Question 2.
Use Identity 1C to calculate 98 × 102 and 45 × 55.
Solution:
We have, 98 × 102 = (100 – 2) × (100 + 2)
= (100)2 – (2)2 [∵(a – b) (a + b) = a2 – b2]
= 10000 – 4
= 9996
and 45 × 55 = (50 – 5) (50 + 5)
= (50)2 – (5)2 [∵ (a – b) (a + b) = a2 – b2]
= 2500 – 25
= 2475
Question 3.
Show that (a + b) × (a – b) = a2 – b2 geometrically.
Solution:
Let ABCD be a square with length of side be a and take a smaller square in this square at one corner with length of side be b.
Area of square ABCD = a2
Area of square PQRB = b2
Remaining area (APQRCD) = a2 – b2
⇒Area of rectangle APSD + Area of rectangle QRCS
= a2 – b2
⇒ AP × AD + QR × RC = a2 – b2
⇒ (a – b) a + b × (a – b) = a2 – b2
⇒ (a -b) (a+ b) = a2 – b2
Figure it Out (NCERT Page 149)
Question 1.
Which is greater (a – b)2 or (b-a)2? Justify your answer.
Solution:
We have,(a – b)2 = a2 – 2ab + b2 [∵ (x – y)2 = x2 -2xy + y2]
Also, (b – a)2 = b2 – 2ba + a2
= a2 – 2 ab + b2 [∵ ab = ba]
Hence, (a – b)2 and (b – a)2 are equal.
Question 2.
Express 100 as the difference of two squares.
Solution:
Let the two numbers be a and b.
Difference of two squares = a2 – b2 = 100
⇒ (a – b) (a + b) = 100 [∵ (a – b) (a + b) = a2 – b2]
We can choose factors of 100 that can be expressed as (a – b) and (a + b).
Let a – b = 2 …(i)
and a + b = 50 …(ii)
On adding both Eqs. (i) and (ii), we get a – b + a + b = 2 + 50
⇒ 2a = 52
⇒ a = 26
Now, 26 + b = 50
⇒ b = 50 – 26 = 24
∴ 100 = 262 – 242
Question 3.
Find 4062, 722, 1452, 10972 and 1242 using the identities you have learnt so far.
Solution:
4062 = (400 + 6)2
= (400)2 + 2 × 400 × 6 + (6)2 [∵(a + b)2 = a2 + 2ab + b2]
= 160000 + 4800 + 36
= 164836
722 = (70 + 2)2
= (70)2 + 2 × 70 × 2 + (2)2
= 4900 + 280 + 4 = 5188
1452 = (150 – 5)2
= (150)2 – 2 × 150 × 5 + (5)2 [∵(a – b)2 = a2 – 2ab + b2]
= 22500 – 1500 + 25
= 22525 -1500
= 21025
10972 = (1100 – 3)2 = (1100)2 – 2 × 1100 × 3 + (3)2 [∵(a – b)2 = a2 – 2ab + b2]
= 1210000 – 6600 + 9
= 1210009 – 6600
= 1203409
1242 = (120 + 4)2 = (120)2 + 2 × 120 × 4 + (4)2 [∵(a + b)2 = a2 + 2ab + b2]
= 14400 + 960 + 16
= 15376
Question 4.
Do Patterns 1 and 2 hold only for counting numbers? Do they hold for negative integers as well? What about fractions? Justify your answer.
Solution:
We have, pattern 1 is
2 (a2 + b2) = (a + b)2 + (a – b)2 …(i)
and pattern 2 is
(a + b) x (a – b) = a2 – b2 …(ii)
For negative integers,
Let a = – 3, b = 2
On substituting a = – 3 and b = 2 in Eq. (i), we get
2((-3)2 + (2)2) = (-3 + 2)2 + (-3 – 2)2
⇒ 2(9 + 4) = (-1)2 + (- 5)2
⇒ 2 × 13 = 1 + 25
⇒ 26 = 26, which is true.
Again, on substituting a = -3 and b = 2 in Eq. (ii), we get
(-3 + 2) × (-3 – 2) = (-3)2 – (2)2
⇒ – 1 × (-5) = 9 – 4
⇒ 5 = 5, which is true.
So, it also holds for negative integers.
For fractions,
Let a = \(\frac{1}{2}\), b = \(\frac{1}{2}\)
On substituting a = \(\frac{1}{2}\) and b = \(\frac{1}{2}\) in Eq. (i), we get
Again, on substituting a = \(\frac{1}{2}\) and b = \(\frac{1}{3}\) in Eq.(ii), we get
which is true.
So, it is holds for fractions.
So, the patterns 1 and 2 holds for negative integers as well as fractions.
Mind The Mistake, Mend The Mistake (NCERT Page 150)
Question 1.
We have expanded and simplified some algebraic expression below to their simplest forms.
(i) Check each of the simplifications and see if there is a mistake.
(ii) If there is a mistake, try to explain what could have gone wrong.
(iii) Then write the correct expression.
Solution:
(a) The multiplication was done incorrectly and the terms were not combined properly.
-3p (-5p + 2q) = – 3p × (-5p) – 3p × 2q
= 15p2 – 6pq, which is the correct expression.
(b) The combination and multiplication of terms are incorrect.
2(x – 1) + 3(x + 4) = 2x – 2 + 3x + 12
= 5x + 10, which is the correct expression.
(c) In y + 2 (y + 2), the bracket is missing
(y + 2) (y + 2) = (y + 2)2
= y2 + 4y + 4 [∵ (a + b)2 = a2 + 2ab + b2]
(d) The squaring of 5m + 6M is incorrect.
(5m + 6n)2 = (5m)2 + 2 × 5m × 6n + (6n)2
= 25m2 + 60mn + 36n2
(e) There is no mistake.
(f) The multiplication is not performed correctly.
3a(2b × 3c) = 3a × 6bc = 18abc
(g) There is no mistake.
(h) These are not like terms.
So, they cannot be combined by addition.
5w2 + 6w is itself the correct expression.
(i) Only like terms can be combined.
The terms 6a2b and 6ab1 are not like terms and cannot the added together
So, 2a3 + 3a3 + 6a2b + 6ab2 = 5a3 + 6a2b + 6ab2 is the correct expression.
(j) There is no mistake.
(k) The expansion of (a+ 2) (b + 4) is incomplete.
(a + 2) (b + 4) = (a + 2)b + (a + 2)4
= ab + 2b + 4a + 8
(l) There is no mistake.
Intext (NCERT Rage 152-153)
Question 1.
Use this formula to find the number of circles in Step 15.
Solution:
We know that number of circles in step k = k2 + 2k
Number of circles in step 15 = 152 + 2 × 15
= 225 + 30
= 255
Question 2.
Consider the pattern made of square tiles in the picture below.
Solution:
(i) Figure 1
A 3 × 3 square with a 1 × 1 hole in the centre.
∴ Number of square tiles = 32 – 12
= 9 – 1
= 8
Figure 2
A 4 × 4 square with a 2 × 2 hole in the centre.
∴ Number of square tiles = 42 – 22
= 16 – 4
= 12
Figure 3
A 5 × 5 square with 3×3 hole in the centre.
∴ Number of square tiles = 52 – 32
= 25 – 9
= 16
(ii) Number of square tiles in step 4 = 62 – 42
= 36 – 16
= 20
Number of square tiles in step 10 = 122 – 102
= 144 – 100
= 44
(iii) Method 1
In step M, the outer square has side length (n + 2) and the inner hole has side length n.
∴ Number of tiles = (n + 2)2 – n2
= n2 + 4n + 4 – n2 [∵ (a + b)2 = a2 + 2ab + b2]
= 4n + 4,
which is the required algebraic expressions.
Question 3.
By expanding both expressions, check that (m + n)2 – 4mn = (n – m)2.
Solution:
LHS = (m + n)2 – 4mn
= m2 + 2mn + n2 – 4mn [∵ (a + b)2 = a2 + 2ab + b2]
= m2 – 2mn + n2
RHS = (n – m)2
= n2 – 2nm + m2 [∵ (a – b)2 = a2 – 2ab + b2]
LHS = RHS
So, (m + n)2 – 4mn = (n – m)2
Intext (NCERT Page 154)
Question 1.
By expanding the expressions
x2 – xy, x(x + 2y) – 3xy and x(x – y) verify that all three expressions are equivalent. If x = 8 and y = 3, find the area of the shaded region.
Solution:
Given, expressions are x2 – xy, x(x + 2y) – 3xy and x(x – y).
Now, x(x + 2 y) – 3 xy = x × x + x × 2 y – 3 × y
[by distributive property]
= x2 + 2xy – 3xy
= x2 – xy
Also, x(x – y) = x × x – x × y
[by distributive properly]
= x2 – xy
Hence, all three expression are equivalent.
Now, shaded area = x2 – xy
= (8)2 – 8 × 3 [∵ x = 8, y = 3, given]
= 64 – 24
= 40 sq units
Question 2.
Write an expression for the area of the dashed region in the figure below. Use more than one method to arrive at the answer. Substitute p = 6, r = 35 and s = 9 and calculate the area.
Solution:
Method I
We have, AB = s,
BC = p, RC = r and AP = r
∴ PD = AD – AP
= p – r [∵ AD = BC]
and DR = DC – RC
= s – r [∵ DC = AB]
Now, required area = PD × DR
= (p – r)(s – r)
Here, p = 6, r = 3.5 and s = 9
∴ Required area = (6 – 3.5) (9 – 3.5)
= 2.5 × 5.5
= 13.75 sq units
Method II
CT = BC – TB = p – r [∵ BT = AP]
Required area = area of rectangle ABCD – area of rectangle ABTP – area of rectangle TCRQ
= AB × BC – AB × AP – CT × RC
= sp – sr – (p – r)r
= s(p – r) – (p – r)r
Here, p = 6, r = 3.5 and s = 9
Required area = 9(6 – 3.5) – (6 – 35) 3.5
= 9 × 2.5 – 25 × 35
= 22.5 – 8.75
= 13.75 sq units
Figure it Out (NCERT Page 154 – 156)
Question 1.
Compute these products using the suggested identity.
(i) 462 using Identity 1A for (a + b)2
Solution:
We have, 462 = (40 + 6)2
= (40)2 + 2 × 40 × 6 + (6)2 [∵ (a + b)2 = a2 + 2ab + b2]
= 1600 + 480 + 36
= 2116
(ii) 397 × 403 using Identity 1C for (a + b) (a – b)
Solution:
We have, 397 × 403 = (400 – 3) (400 + 3)
= (400)2 – (3)2 [∵ (a – b)(a + b) = a2 – b2]
= 160000 – 9
= 159991
(iii) 912 using Identity IB for (a – b)2
Solution:
We have, 912 =(100 – 9)2
= (100)2 – 2 × 100 × 9 + (9)2 [∵ (a – b)2 =a2 – 2ab + b2]
= 10000 – 1800 + 81
= 10081 – 1800
= 8281
(iv) 43 × 45 using Identity 1C for (a + b) (a – b)
Solution:
43 × 45 = (44 – 1) (44 + 1)
= (44)2 – (1)2 [∵ (a -b)(a + b) = a2 – b2]
= (40 + 4)2 – 1
= (40)2 + 2 × 40 × 4 + (4)2 – 1
= 1600 + 320 + 16 – 1
= 1936 – 1
= 1935
Question 2.
Use either a suitable identity or the distributive property to find each of the following products.
(i) (p – 1) (p + 11)
Solution:
We have, (p – 1) (p + 11) = p(p + 11) – 1 (p + 11)
[by distributive property] = p2 + 11p – p – 11
= p2 + 10p – 11
(ii) (3a – 9b) (3a + 9b)
Solution:
We have, (3a – 9b) (3a + 9b)
= (3a)2 – (9b)2
= 9a2 – 81b2 [∵ (a + b) (a – b) = a2 – b2]
(iii) – (2y + 5) (3y + 4)
Solution:
We have, – (2y + 5) (3y + 4) = (- 2y – 5) (3y + 4)
= -2y (3y + 4) – 5 (3y + 4) [by distributive property]
= -6y2 – 8y – 15y – 20
= – 6y2 – 23y – 20
(iv) (6x + 5y)2
Solution:
We have, (6x + 5y)2
= (6x)2 + 2 × 6x × 5y + (5y)2
[∵ (a + b)2 = a2 + 2ab + b2]
= 36x2 + 60xy + 25y2
(v) (2x – \(\frac{1}{2}\))2
Solution:
We have, (2x – \(\frac{1}{2}\))2
= (2x)2 – 2 × 2x × \(\frac{1}{2}+\left(\frac{1}{2}\right)^2\)
[∵ (a + b)2 = a2 + 2ab + b2]
= 4x2 – 2x + (\(\frac{1}{2}\))2
(vi) (7p) × (3r) × (p + 2)
Solution:
We have, (7p) × (3r) × (p + 2) = 21pr × (p + 2)
= 21pr × p + 21pr × 2
= 21p2r + 42pr
Question 3.
For each statement identify the appropriate algebraic expressions.
(i) Two more than a square number.
Solution:
Let s be the number.
∴ Square of s = s2
Two more than s2 = s2 + 1
Required expression is s2 + 2.
(ii) The sum of the squares of two consecutive numbers
Solution:
Let the first numbertJefw.
The second number = m + 1
Now, square of first number = m2
and square of second number = (m+ 1)2
∴ Sum of their squares = m2 + (m + 1)2, which is the required expression.
Question 4.
Consider any 2 by 2 square of numbers in a calendar, as shown in the figure.
Find products of numbers lying along each diagonal 4 × 12 = 48, 5 × 11 = 55. Do this for the other 2 by 2 squares. What do you observe about the diagonal products? Explain why this happens.
Hint: Label the numbers in each 2 by 2 square is
Solution:
Let the top left number in a 2 × 2 square be a.
Then, the numbers in the square can be represented as
| a | (a + 1) |
| a + 7 | (a + 8) |
Product of diagonal (top left to bottom right)
= a × (a + 8)
= a2 + 8a
Product of diagonal (top right to bottom left)
= (a + 1) × (a + 7)
= a2 + 7a + a + 7
= a2 + 8a + 7
∴ The product of diagonal (top right to bottom left) is always greater than the product of diagonal (top left to bottom right) by 7.
This pattern occurs because of the fixed structure of a calender.
Question 5.
Verify which of the following statements are true.
(i) (k + 1) (k + 2) – (k + 3) is always 2.
Solution:
We have, (k + 1) (k + 2) – (k + 3)
= k(k + 2) + 1 (k + 2) – (k + 3)
= k2 + 2k + k + 2 – k – 3
= k2 + 3k + 2 – k – 3
= k2 + 2k – 1
So, this expression is not always equal to 2.
For example,
if k = 0 then
k2 + 2k – 1 = 02 + 2 × 0 – 1 = -1,
if k = 1 then k2 + 2k – 1
= 12 + 2 × 1 – 1
= 2,
if k = 2 then
k2 + 2k – 1 = 22 + 2 × 2 – 1
= 4 + 4 – 1
= 7
So, given statement is false.
(ii) (2q +1) (2q – 3) is a multiple of 4.
Solution:
We have, (2q + 1) (2q – 3)
= 2q(2q – 3) + 1 (2q – 3)
= 4q2 – 6q + 2q – 3
= 4q2 – 4q – 3
= 4q2 – 4q – 4 + 4 – 3
= 4(q2 – q – 1) + 1
Since, 4(q2 – q – 1) is a multiple of 4.
The expression 4q2 – 4q – 3 will always leave a remainder 1, when divided by 4.
So, the given statement is false.
(iii) Squares of even numbers are multiples of 4 and squares of odd numbers are 1 more than multiples of 8.
Solution:
Let an even number be represented as 2n.
Square of 2n = (2n)2 = 4n2
Since, n2 is an integer.
So, 4n2 is always a multiple of 4.
Let an odd number be represented as 2n + 1
∴ Square of (2n + 1) = (2n + 1)2
= (2n)2 + 2 × 2n × 1 + (1)2
= 4n2 + 4n + 1
= 4n (n + 1) + 1
Since, n(n + 1) is always an even number.
∴ Either n or n + 1 is even.
Let n(n + 1) = 2p for some integer p.
Then, 4 n(n + 1) = 4(2p) + 1 = 8p + 1
This show that the square of an odd number is 1 more than a multiple of 8.
So, the given statement is true.
(iv) (6n + 2)2 -(4n + 3)2 is 5 less than a square number.
Solution:
We have, (6n + 2)2 – (4n + 3)2
= (6n)2 + 2 × 6n × 2 + (2)2
– [(4n)2 + 2 × 4n × 3 + (3)2]
= 36n2 + 24n + 4 – [16n2 + 24n + 9]
= 36n2 + 24n + 4 – 16n2 – 24n – 9
= 20n2 – 5
= 5(4n2 – 1)
This expression is not generally 5 less than a square number.
For example, if n = 1 then
5 (4(1)2 – 1) = 5 (4 – 1) = 5 × 3 = 15, which is not 5 less than a square number.
So, the given statement is false.
Question 6.
A number leaves-a remainder of 3 when divided by 7, and another number leaves a remainder of 5 when divided by 7. What is the remainder when their sum, difference and product are divided by 7?
Solution:
Let the first number be A.
When divided by 7, remainder is 3
∴ A = 7p + 3, where p is any whole number.
Let the second number be B.
When divided by 7, remainder is 5
∴ B = 7q + 5, where q is any whole number.
Now, sum = A + B
= 7 p + 3 + 7 q + 5
= 7p + 2 + 1 + 7q + 5
= (7p + 7q + 7) + 1
= 7(p + q + 1) + 1
When sum is divided by 7 then remainder is 1.
Difference = A – B
= 7p + 3 – (7q + 5)
= 7p + 3 – 7q – 5
= 7p – 7q – 2
= 7p – 7q – 7 + 7- 2
= 7(p – q – 1) + 5
When difference is divided by 7 then remainder is 5. Product =AxB
= (7p + 3) (7q + 5) = 49pq + 35p + 21q + 15
= 49 pq + 35 p + 21q + 14 + 1
= 7(7 pq + 5p + 3q + 2) + 1
When product is divided by 7 then remainder is 1.
Question 7.
Choose three consecutive numbers, square the middle one and subtract the product of the other two. Repeat the same with other sets of numbers. What pattern do you notice? How do we write this as an algebraic equation? Expand both sides of the equation to check that it is a true identity.
Solution:
Let us take three consecutive numbers.
Example 1: 3, 4, 5
Square the middle one = 42 =16
Product of other two = 3 × 5 = 15
Difference = 16 – 15 = 1
Example 2: 7, 8, 9
Square of middle one = 82 =64
Product of other two = 7 × 9 = 63
Difference = 64 – 63 = 1
Whenever we choose three consecutive numbers, square the middle and subtract the product of the other two the result is always 1.
Let the three consecutive numbers be x – 1, x, x + 1
Square of middle one = x2
Product of other two = (x – 1) (x + 1) = x2 – 1
Difference = x2 – (x2 – 1)
= x2 – x2 + 1
= 1 ,
Question 8.
What is the algebraic expression describing the following steps-add any two numbers. Multiply this by half of the sum of the two numbers? Prove that this result will be half of the square of the sum of the two numbers.
Solution:
Let a and b be the two numbers.
Sum of a and b = a + b, \(\frac{1}{2}\) of sum of a and b = \(\frac{1}{2}\) (a + b)
Required expression is (a + b) x \(\frac{1}{2}\) (a + b) = \(\frac{1}{2}\) (a + b)2
The result is \(\frac{1}{2}\)(a+ b)2, which is half of the square of the 2 sum of the two numbers.
Question 9.
Which is larger? Find out without fully computing the product.
(i) 14 × 26 or 16 × 24
Solution:
Let A= 14 × 26 and B = 16 × 24
Now, B = 16 × 24 = (14 + 2) × (26 – 2)
= 14 × 26 – 14 × 2 + 2 × 26 – 2 × 2
= 14 × 26 – 28 + 52 – 4
= 14 × 26 + 20
∴ B = A + 20
Hence, B is greater than A by 20 i.e. 16 × 24 is greater than 14 × 26 by 20.
(ii) 25 × 75 or 26 × 74
Solution:
Let A = 25 × 75 and B = 26 × 74
Now,B = 26 × 74 = (25 + 1) (75 – 1)
= 25 × 75 – 25 × 1 + 1 × 75 – 1 × 1
= 25 × 75 – 25 + 75 – 1
= 25 × 75 + 49
B = A + 49
Hence, B is greater than A by 49 i.e. 26 × 74 is greater than 25 × 75 by 49.
Question 10.
A tiny park is coming up in Dhauli. The plan is shown in the figure. The two square plots, each of area g2 sq. ft., will have a green cover. All the remaining area is a walking path w ft. wide that needs to be tiled. Write an expression for the area that needs to be tiled.
Solution:
Given, area of each square plot = g2 sq. ft
∴ Side of each square pot = g ft
Now, length of the park = w + g + w + w + g + w = 2g + 4w
and breadth of the park = w + g + w = g + 2w
Total area of the park = (2g + 4w) (g + 2w)
= 2 g2 + 4 gw + 4 gw + 8w2
= 2g2 + 8w2 + 8 gw
Area of walking path = area of park – 2 × area of square plot
= 2g2 + 8w2 + 8gw – 2g2
= (8w2 + 8gw) sq ft
Question 11.
For each pattern shown below,
(i) Draw the next figure in the sequence.
(ii) How many basic units are there in Step 10?
(iii) Write an expression to describe the number of basic units in Stepy.
Solution:
(i) For 1st pattern :
The next figure in the sequence is given below.
(ii) Looking at steps 1 to 3.
For horizontal blocks (or units)
3 × 1 (step 1)
4 × 2 (step 2)
5 × 3 (step 3)
(n + 2) × n (step n)
For vertical blocks (or units) Since, vertical units are equally repeating two times at both sides of horizontal units.
So, we will draw a pattern for either side then we will multiply by 2 for counting total vertical units.
For either side
So, total number ci rtical units at both sides
= 2 x [3 + \(\frac{n(n-1)}{2}\)]
= 6 + n(n – 1)
Therefore, total number of basic units in step 10
= number of horizontal units in step 10 + number of vertical units in step 10
= (10 + 2) × 10 + [6 + 10 × (10 – 1)]
= 120 + [6 + 90]
= 120 + 96
= 216
(iii) Total number of basic units in stepy
= the number of horizontal units in step y + the number of vertical units in step
y – (y + 2) × y + [6 + y(y – 1)]
= (y + 2)y + y(y – 1) + 6
= y[y + 2 + y – 1] + 6
= y(2y + 1) + 6
For IInd pattern:
(ii) Looking at step 1 to step 3.
(2)2 + 1 (step 1)
(3)2 + 2 (step 2)
(4)2 + 3 (step 3)
(n + 1)2 + n (step n)
Therefore total number of basic units in step 10
=(10 + 1)2 + 10
= (11)2 + 10
= 121 + 10
= 131
(iii) Total number of basic units in step y = (y + 1)2 + y
= y2 +1 + 2 y + y
= y2 + 3y + 1
= y(y + 3) + 1