Category: Class 8

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.3 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 12th Lesson Factorisation Exercise 12.3

Question 1.
Carry out the following divisions
(i) 48a³ by 6a
(ii) 14x³ by 42x³
(iii) 72a³b⁴c⁵ by 8ab²c³
(iv) 11xy²z³ by 55xyz
(v) -54l⁴m³n² by 9l²m²n²
Solution:
(i) 48a³ by 6a
48a³ ÷ 6a
= \(\frac{6 \times 8 \times a \times a^{2}}{6 \times a}\)
= 8a²

(ii) 14x³ by 42x³
= 14x³ ÷ 42x³

(iii) 72a³b⁴c⁵ by 8ab²c³

(iv) 11xy²z³ by 55xyz
11xy²z³ ÷ 55xyz

(v) -54l⁴m³n² by 9l²m²n²
-54l⁴m³n² ÷ 9l²m²n²

= -6l²m

Question 2.
Divide the given polynomial by the given monomial
(i) (3x² – 2x) ÷ x
(ii) (5a³b – 7ab³) ÷ ab
(iii) (25x⁵ – 15x⁴) ÷ 5x³
(iv) (4l⁵ – 6l⁴ + 8l³) ÷ 2l²
(v) 15 (a³b²c² – a²b³c² + a²b²c³ ) ÷ 3abc
(vi) 3p³– 9p²q – 6pq²) ÷ (-3p)
(vii) (\(\frac{2}{3}\) a² b² c²+ \(\frac{4}{3}\) a b² c³) ÷ \(\frac{1}{2}\)abc
Solution:
(i) (3x² – 2x) ÷ x

(ii) (5a³b – 7ab³) ÷ ab

(iii) (25x⁵ – 15x⁴) ÷ 5x³

= 5x² – 3x (or) x(5x – 3)

(iv) (4l⁵ – 6l⁴ + 8l³) ÷ 2l²

= 2l² – 3l² + 4l = l(2l² – 3l + 4)

(v) 15 (a³ b² c² – a² b³ c² + a² b² c³ ) ÷ 3abc

= 5[a x abc – b x abc + c x abc ]
= 5abc [a – b + c]

(vi) 3p³– 9p²q – 6pq²) ÷ (-3p)

= -[p² – 3pq – 2q²]
= 2² + 3pq – p²

(vii) (\(\frac{2}{3}\) a²b²c²+ \(\frac{4}{3}\) ab²c³) ÷ \(\frac{1}{2}\)abc

Question 3.
Workout the following divisions:
(i) (49x -63) ÷ 7
(ii) 12x (8x – 20,) ÷ 4(2x – 5)
(iii) 11a³ b³ (7c – 35) ÷ 3a² b² (c – 5)
(iv) 54lmn (l + m) (m + n) (n + l) ÷ 8 lmn (l + m) (n +l)
(v) 36(x + 4)(x² + 7x + 10) ÷ 9(x + 4)
(vi) a(a+1)(a+2)(a + 3) ÷ a(a + 3)
Solution:
(i) (49x -63) ÷ 7

(ii) 12x (8x – 20,) ÷ 4(2x – 5)

(iii) 11a³ b³ (7c – 35) ÷ 3a² b² (c – 5)

(iv) 54lmn (l + m) (m + n) (n + l) ÷ 8 lmn (l + m) (n +l)

(v) 36(x + 4)(x² + 7x + 10) ÷ 9(x + 4)

4 ( x² + 7x + 10)
= 4 ( x² + 5x + 2x + 10)
= 4 [x( x + 5) +2(x + 5)]
= 4( x + 5) (x + 2)

(vi) a(a+1)(a+2)(a + 3) ÷ a(a + 3)

= ( a + 1)(a + 2)

Question 4.
Factorize the expressions and divide them as directed:
(i) (x² + 7x + 12) ÷ (x + 3)
(ii) (x² – 8x + 12) ÷ (x – 6)
(iii) (p² + 5p + 4,) (p + l)
(iv) 15ab(a² – 7a + 10) ÷ 3b(a – 2)
(v) 151m (2p² – 2q²) ÷ 3l(p + q)
(vi) 26z³(32z² – 18,) ÷ 13z² (4z – 3)
Solution:
(i) (x² + 7x + 12) ÷ (x + 3)
(x² + 7x + 12) ÷ (x + 3)
x² + 7x + 12 = x² + 3x + 4x + 12
= x(x + 3) + 4(x + 3)
= (x + 3) (x + 4)

(ii) (x² – 8x + 12) ÷ (x – 6)
(x² – 8x + 12) ÷ (x – 6)
x² – 8x + 12 = x² – 6x – 2x + 12
= x(x – 6) – 2(x – 6)
= (x – 6) (x – 2)
∴ (x² – 8x + 12) 4 (x – 6)
= \(\frac{(x-6)(x-2)}{(x-6)}\) = x – 2

(iii) (p² + 5p + 4,) (p + 1)
p² + 5p + 4 = p² + p + 4p + 4
= p(p + 1) + 4(p + 1)
= (p + 1) (p + 4)
(p² + 5p + 4) ÷ (p + 1)
= \(\frac{(p+1)(p+4)}{(p+1)}\) = p + 4

(iv) 15ab(a² – 7a + 10) ÷ 3b(a – 2)
15ab (a² – 7a + 10) ÷ 3b (a – 2)
15ab (a² – 7a + 10) = 15ab (a² – 5a – 2a + 10)
= 15ab [(a² – 2a) – (5a -10)]
= 15ab [a(a – 2) – 5(a – 2)]
= 15ab(a – 2)(a – 5)
∴ 15ab (a² – 7a + 10) ÷ 3b (a – 2)

(v) 151m (2p² – 2q²) ÷ 3l(p + q)
15lm (2p² – 2q²) ÷ 3l (p + q)
15lm (2p² – 2q²) = 15lm x 2(p² – q²)
= 30lm (p + q) (p – q)
∴ 15lm(2p² – 2q²) ÷ 3l(p + q)

(vi) 26z³(32z² – 18,) ÷ 13z² (4z – 3)
26z³(32z² – 18) ÷ 13z² (4z – 3)
26z³(32z² – 18) = 26z³ (2 x 16z² – 2 x 9)
= 26z³ x 2 [16z³ – 9]
= 52z³ [(4z)³ – (3)³]
= 52z³ (4z + 3) (4z – 3)
∴ 26z³ (32z² – 18) ÷ 13z² (4z – 3)

AP State Syllabus 8th Class Maths Solutions 5th Lesson Comparing Quantities Using Proportion InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 5 Comparing Quantities Using Proportion InText Questions and Answers.

8th Class Maths 5th Lesson Comparing Quantities Using Proportion InText Questions and Answers

Do this

Question 1.
How much compound interest is earned by investing Rs. 20000 for 6 years at 5% per annum compounded annually? (Page No. 114)
Answer:
P = Rs. 20,000; R = 5%; n = 6 years

∴ Compound Interest = Amount – Principal = 26802 – 20,000
∴ C.I. = Rs. 6802 /-

Question 2.
Find compound interest on Rs. 12600 for 2 years at 10% per annum compounded annually.    (Page No. 114)
Answer:
P = Rs. 12,600; R = 10%; n = 2 years

∴ Compound Interest = Amount – Principal = 15,246 – 12,600
∴ C.I. = Rs. 2646 /-

Question 3.
Find the number of conversion times the interest is compounded and rate for each.
i) A sum taken for 1\(\frac{1}{2}\) years at 8% per annum is compounded half yearly.
ii) A sum taken for 2 years at 4% per annum is compounded half yearly.     (Page No. 115)
Answer:
Compound interest will be calculated for every 6 months.
There will be 3 periods in 1\(\frac{1}{2}\) year.
∴ n = 3
∴ Rate of interest for half yearly = \(\frac{1}{2}\) × 8% = 4%
∴ R = 4%; n = 3
ii) C.I. should be calculated for every 6 months.
There will be 4 time periods in 2 years.
∴ n = 4
∴ Rate of interest for half yearly = \(\frac{1}{2}\) × 4% = 2%
∴ n = 4 ; R = 2%

Try These

Question 1.
Find the ratio of gear of your bicycle.       (Page No. 96)

Count the number of teeth on the chain wheel and the number of teeth for the sprocket wheel.
{number of teeth on the chain wheel} : {number of teeth of sprocket wheel}
This is called gear ratio. Write how many times sprocket wheel turns for every time the chain wheel rotates once.
Answer:
The ratio between the rotations of chain wheel and sprocket wheel is 4 : 1.

Question 2.
Collect newspaper cuttings related to percentages of any five different situations.  (Page No. 96)
Answer:
1) Bharti to sell 5% stake for $ 1.2b:
New Delhi, May 3: The country’s largest telecom operator Bharti Airtel said on Friday that it will sell 5 per cent stake to Doha – based Qatar Foundation Endowment (QFE) for $1.26 billion (Rs. 6,796 crores) to fund its future growth plans.
The deal will bring cash for the company at a time when its balance sheet is stretched and there is threat of Bharti Airtel having to pay hefty fees to regulatory authorities as government is re-looking at past policies.

2) Indian Firms Mop – Up Down By 36% In FY13:
New Delhi: Indian companies raised nearly Rs. 31,000 crore from the public issuance of equity and debt in 2012 – 13, a slump of 36 per cent from the preceding year.
According to latest data available with market regulator Sebi (Securities and Exchange Board of India), a total of Rs. 30,859 crore worth of fresh capital were mopped – up from equity and debt market during 2012 – 13, which was way below than Rs. 48,468 crore garnered in 2011 -12. Going by the statistics, it was mostly debt market that was leveraged to meet the funding requirements of businesses in the past fiscal as compared to capital raised through sale of shares through instruments like initial public offering (IPO) and rights issue. A total of Rs. 15,386 crore were raised from the debt market via 11 issues in 2012 – 13, much lower than Rs. 35,611 crore garnered through 20 issues in the preceding fiscal.

3) IT – ITeS sector employs 2.97m people in FY13:
New Delhi: The total number of professionals working in India’s $100 billion IT – information technology enabled services (IT – ITeS) sector grew by 7 per cent to 2.97 million in the last fiscal, Parliament was informed on Friday.
The IT – ITeS sector, which contributes about 8 per cent to the country’s economy, provided employment to 2.77 million professionals in 2011 -12 fiscal, minister of state for communications and IT Milind Deora said. “The Indian IT – ITeS industry has been progressively growing and is able to secure new projects from various foreign coun¬tries,” Mr Deora said. During the 2012 -13 fiscal, 6,40,000 professionals were employed in the domestic market.

4) For RBI, it’s not all is well yet:
Slashes repo rate by 0.25%; rules out any more cuts; raises red flag on CAD DC Correspondent Mumbai, May 3:
The RBI cut the repo rate (rate at which it lends to banks) by a quarter per cent on Friday to 7.25 per cent from 7.75 per cent, but this will not be passed on to the consumers by way of lower personal loans for housing etc., immediately according to bankers.
It also raised the growth rate from 5.2 per cent projected in January to 5.7 per cent for 2013 -14 and lowered the inflation rate to 5.5 per cent for the year.
RBI governor Dr D. Subbarao said based on the current and prospective assessment of various economic factors and the dismal 4.5 per cent lowest growth rate in the last quarter, it was decided to cut the policy rate by 25 basis points.

5) Markets sink on RBI’s Bearish outlook on rate:
DC Correspondent Mumbai, May 3:
In a highly volatile trading session, the markets retreated from their three month high led by interest rate sensitive banking, auto and real estate sector stocks after the Reserve Bank of India (RBI) cautioned that the room for further monetary policy easing is limited.
The Sensex closed 19,575.64, sliding 160.13 points or 0.81 per cent while the Nifty dropped 55.35 points or 0.92 per cent to end the week at 5,944.

Question 3.
Find the compound ratios of the following. (Page No. 99)
a) 3 : 4 and 2 : 3
b) 4 : 5 and 4 : 5
c) 5 : 7 and 2 : 9
Answer:
Compound ratio of a : b and c : d is ac : bd.

Question 4.
Give examples for compound ratio from daily life.     (Page No. 99)
Answer:
Examples for compound ratio from daily life:
i) To compare the ratio of tickets of 8th class students (Boys & Girls) is 3:4 and the ratio of tickets of 7th class students is 4 : 5.
ii) The comparision between two situations is 4 men can do a piece of work in 12 days, the same work 6 men can do in 8 days.
iii) Time – distance – speed.
iv) Men – days – their capacities etc.

Question 5.
Fill the selling price for each.     (Page No. 104)

Answer:

Question 6.
i) Estimate 20% of Rs. 357.30 ii) Estimate 15% of Rs. 375.50      (Page No. 105)
Answer:

ii) 15% of 375.50 = \(\frac{15}{100}\) × 375.50 = 15 × 3.7550 = Rs. 56.325

Question 7.
Complete the table.     (Page No. 105)

Answer:

Think, discuss and write

Question 1.
Two times a number is 100% increase in the number. If we take half the number what would be the decrease in percent?    (Page No. 101)
Answer:
Increase percent of 2 times of a number = \(\frac{(2-1)}{1}\) × 100 = 1 × 100 = 100%
Half of the number = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
Decrease in percent = \(\frac{\frac{1}{2}}{1}\) × 100 = \(\frac{1}{2}\) × 100 = 50%

Question 2.
By what percent is Rs. 2000 less than Rs. 2400? Is it the same as the percent by which Rs. 2400 is more than Rs. 2000? (Page No. 101)
Answer:
Decrease in percent of Rs. 2000 less than Rs. 2400

Increase in percent of Rs. 2400 more than Rs. 2000

Question 3.
Preethi went to a shop to buy a dress. Its marked price is Rs. 2500. Shop owner gave 5% discount on it. On further insistence, he gave 3% more discount. What will be the final discount she obtained? Will it be equal to a single discount of 8%? Think, discuss with your friends and write it in your notebook. (Page No. 105)
Answer:
Marked price of a dress selected by Preethi = Rs. 2500
After allowing 5% of discount then S.P = M.P. – Discount%
= 2500 – \(\frac{5}{100}\) × 2500 = 2500 – 125 = Rs. 2375
Again 3% discount is allowed on Rs. 2375 then
S.P = 2375 – 3% of 2375
= 2375 – \(\frac{3}{100}\) × 2375 = 2375 – 71.25 = Rs. 2303.75
If 8% discount is allowed then S.P =

The S.P’s of both cases are not equal.
Discount on 5% + Discount on 3% = 125 + 71.25 = Rs. 196.25
Discount on 8% = Rs. 200
∴ Discounts are not equal which are obtained by Preethi.

Question 4.
What happens if cost price = selling price. Do we get any such situations in our daily life?
It is easy to find profit % or loss% in the above situations. But it will be more meaningful if we express them in percentages. Profit % is an example of increase percent of cost price and loss % is an example of decrease percent of cost price. (Page No. 106)
Answer:
If selling price is equal to cost price then either profit or loss will not be occurred.
In our daily life S.P. will not be equal to C.P. Then profit or loss will be occurred.
∴ Profit % = \(\frac{\text { Profit }}{\text { C.P. }}\) × 100;
Loss % = \(\frac{\text { Loss }}{\text { C.P. }}\) × 100.

Question 5.
A shop keeper sold two TV sets at Rs. 9,900 each. He sold one at a profit of 10% and the other at a loss of 10%. Oh the whole whether he gets profit or loss? If so what is its percentage? (Page No. 108)
Answer:
S.P of each T.V = Rs. 9,900
S.P of both T.Vs = 2 × 9,900 = Rs. 19,800
10% profit is allowed on first then C.P. =

10% loss is allowed on second then C.P.

C.P. of both T.V.’s = 9000 + 11000 = Rs. 20,000
Here C.P > S.P then loss will be occurred.
∴ Loss = C.P – S.P = 20000 – 19,800 = 200

Question 6.
What will happen if interest is compounded quarterly? How many conversion periods will be there? What about the quarter year rate – how much will it be of the annual rate? Discuss with your friends. (Page No. 115)
Answer:
Here C.I will be calculated for every 3 months. So, 4 time periods will be occurred in 1 year.
Rate of Interest (R) = \(\frac{R}{4}\) [∵ \(\frac{12}{3}\) = 4]

A = P\(\left[1+\frac{R}{400}\right]^{4}\)

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 12th Lesson Factorisation Exercise 12.2

Question 1.
Factorise the following expression
i) a² + 10a +25
ii) l² – 16l + 64
iii) 36x² + 96xy + 64y²
iv) 25x² + 9y² – 30xy
v) 25m²– 40mn + 1 6n²
vi) 81x² – 198 xy + 12ly²
vii) (x+y)² – 4xy
(Hint : first expand ( x + y)² )
viii) l⁴ + 4l²m² + 4m⁴
Solution:
i) a² + 10a +25
= (a)² + 2 × a × 5 + (5)²
It is in the form of a² + 2ab + b²
a² + 2ab + b²= (a + b)²
∴ a² + 10a + 25 = (a + 5)² = (a + 5) (a + 5)

ii) l² – 16l + 64
l² – 16l + 64
= (l)² – 2 × l × 8 + (8)²
It is in the form of a² – 2ab + b²
a² – 2ab + b² = (a – b)²
∴ l² – 16l + 64 = (l – 8)² = (l – 8) (l – 8)

iii) 36x² + 96xy + 64y²
36x² + 96xy + 64y²
= (6x)² + 2 × 6x × 8y + (8y)²
It is in the form of a² + 2ab + b²
a² + 2ab + b² = (a + b)²
∴ 36x² + 96xy + 64y²
= (6x + 8y)² = (6x + 8y) (6x + 8y)

iv) 25x² + 9y² – 30xy
25x² + 9y² – 30xy
= (5x)² + (3y)² – 2 × 5x × 3y
It is in the form of a² + b² – 2ab
a² + b² – 2ab = (a – b)²
∴ 25x² + 9y² – 30xy
= (5x – 3y)² = (5x – 3y) (5x – 3y)

v) 25m²– 40mn + 1 6n²
25m² – 40mn + 16n²
= (5m)² – 2 × 5m × 4n + (4n)²
It is in the form of a² – 2ab + b²
a² – 2ab + b² = (a – b)²
∴ 25m² – 40mn + 16n²
= (5m – 4n)²
= (5m – 4n) (5m – 4n)

vi) 81x² – 198 xy + 12ly²
81x² – 198xy + 121y²
= (9x)² – 2 × 9x × 11y + (11y)²
It is in the form of a² – 2ab + b²
a² – 2ab + b² = (a – b)²
∴ 81x² – 198xy + 121y²
= (9x – 11y)² – (9x – 11y) (9x – 11y)

vii) (x+y)² – 4xy
(Hint : first expand ( x + y)² )
= (x + y)² – 4xy
= x² + y² + 2xy – 4xy
= x² + y² – 2xy = (x – y)² = (x – y)(x – y)

viii) l⁴ + 4l²m² + 4m⁴
l⁴ + 4l²m² + 4m⁴
= (l²)² + 2 × l² × 2m² + (2m²)²
It is in the form of a² + 2ab + b²
a² + 2ab + b² = (a – b)²
∴ l⁴ + 4l²m² + 4m⁴
= (l² + 2m²)² = (l² + 2m²) (l² + 2m²)

Question 2.
Factorise the following
i) x² – 36
ii) 49x² – 25y²
iii) m² – 121
iv) 81 – 64x²
v) x²y² – 64
vi) 6x² – 54
vii) x² – 81
viii) 2x -32 x⁵
ix) 81x⁴ – 121x²
x) (p² – 2pq + q²)-r²
xi) (x+y)² – (x-y)²
Solution:
i) x² – 36
x² – 36
⇒ (x)² – (6)² is in the form of a² – b²
a² – b² = (a + b) (a – b)
∴ x² – 36 = (x + 6) (x – 6)

ii) 49x² – 25y²
= (7x)² – (5y)²
= (7x + 5y) (7x – 5y)

iii) m² – 121
m² -121
= (m)² – (11)²
= (m + 11) (m – 11)

iv) 81 – 64x²
81 – 64x²
= (9)² – (8x)²
= (9 + 8x) (9 – 8x)

v) x²y² – 64
= (xy)² – (8)²
= (xy + 8)(xy – 8)

vi) 6x² – 54
6x² – 54
= 6x² – 6 x 9 ‘
= 6(x² – 9)
= 6[(x)² – (3)²]
= 6(x + 3) (x – 3)

vii) x² – 81
x² – 81
= x² – 9²
= (x + 9 )(x – 9)

viii) 2x – 32 x⁵
2x – 32 x⁵
= 2x – 2x x 16x⁴
= 2 x (1 – 16x⁴)
= 2x [1²) – (4x²)²]
= 2x (1 + 4x²) (1 – 4x²)
= 2x (1 + 4x²) [(1⁵ – (2x)²]
= 2x (1 + 4x²) (1 + 2x) (1 – 2x)

ix) 81x⁴ – 121x²
81x⁴ – 121x²
– x² (81² – 121)
= x²[(9x)² – (11)²]
= x² (9x + 11) (9x -11)

x) (p² – 2pq + q²)-r²
(p² – 2pq + q²) – r²
= (p – q)² – (r)² [∵ p² – 2pq + q² = (p – q)²]
= (p – q + r) (p – q – r)

xi) (x + y)² – (x – y)²
(x + y)² – (x – y)²
It is in the form of a² – b²
a = x + y, b = x- y
∴ a² – b² =(a + b)(a-b)
= (x + y + x – y) [x + y- (x – y)]
= 2x [x + y-x + y]
= 2x x 2y = 4xy

Question 3.
Factorise the expressions
(i) lx² + mx
(ii) 7y² + 35Z²
(iii) 3x⁴ + 6x³y + 9x²Z
(iv) x² – ax – bx + ab
(v) 3ax – 6ay – 8by + 4bx
(vi) mn + m + n + 1
(vii) 6ab – b² + 12ac – 2bc
(viii) p²q – pr² – pq + r²
(ix) x (y + z) -5 (y + z)

(i) lx² + mx
lx² + mx
= l × x × x + m × x = x(lx + m)

(ii) 7y² + 35z²
7y²+ 35z²
= 7 × y² + 7 × 5 × z²
= 7(y² + 5z²)

(iii) 3x⁴ + 6x³y + 9x²Z
3x⁴ + 6x³y + 9x²Z
= 3 × x² × x² + 3 × 2 × x × x² × y + 3 × 3 × x² × z
= 3x² (x² + 2xy + 3z)

(iv) x² – ax – bx + ab
x² – ax – bx + ab
= (x² – ax) – (bx – ab)
= x(x – a) – b(x – a)
= (x – a) (x – b)

(v) 3ax – 6ay – 8by + 4bx
3ax – 6ay – 8by + 4bx
= (3ax – 6ay) + (4bx – 8by)
= 3a (x – 2y) + 4b (x – 2y)
= (x – 2y) (3a + 4b)

(vi) mn + m + n + 1
mn + m + n + 1
= (mn + m) + (n + 1)
= m (n + 1) + (n + 1)
= (n + 1) (m + 1)

(vii) 6ab – b² + 12ac – 2bc
6ab – b² + 12ac – 2bc
= (6ab – b²) + (12ac – 2bc)
= (6 × a× b – b × b) + (6 × 2 × a × c – 2 × b × c)
= b [6a – b] + 2c [6a – b]
= (6a – b) (b + 2c)

(viii) p²q – pr² – pq + r²
p²q – pr² – pq + r²
= (p²q – pr²) – (pq – r²)
= (p × p × q – p × r × r) – (pq – r²)
= P(pq – r²) – (pq – r²) × 1
= (pq – r²)(p – 1)

(ix) x (y + z) -5 (y + z)
= x(y + z) – 5(y + z)
= (y + z) (x – 5)

Question 4.
Factorise the following
(i) x⁴ – y⁴
(ii) a⁴ – (b + c)⁴
(iii) l² – (m – n)²
(iv) 49x² – \(\frac{16}{25}\)
(v) x⁴ – 2x²y² + y⁴
(vi) 4 (a + b)² – 9 (a – b)²
Solution:
= (x²)² – (y²)² is in the form of a² – b²
a² – b² = (a + b) (a – b)
x⁴ – y⁴ = (x² + y²)(x² – y²)
= (x² + y²)(x + y)(x – y)

(ii) a⁴ – (b + c)⁴
a⁴ – (b + c)⁴
= (a²)² – [(b + c)²]²
= [a² + (b + c)²] [a² – (b + c)²] ,
= [a² + (b + c)²] (a + b + c) [a – (b + c)]
= [a² + (b + c)²] (a + b + c) (a – b – c)

(iii) l² – (m – n)²
l² – (m – n)²
= (l)² – (m – n)²
= [l + m – n] [l – (m – n)]
= [l + m -n] [l – m + n]

(iv) 49x² – \(\frac{16}{25}\)
= (7x)² – (\(\frac{4}{5}\))²
= (7x+ (\(\frac{4}{5}\)) (7x – (\(\frac{4}{5}\))

(v) x⁴ – 2x² y² + y⁴
= (x² )² – 2x² y² + (y² )²
It is in the form of a² – 2ab + b²
a² – 2ab + b² = (a – b)²
∴ x⁴ – 2x² y² + y⁴ = (x² – y² )²
= [(x)² – (y)² ]²
= [(x + y) (x – y)]²
= (x + y)² (x – y)²
[∵ (ab)^m = a ^m . bⁿ ]

(vi) 4 (a + b)² – 9 (a – b)²
4 (a + b)² – 9 (a – b)²
= [2(a + b)]² – [3(a – b)]²
= [2(a + b) + 3(a- b)] [2(a + b)-3(a- b)]
= (2a + 2b + 3a – 3b) (2a + 2b – 3a + 3b)
= (5a – b) (5b – a)

Question 5.
Factorise the following expressions
(i) a²+ 10a + 24
(ii) x² +9x + 18
(iii) p² – 10q + 21
(iv) x² – 4x – 32
Solution:
(i) a²+ 10a + 24
a² + 10a + 24 .
= a² + 6a + 4a + 24
= a x a + 6a + 4a + 6 × 4
= a(a + 6) + 4(a + 6)
= (a + 6) (a + 4) (or)
a² + 10a + 24

∴ a² + 10a + 24 = (a + 6) (a + 4)

(ii) x² + 9x + 18
x² + 9x + 18
= (x + 3) (x + 6)

∴ x² + 9x + 18 = (x + 3) (x + 6)

(iii) p² – 10q + 21
p² – 10p + 21
= (P – 7) (p – 3)

∴ p² – 10p + 21 = (p – 7)(p – 3)

(iv) x² – 4x – 32
x² – 4x – 32
= (x – 8) (x + 4)

∴ x² – 4x – 32 = (x – 8) (x + 4)

Question 6.
The lengths of the sides of a triangle are integrals, and its area is also integer. One side is 21 and the perimeter is 48. Find the shortest side.
Solution:
Perimeter of a triangle
= AB + BC + CA = 48
⇒ c + a + b = 48

The solutions of Harmeet, Rosy are wrong.

∴ Srikar had done it correctly.
⇒ 21 + a + b = 48
⇒ a + b = 48 – 21 = 27
∴ The lengths of a, b should be 10, 17
∴ a + b > c [the sum of any two sides of a triangle is greater than the 3rd side]
∴ 10 + 17 > 2
27 > 21 (T).
∴ The length of the shortest side is 10 cm.

Question 7.
Find the values of ‘m’ for which x² + 3xy + x + my – in has two linear factors in x and y, with integer coefficients.
Solution:
Given equation is x² + 3xy + x + my – m ……….(1)
Let the two linear equations in x and y be (x + 3y + a) and (x + 0y + b).
Then (x + 3y + a) (x + 0y + b)
= x² + 0xy + bx + 3xy + 0y² + 3by + ax + 0y + ab
= x² + bx + ax + 3xy + 3by + ab ………….. (2)
Comparing equation (2) with (1),
x² + 3xy + x + my – m
= x² + (a + b)x + 3xy + 3by + ab
Equating the like terms on both sides,
ab = – m ………….. (3)
(a + b)x = x ⇒ a + b = 1 ……………. (4)
3by = my ⇒ 3b = m ⇒ b = \(\frac{\mathrm{m}}{3}\)
Substitute ‘b’ value in equation (4),
a = \(1-\frac{m}{3}=\frac{3-m}{3}\)
ab = -m
[ ∵ from (3)]
put a & b value then ,
\(\left(\frac{3-m}{3}\right)\left(\frac{m}{3}\right)\) = -m
\(\frac{3 \mathrm{~m}-\mathrm{m}^{2}}{9}\)= -m
⇒ 3m – m² = – 9m
⇒ m² – 12m = 0
⇒ m(m – 12) = 0
⇒ m = 0 (or) m = 12
lf m = 12

∴ b = \(\frac{12}{3}\) = 4&a = \(\frac{3-\mathrm{m}}{3}=\frac{3-12}{3}\)
= \(\frac{-9}{3}\) = -3
∴ Linear factors are (x + 3y – 3), (x + 4) If m = 0
b = \(\frac{0}{3}\) = 0 & a = \(\frac{3-0}{3}=\frac{3}{3}\) = 1
∴ Linear factors are (x + 3y + 1), x.

AP State Syllabus 8th Class Maths Solutions 4th Lesson Exponents and Powers InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 4 Exponents and Powers InText Questions and Answers.

8th Class Maths 4th Lesson Exponents and Powers InText Questions and Answers

Do this

Question 1.
Simplify the following.   (Page No. 81)
i) 3⁷ × 3³
ii) 4 × 4 × 4 × 4 × 4
iii) 3⁴ × 4³
Answer:
(i) 3⁷ × 3³ = 3⁷ + 3³ = 3¹⁰       [∵ a^m × aⁿ = a^m+n]
(ii) 4 × 4 × 4 × 4 × 4 = 4⁵      [∵ a × a × a × ……. m times = a^m]
(iii) 3⁴ × 4³ = 3⁴⁺³ = 3⁷      [∵ a^m × aⁿ = a^m+n]

Question 2.
The distance between Hyderabad and Delhi is 1674.9 km by rail. How would you express this in centimetres? Also express this in the scientific form.     (Page No. 81)
Answer:
Distance from Hyderabad to Delhi is
= 1674.9 km = 1674.9 × 1000 m = 1674900 mts
= 1674900 × 100 cm
= 167490000 cm
= 16749 × 10⁴ cm

Question 3.
What is 10^-10 equal to?     (Page No. 83)
Answer:
10^-10 = \(\frac{1}{10^{10}}\)      [∵ a^-n = \(\frac{1}{a^{n}}\)]

Question 4.
Find the multiplicative inverse of the following. (Page No. 83)
Answer:

Question 5.
Expand the following numbers using exponents. (Page No. 84)
Answer:
i) 543.67
= (5 × 100) + (4 × 10) + (3 × 10⁰) + \(\left(\frac{6}{10}\right)\) + \(\left(\frac{7}{10^{2}}\right)\)
= (5 × 10²) + (4 × 10) + (3 × 10⁰) + (6 × 10^-1) + (7 × 10^-2)   [∵ aⁿ = a^-n]

ii) 7054.243
= (7 × 1000) + (0 × 100) + (5 × 10) + (4 × 10⁰) + \(\left(\frac{2}{10}\right)\) + \(\left(\frac{4}{100}\right)\) + \(\left(\frac{3}{1000}\right)\)
= (7 × 10³) + (0 × 10²) + (5 × 10¹) + (4 × 10⁰) + (2 × 10^-1) + (4 × 10^-2) + (3 × 10^-3)

iii) 6540.305
= (6 × 1000) + (5 × 100) + (4 × 10) + (0 × 10⁰) + \(\left(\frac{3}{10}\right)\) + \(\left(\frac{0}{100}\right)\) + \(\left(\frac{5}{1000}\right)\)
= (6 × 10³) + (5 × 10²) + (4 × 10¹) + (0 × 10⁰) + (3 × 10^-1) + (0 × 10^-2) + (5 × 10^-3)

iv) 6523.450
= (6 × 1000) + (5 × 100) + (2 × 10) + (3 × 10⁰) + \(\left(\frac{4}{10}\right)\) + \(\left(\frac{5}{100}\right)\) + \(\left(\frac{0}{1000}\right)\)
= (6 × 10³) + (5 × 10²) + (2 × 10¹) + (3 × 10⁰) + (4 × 10^-1) + (5 × 10^-2) + (0 × 10^-3)

Question 6.
Simplify and express the following as single exponent.    (Page No. 85)
(i) 2^-3 × 2^-2
(ii) 7^-2 × 7⁵
(iii) 3⁴ × 3^-5
(iv) 7⁵ × 7^-4 × 7^-6
(v) m⁵ × m^-10
(vi) (-5)^-3 × (-5)^-4
Answer:
(i) 2^-3 × 2^-2 = 2^(-3)+(-2) = 2^-5 = \(=\frac{1}{2^{5}}\) = \(\frac{1}{2 \times 2 \times 2 \times 2 \times 2}\) = \(\frac{1}{32}\) [∵ a^m × aⁿ = a^m+n]
(ii) 7^-2 × 7⁵ = 7^-2+5 = 7³ = 343
(iii) 3⁴ × 3^-5 = 3^4+(-5) = 3^-1 = \(\frac{1}{3}\) [∵ a^-n = \(\frac{1}{a^{n}}\)]
(iv) 7⁵ × 7^-4 × 7^-6 = 7^5+(-4)+(-6) = 7^5-10 = 7^-5 = \(=\frac{1}{7^{5}}\)
(v) m⁵ × m^-10 = m^5+(-10) = m^-5 = \(=\frac{1}{m^{5}}\)
(vi) (-5)^-3 × (-5)^-4 = (-5)^(-3)+(-4) = (-5)^-7 = \(\frac{1}{(-5)^{7}}\) = –\(\frac{1}{5^{7}}\)

Question 7.
Change the numbers into standard form and rewrite the statements.      (Page No. 93)
i) The distance from the Sun to Earth is 149,600,000,000 m
Answer:
149,600,000,000 m = 1496 × 10⁸ m

ii) The average radius of the Sun is 695000 km
Answer:
695000 km = 695 × 10³ km

iii) The thickness of human hair is in the range of 0.005 to 0.001 cm.
Answer:
0.005 to 0.001 cm
= \(\frac{5}{1000}\) to \(\frac{1}{1000}\) cm = 5 × 10^-3 to 1 × 10^-3 cm

iv) The height of Mount Everest is 8848 m
Answer:
8848 m, itself is a standard form.

Question 8.
Write the following numbers in the standard form.      (Page No. 93)
The standard form of the following numbers are
Answer:
(i) 0.0000456 = \(\frac{456}{10000000}\) = 456 × 10^-7
(ii) 0.000000529 = \(\frac{529}{1000000000}\) = 529 × 10⁹
(iii) 0.0000000085 = \(\frac{85}{10000000000}\) = 85 × 10¹⁰
(iv) 6020000000 = 602 × 10000000 = 602 × 10⁷
(v) 35400000000 = 354 × 100000000 = 354 × 10⁸
(vi) 0.000437 × 10⁴ = \(\frac{437}{1000000}\) × 10⁴
= 437 × 10^-6 × 10⁴
= 437 × 10^(-6)+4
= 437 × 10^-2

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 12th Lesson Factorisation Exercise 12.1

Question 1.
Find the common factors of the given terms in each.

(i) 8x, 24
(ii) 3a, 2lab
(iii) 7xy, 35x²y³
(iv) 4m², 6m², 8m³
(v) 15p, 20qr, 25rp
(vi) 4x², 6xy, 8y²x
(vii) 12 x²y, 18xy²
Solution:
8x = 2 × 2 × 2 × x
24 = 8 × 3 = 2 × 2 × 2 × 3
∴ Common factors of 8x, 24 = 2, 4, 8.

ii) 3a, 2lab
3a = 3 × a
21ab = 7 × 3 × a × b
∴ Common factors of 3a, 21ab = 3, a, 3a.

iii) 7xy, 35x²y³
7xy = 7 × x × y
35x²y³ = 7 × 5 × x × x × y × y × y
∴ Common factors of 7xy, 35x²y³
= 7, x, y, 7x, 7y, xy, 7xy.

iv) 4m², 6m², 8m³
4m² = 2 × 2 × m × m
6m² = 2 × 3 × m × m
8m³ = 2 × 2 × 2 × m × m × m
∴ Common factors of 4m² , 6m² , 8m³
= 2, m, m², 2m, 2m².

v) 15p, 20qr, 25rp
15p = 3 × 5 × p
20qr = 4 × 5 × q × r
25rp = 5 × 5 × r × p
∴ Common factors of 15p, 20qr, 25rp = 5.

vi) 4x², 6xy, 8y²x
4x² = 2 × 2 × x × x
6xy = 2 × 3 × x × y
8y²x = 2 × 2 × 2 × y × y × x
∴ Common factors of 4x², 6xy, 8xy² = 2, x, 2x.

vii) 12x²y, 18xy²
12x²2y = 2 × 2 × 3 × x × x × y
18xy² = 3 × 3 × 2 × x × y × y
∴ Common factors of 12x²y, 18xy²
= 2,3, x, y, 6, xy, 6x, 6y, 2x, 2y, 3x, 3y, 6xy.

Question 2.
Factorise the following expressions
(i) 5x² – 25xy
(ii) 9a² – 6ax
(iii) 7p² + 49pq
(iv) 36 a²b – 60 a²bc
(v) 3a²bc + 6ab²c + 9abc²
(vi) 4p² + 5pq – 6pq²
(vii) ut + at²
Solution:
(i) 5x² – 25xy
= 5 x × x × – 5 × 5 × x × y
= 5 × x [x – 5 × y]
= 5x [x – 5y]

ii) 9a² – 6ax
= 3 × 3 × a × a – 2 × 3 × a × x
= 3a [3a – 2x]

iii) 7p² + 49pq
= 7 × p × p +7 × 7 × p × q
= 7p[p + 7q]

iv) 36a²b – 60a²bc
= 2 × 2 × 3 × 3 × a × a × b – 2 × 2 × 3 × 5 × a × a × b × c
= 2 × 2 × 3 × a × a × b[3 – 5c]
= 12a²b [3 – 5c]

v) 3a²bc + 6ab²c + 9abc²
= 3 × a × a × b × c + 3 × 2 × a × b × b × c + 3 × 3 × a × b × c × c
= 3abc [a + 2b + 3c]

vi) 4p² + 5pq – 6pq²
= 2 × 2 × p × p + 5 × p × q – 2 × 3 × p × q × q
= p [4p + 5q – 6q²]

vii) ut + at²
= u × t + a × t × t = t [u + at]

Question 3.
Factorise the following:
(i) 3ax – 6xy + 8by – 4bx
(ii) x³ + 2x² + 5x + 10
(iii) m² – mn + 4m – 4n
(iv) a³ – a²b² – ab + b³
(v) p²q – pr² – pq + r²
Solution:
i) 3ax – 6xy + 8by – 4ab
= (3ax – 6xy) – (4ab – 8by)
= (3 × a × x – 2 × 3 × x × y)
– (4 ×a × b – 4 × 2 × b × y)
= 3x(a – 2y) – 4b(a – 2y)
= (a – 2y)(3x – 4b)

ii) x³ + 2x² + 5x + 10
= (x³ + 2x²) + (5x +10)
= (x² × x + 2 × x²) + (5 × x + 5 × 2)
= x²(x + 2) + 5(x + 2)
= (x + 2) (x² + 5)

iii) m² – mn + 4m – 4n
= (m² – mn) + (4m – 4n)
= (m × m – m × n) + (4 × m – 4 × n)
= m(m – n) + 4(m – n)
= (m – n) (m + 4)

iv) a³ – a²b² – ab + b³
= (a³ – a²b²) – (ab – b³)
= (a² × a – a² × b²) – (a × b – b × b²)
= a²(a – b²) – b(a – b²)
= (a – b²) (a² – b)

v) p²1 – pr² – pq + r²
= (p²q – pr²) – (pq – r²)
= (p × p × q – p × r × r) – (pq – r²)
= p(pq – r²) – (pq – r²) × 1
= (p – 1) (pq – r²)

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.5 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions Exercise 11.5

Question 1.
Verify the identity (a + b)² ≡ a² + 2ab + b² geometrically by taking
(i) a = 2 units, b = 4 units
(ii) a = 3 units, b = 1 unit
(iii) a = 5 units, b = 2 unit
Solution:
(i)

= 4 × 4 + 4 × 2 + 2 × 2 + 4 × 2
= 16+ 8 + 4 + 8 = 36 sq.units
[∵ (2 + 4)² = 6² = 36]

(ii)

Area of a square AEGI
= area of square ABCD + area of rectangle CDEF + area of square CFGH + area of rectangle BIHC.
= 3 × 3 + 3 × 1 + 1 × 1+3 × 1
= 9 + 3 + 1 + 3
= 16 sq. units
[∵ (3 + 1)2 = 4² = 16]

(iii)

= 5 × 5 + 2 × 5 + 2 × 2 + 5 × 2
= 25 + 10 + 4 + 10
= 49 sq.units
[∵ (5 + 2)² = 7² = 49]

Question 2.
Verify the identity (a – b)² ≡ a² – 2ab+ b² geometrically by taking
(i) a = 3 units, b= 1 unit
(ii) a = 5 units, b = 2 units
Solution:
(i)

Area of AIFE + Area of FGCH = (a – b)² = a² – 2ab + b² [area of AIFE – area of IBGF – area of EFHD + area of FGCH]
= 3 × 3 – 1 × 3 – 3 × 1 + 1 × 1
= 9 – 3 – 3 + 1 = 4
∴ (a – b)² = 4 sq. units
[∵ (3 – 1 )² = 2² = 4]

ii)

∴ (a – b)² = a² – 2ab + b²
Area of ABCD + Area of CYZS = a² – 2ab + b²
area of ABCD – area of BXYC – area of DCST + area of CYZS
=5 × 5 – 2 × 5 – 2 × 5 + 2 × 2
= 25 – 10 – 10 + 4
= 9 sq.units
[∵ (5 – 2)² = (3)² = 9]

Question 3.
Verify the identity(a + b)(a – b) ≡ a² – b² geometrically by taking
(i) a = 3 units, b = 2 units
(ii) a = 2 units, b = 1 unit
Solution:
i)

a² – b² = Area of Fig I. + Area of Fig II
= a(a – b) + b(a – b)
= (a – b) (a + b)
= 3 × 3 – 2 × 2
a² – b² = 9 – 4= 5sq . units
[ ∵ 3² – 2² = 9 – 4 = 5]

ii)

a² – b² = Area of Fig I. + Area of Fig II
= a(a – b) + b(a – b)
= (a + b) (a – b)
=(2 + 1)(2 – 1)
= 3 × 1 = 3
a² – b² = 3 sq. units
[∵ (2² – 1²) = 4 – 1 = 3]

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.4 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions Exercise 11.4

Question 1.
Select a suitable identity and find the following products
(i) (3k + 4l)(3k + 4l)
(ii) (ax² + by²)(ax² + by²)
(iii) (7d – 9e)(7d – 9e)
(iv) (m² – n²)(m² + n²)
(v) (3t + 9s) (3t – 9s)
(vi) (kl – mn) (kl + mn)
(vii) (6x + 5)(6x + 6)
(viii) (2b – a)(2b +c)
Solution:
(3k + 4l) (3k 4l) = (3k + 4l)² is in the form of (a + b)².
=(3k)² + 2 × 3k × 4l+ (4l)² [ (a+ b)² = a² + 2ab + b²
= 3k × 3k + 24kl + 4l × 4l
= 9k² + 24kl + 16l²

ii) (ax² + by²) (ax² + by²) = (ax² + by²)² is in the form of (a + b)².
= (ax²)² + 2 × ax² × by² + (by²)² [ ∵ (a + b)² = a² + 2ab + b²]
= ax² × ax² + 2abx²y² + by² × by²
= a²x⁴ + 2ab x²y² + b²y⁴

iii) (7d – 9e) (7d – 9e)
= (7d – 9e)² is in the form of (a – b)².
= (7d)² – 2 × 7d × 9e + (9e)² [ ∵ (a – b)² = a² – 2ab + b²]
= 7d × 7d – 126de + 9e × 9e
= 49d² – 126de + 81e²

iv) (m² – n²) (m² + n²) is in the form of (a + b) (a – b).
∴ (a + b) (a – b) = a² – b²
∴ (m² + n²) (m² – n²) = (m²)² – (n²)² = m⁴ – n⁴

v) (3t + 9s) (3t – 9s) = (3t)² – (9s)² [ ∵ (a + b) (a – b) = a² – b² ]
= 3t × 3t – 9s × 9s
= 9t² – 81s²

vi) (kl – mn) (kl + mn) = (kl)² – (mn)² [ ∵(a + b) (a – b) = a² – b² ]
= kl × kl – mn × mn
= k²l² – m²n²

vii) (6x + 5) (6x + 6) is in the form of
(ax + b) (ax + c).
(ax + b) (ax + c) = a²x² + ax(b + c) + bc
(6x + 5) (6x + 6) = (6)²x² + 6x (5 + 6) + 5 × 6
= 36x² + 6x × 11 + 30
= 36x² + 66x + 30

viii) (2b – a) (2b + c) is in the form of (ax – b) (ax + c).
(ax – b) (ax + c) = a²x² + ax(c – b) – cb
(2b – a) (2b + c) = (2)²(b)² + 2b (c – a) – ca
= 4b² + 2bc – 2ab – ca

Question 2.
Evaluate the following by using suitable identities:
(i) 304²
(ii) 509²
(iii) 992²
(iv) 799²
(v) 304 × 296
(vi) 83 × 77
(vii) 109 × 108
(viii) 204 × 206
Solution:
i) 304² = (300 + 4)² is in the form of (a + b)².
∵ (a+b)² = a² + 2ab + b²
a = 300, b = 4
(300 + 4)² = (300)² + 2 × 300 × 4 + (4)²
= 300 × 300+ 2400 + 4 × 4
= 90,000 + 2400 + 16
= 92,416

ii) 509² = (500 + 9)²
a  = 500, b = 9
= (500)² + 2 × 500 × 9 + (9)²
[ ∵ (a + b)² = a² + 2ab + b²]
= 500 × 500 + 9000 + 9 × 9
= 2,50,000 + 9000 + 81
= 2,59,081

iii) 992² = (1000 – 8)²
a = 1000, b = 8
= (1000)² – 2 × 1000 × 8 + (8)² [∵ (a-b)² = a² – 2ab + b²]
= 1000 × 1000 – 16,000 + 8 × 8
= 10,00,000 – 16000 + 64
= 10,00,064 – 1600
= 9,98,464

iv) 799² = (800 – 1)²
a = 800, b = 1
= (800)² – 2 × 800 × 1 + (1)²
= 800 × 800 – 1600 + 1
= 6,40,000 – 1600 + 1
= 6,40,001 – 1600
= 6,38,401

v) 304 × 296 = (300 + 4) (300 – 4) is in the form of (a + b) (a – b).
(a + b) (a – b) = a² – b²
∴ (300 + 4) (300 – 4) = (300)² – (4)²
= 300 × 300 – 4 × 4
= 90,000 – 16
= 89,984

vi) 83 × 77 = (80 + 3) (80 – 3)
= (80)² – (3)² [ ∵ (a + b) (a – b) = a² – b²]
= 80 × 80 – 3 × 3
= 6400 – 9
= 6391

vii) 109 × 108 = (100 + 9) (100 + 8)
= (100)² + (9 + 8)100 + 9 × 8
= 10,000 + 1700 + 72
= 11,772

viii) 204 × 206 = (205 – 1) (205 + 1)
= (205)² – (1)² [∵ (a + b)(a-b) = a² – b²]
= 205 × 205 – 1 × 1
= 42,025 -1
= 42,024

AP State Syllabus 8th Class Maths Solutions 3rd Lesson Construction of Quadrilaterals InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals InText Questions and Answers.

8th Class Maths 3rd Lesson Construction of Quadrilaterals InText Questions and Answers

Do this

Question 1.
Take a pair of sticks of equal length, say 8 cm. Take another pair of sticks of equal length, say 6 cm. Arrange them suitably to get a rectangle of length 8 cm and breadth 6 cm. This rectangle is created with the 4 available measurements. Now just push along the breadth of the rectangle. Does it still look alike? You will get a new shape of a rectangle Fig (ii), observe that the rectangle has now become a parallelogram. Have you altered the lengths of the sticks? No! The measurements of sides remain the same.

Give another push to the newly obtained shape in the opposite direction; what do you get? You again get a parallelogram again, which is altogether different Fig (iii). Yet the four measurements remain the same. This shows that 4 measurements of a quadrilateral cannot determine its uniqueness. So, how many measurements determine a unique quadrilateral? Let us go back to the activity!
You have constructed a rectangle with two sticks each of length 8 cm and other two sticks each of length 6 cm. Now introduce another stick of length equal to BD and put it along BD (Fig iv). If you push the breadth now, does the shape change? No!
It cannot, without making the figure open. The introduction of the fifth stick has fixed the rectangle uniquely, i.e., there is no other quadrilateral (with the given lengths of sides) possible now. Thus, we observe that five measurements can determine a quadrilateral uniquely. But will any five measurements (of sides and angles) be sufficient to draw a unique quadrilateral? (Page No. 60)
Answer:
Yes, any 5 individual measurements are needed to construct a quadrilateral.

Question 2.
Equipment (Page No. 61)
You need: a ruler, a set square, a protractor.
Remember: To check if the lines are parallel.
Slide set square from the first line to the second line as shown in adjacent figures.

Now let us investigate the following using proper instruments. For each quadrilateral,
a) Check to see if opposite sides are parallel.
b) Measure each angle.
c) Measure the length of each side.

Record your observations and complete the table below.
Answer:

Question 3.
Can you draw the angle of 60°?    (Page No. 63)

Answer:
Using a scale and compass,
we can construct 60°.

Question 4.
Construct the parallelogram above (Refer text book page no: 75) BELT by using other properties of parallelogram. (Page No. 75)
Answer:

We can construct a parallelogram using the measurements of a side, a diagonal and an angle.
BE = 5 cm ⇒ LT = 5 cm
∠B = 110° ⇒ ∠E = 180° – 110° = 70°
TE= 7.2 cm

Try these

Question 1.
Can you draw a parallelogram BATS where BA = 5 cm, AT = 6 cm and AS = 6.5 cm?    (Page No. 70)
Answer:

In a parallelogram BATS, opposite sides are equal.
BA = ST = 5 cms
AT = BS 6 cms
AS = 6.5 cms

∴ So, we can construct BATS parallelogram. It needs only three measurements.

Question 2.
A student attempted to draw a quadrilateral PLAY given that PL = 3 cm, LA = 4 cm, AY = 4.5 cm, PY = 2 cm and LY = 6 cm. But he was not able to draw it why ?
Try to draw the quadrilateral yourself and give reason. (Page No. 70)
Answer:
In a quadrilateral PLAY
PL = 3 cm LA = 4 cm AY = 4.5 cm
PY = 2 cm LY = 6 cm
Here YP + PL < YL [∵ 2 + 3 < 6 ⇒ 5 < 6]

But in a △YPL, the sum of two sides is less than the third side.
∴ We are unable to construct a quadrilateral PLAY [∵ YL > YP]
[∵ The arcs do not intersect which are drawn from L and P, also Y, P, L are collinear points]

Think, discuss and write

Question 1.
Is every rectangle a parallelogram? Is every parallelogram a rectangle?    (Page No. 63)
Answer:
Yes, every rectangle is a parallelogram. But every parallelogram is not a rectangle.

Question 2.
Uma has made a sweet chikki. She wanted it to be rectangular. In how many different ways can she verify that it is rectangular?    (Page No. 63)
Answer:
If the sweet chikki is to made into a rectangular shape, she has to verify the following shapes:

1. Quadrilateral
2. Trapezium
3. Parallelogram

Question 3.
Can you draw the quadrilateral ABCD with AB = 4.5 cm, BC = 5.2 cm, CD = 4.8 cm and diagonals AC = 5 cm, BD = 5.4 cm by constructing △ABD first and then fourth vertex ‘C’ ? Give reason.       (Page No. 72)
Answer:
We cannot construct △ABD. So, if we start first from △ABD, it is impossible to construct □ ABCD.
[∵ The length of \(\overline{\mathrm{AD}}\) is not given]

Question 4.
Construct a quadrilateral PQRS with PQ = 3 cm, RS = 3 cm, PS = 7.5 cm, PR = 8 cm and SQ = 4 cm. Justify your result.      (Page No. 72)
Answer:

PQ = 3 cm
RS = 3 cm
PS = 7.5 cm
PR = 8 cm
SQ = 4 cm
With the given measurements △PQS is not possible to construct.
∵ PQ + QS < PS
The arcs which drawn from P and Q are not intersecting.
∴ We can’t obtain vertex ‘S’.
∴ Without vertex ‘S’ we can’t get a quadrilateral PQRS.

Question 5.
Can you construct the quadrilateral PQRS, if we have an angle of 100° at P instead of 75°? Give reason. (Page No. 74)
Answer:

PQ = 4 cm,
QR = 4.8 cm,
ZP = 100°,
ZQ = 100°,
ZR = 120°
∴ We can construct a quadrilateral with the given measurements.
Since the sum of 4 angles is equal to 360°.

Question 6.
Can you construct the quadrilateral PLAN if PL = 6 cm, ∠A = 9.5 cm, ∠P = 75°, ∠L = 15° and ∠A = 140°?
(Draw a rough sketch in each case and analyse the figure). State the reasons for your conclusion. (Page No. 74)
Answer:
PL = 6 cm, ∠A = 9.5 cm, ∠P = 75°, ∠L = 15°, ∠A = 140°

∴ With the given measurements it is not possible to construct a quadrilateral.

Question 7.
Do you construct the given quadrilateral ABCD with AB = 5 cm, BC = 4.5 cm, CD = 6 cm, ∠B = 100°, ∠C = 75° by taking BC as base instead of AB? If so, draw a rough sketch and explain the various steps involved in the construction. (Page No. 77)
Answer:
AB = 5 cm, BC = 4.5 cm, CD = 6 cm, ∠B = 100°, ∠C = 75°

Construction Steps:

1. Construct a line segment with radius 4.5 cms as \(\overline{\mathrm{BC}}\)
2. With the centres B and C draw two rays with 100°, 75° respectively.
3. With the centres B and C, two arcs are drawn with radius 5 cm and 6 cm respectively. The arcs and the rays are intersected.
4. Let the intersecting points be keep as A, D.
5. Join A, D.
6. ∴ ABCD quadrilateral is formed.
   

Question 8.
Can you construct the given AC = 4.5 cm and BD = 6 cm quadrilateral (rhombus) taking BD as a base instead of AC? If not give reason. (Page No. 79)
Answer:

We can construct a rhombus taking BD as base instead of base AC.

Question 9.
Suppose the two diagonals of this rhombus are equal in length, what figure do you obtain? Draw a rough sketch for it. State reasons. (Page No. 79)
Answer:
In a rhombus if the two diagonals are equal then it becomes a square.
∴ ABCD is a square.
[∵ AB = BC = CD = DA Also AC = BD]

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.3 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions Exercise 11.3

Question 1.
Multiply the binomials:
(i) 2a – 9 and 3a + 4
(ii) x – 2y and 2x – y
(iii) kl + lm and k – l
(iv) m² – n² and m + n
Solution:
i) 2a – 9 and 3a + 4
(2a – 9) (3a + 4) = 2a (3a + 4) – 9(3a + 4)
= 6a² + 8a – 27a – 36
= 6a² – 19a – 36

ii) x – 2y and 2x – y
(x – 2y) (2x – y) = x(2x – y) – 2y(2x – y)
= 2x² – xy – 4xy + 2y²
= 2x² – 5xy + 2y²

iii) kl + lm and k – l
(kl + lm) (k – l) = kl(k – l) + lm(k – l)
= k²l – l²k + klm – l²m

iv) m² – n² and m + n
(m² – n²) (m + n) = m²(m + n) – n²(m + n)
= m³ + m²n – n²m – n³

Question 2.
Find the product:
(i) (x + y)(2x – 5y + 3xy)
(ii) (mn – kl + km) (kl – lm)
(iii) (a – 2b + 3c)(ab² – a²b)
(iv) (p³ + q³)(p – 5q+6r)
Solution:
i) (x + y) (2x – 5y + 3xy)
= x(2x – 5y + 3xy) + y(2x – 5y + 3xy)
= 2x² – 5xy + 3x²y + 2xy – 5y² + 3xy²
= 2x² – 5y² – 3xy + 3x²y + 3xy²

ii) (mn – kl + km) (kl – lm)
= kl(mn – kl + km) – lm(mn – kl + km)
= klmn – k²l² + k²lm – lm²n + kl²m – klm²

iii) (a – 2b + 3c) (ab² – a²b) = a(ab² – a²b) – 2b(ab² – a²b) + 3c(ab²– a²b)
= a²b² – a³b – 2ab³ + 2a²b² + 3ab²c – 3a²bc
= 3a²b² – a³b – 2ab³ + 3ab²c – 3a²bc

iv) (p³ + q³) (p – 5q + 6r) = p³(p – 5q + 6r) + q³(p – 5q + 6r)
= p⁴ – 5p³q + 6p³r + pq³ – 5q⁴ + 6rq³
= p⁴ – 5q⁴ – 5p³q + 6p³r + pq³ + 6rq³

Question 3.
Simplify the following:
(i) (x-2y) (y – 3x) + (x+y) (x-3y) – (y – 3x) (4x – 5y)
(ii) (m + n) (m² – mn + n²)
(iii) (a – 2b + 5c) (a – b) – (a – b – c) (2a + 3c) + (6a + b) (2c – 3a – 5b)
(iv) (pq-qr-i-pr) (pq-i-qr) – (pr-i-pq) (p-i-q – r)
Solution:
i) (x – 2y) (y – 3x) + (x + y) (x – 3y) – (y – 3x) (4x – 5y)
= (y – 3x) [x – 2y – (4x – 5y)] + (x + y)(x – 3y)
= (y – 3x) [x – 2y – 4x + 5y] + (x + y) (x – 3y)
= (y – 3x) (3y – 3x) + (x + y) (x – 3y)
= y(3y – 3x) – 3x(3y – 3x) + x(x – 3y) + y(x – 3y)
= 3y² – 3xy – 9xy + 9x² + x² – 3xy + xy – 3y²
= 10x² – 14xy

ii) (m + n) (m²– mn + n²)
= m(m² – mn + n²) + n(m² – mn + n²)
= m³ – m²n + n²m + nm² – mn² + n³
= m³ + n³

iii) (a – 2b + 5c) (a – b) – (a – b – c) (2a + 3c) + (6a + b) (2c – 3a – 5b)
= a(a – 2b + 5c) – b(a – 2b + 5c) – 2a(a – b – c) – 3c(a – b – c) + 6a(2c – 3a – 5b) + b(2c – 3a – 5b)
= a² – 2ab + 5ac – ab + 2b² – 5bc – 2a² + 2ab + 2ac – 3ac + 3bc + 3c² + 12ac – 18a² – 30ab + 2bc – 3ab – 5b²
= – 19a² – 3b² – 34ab + 16ac + 3c²

iv) (pq – qr + pr) (pq + qr) – (pr + pq) (p + q – r)
= pq(pq – qr + pr) + qr(pq – qr + pr) – pr(p + q – r) – pq(p + q – r)
= p²q² – pq²r + p²qr + pq²r – q²r² + pqr² – p²r – pqr + pr² – p²q – pq² + pqr
= p²q² – q²r² + p²qr + pqr² – p²r + pr² – p²q – pq ²

Question 4.
If a, b, care positive real numbers such that \(\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}\) ,find the value of \(\frac{(a+b)(b+c)(c+a)}{a b c}\)
Solution:
\(\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}\) = k then
\(\frac{a+b-c}{c}\) = k ⇒ a + b – c = kc
⇒ a + b = (ck + c) = c(k + 1) …………… (1)
Similarly b + c = a(k + 1) ……………(2)
c + a = b(k + 1) ………………..(3)

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions Exercise 11.2

Question 1.
Complete the table:

Solution:

Question 2.
Simplify: 4y(3y + 4)
Solution:
4y(3y + 4) = 4y × 3y + 4y × 4
= 12y² + 6y

Question 3.
Simplify x(2x² – 7x + 3) and find the values of it for (i) x = 1 and (ii) x = 0
Solution:
x(2x² – 7x + 3)
= x × 2x² – x × 7x + x × 3
= 2x³ – 7x² + 3x
= 3 × 3 – 1 × 3 – 3 × 1 + 1 × 1
=9 – 3 – 3 + 1 = 4
∴(a – b)² = 4sq.units
[∵ (3 – 1)² = 2² = 4]

(ii)

∴ (a – b)² = a² – 2ab + b²
Area of ABCD + Area of CYZS
= a² – 2ab + b²
area of ABCD – area of BXYC – area of DCST + area of CYZS
= 5 × 5 – 2 × 5 – 2 × 5 + 2 × 2
= 25 – 10 – 10 + 4
= 9 sq.units
[∵ (5 – 2)² = (3)² = 9]

Question 4.
Add the product: a(a – b), b(b – c), c(c – a)
Solution:
a(a – b) + b(b – c) + c(c – a)
=a × a – a × b + b × b – b × c + c × c – c × a
=a² – ab + b² – bc + c² – ca
=a² + b² + c² – ab – bc – ca

Question 5.
Add the product: x(x + y – r), y(x – y+r), z(x – y – z)
Solution:
x(x + y – r) +y(x – y + r) + z(x – y – z)
= x² + xy – xr + xy – y² + yr + zx – yz – z²
= x² – y² – z² + 2xy – xr + yr + zx – yz

Question 6.
Subtract the product of 2x(5x – y) from product of 3x(x+2y)
Solution:
3x(x + 2y) – 2x(5x – y)
=(3x × x + 3x × 2y)-(2x × 5x – 2x × y)
= 3x² + 6xy – (10x² – 2xy)
= 3x² + 6xy- 10x² + 2xy
= 8xy – 7x²

Question 7.
Subtract 3k(5k – l + 3rn) from 6k(2k + 3l – 2rn)
Solution:
6k(2k + 3l – 2m) – 3k(5k – l + 3m)
= 12k²+ 18kl – 12km – 15k² + 3kl – 9km
= -3k² + 21kl – 21km

Question 8.
Simplify: a²(a – b + c) + b²(a + b – c) – c²(a – b – c)
Solution:
a²(a – b + c) + b²(a + b – c) – c²(a – b – c)
= a³ – a²b + a²c + ab² + b³ – b²c – ac² + bc² + c³
= a³ + b³ + c³ – a²b + a²c + ab² – b²c – ac² – bc²

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions Exercise 11.1

Question 1.
Find the product of the following pairs:
(i) 6, 7k
(ii) – 31, – 2m
(iii) -5t² – 3t²
(iv) 6n, 3m
(v) – 5p², – 2p
Solution:
The product of 6, 7k = 6 × 7k = 42k
ii) The product of – 3l, – 2m = (- 3l) × (- 2m) = 6/m
iii) The product of – 5t², – 3t² = (- 5t²) × (- 3t²) = 15t⁴
iv) The product of 6n, 3m = 6n × 3m = 18mn
v) The product of – 5p², – 2p = (- 5p²) × (- 2p) = 10p³

Question 2.
Complete the table of the products.

Solution:

Question 3.
Find the volumes of rectangular boxes with given length, breadth and height in the following table.

Solution:

Question 4.
Find the product of the following monomials
(i) xy, x²y , xy, x
(ii) a, b, ab, a³ b, ab³
(iii) kl, lm, km, klm
(iv) pq ,pqr, r
(v) – 3a, 4ab, – 6c, d
Solution:
i) The product of xy, x²y, xy, x = xy × x²y × xy × x
= x⁵ × y³= x⁵y³

ii) The product of a, b, ab, a³b, ab³ = a × b × ab × a³b × ab³
= a⁶ × b⁶ = a⁶ b⁶

iii) The product of kl, lm, km, klm = kl × lm × km × klm
k³ × l³ × m³ =k³l³m³

iv) The product of pq, pqr, r = pq × pqr × r
= p² × q² × r² – p²q²r²

v) The product of – 3a, 4ab, – 6c, d = (- 3a) × 4ab × (- 6c) x d
= + 72a² × b × c × d
= 72a²bcd

Question 5.
If A = xy,B = yz and C = zx, then find ABC=
Solution:
ABC = xy × yz × zx = x²y²z²

Question 6.
If P = 4x², T = 5x and R = 5y, then \(\frac{\mathrm{PTR}}{100}\) =
Solution:
\(\frac{P^{\prime} \Gamma R}{100}=\frac{4 x^{2} \times 5 x \times 5 y}{100}=\frac{100 x^{3} y}{100}\) = x³ y

Question 7.
Write some monomials of your own and find their products.
Solution:
The product of,some monomials is given below :
i) abc × a²bc = a³b²c²
ii) xy × x²z × yz² = x³y²z³
iii) p × q × r = p³q³r³

AP State Syllabus 8th Class Maths Solutions 2nd Lesson Linear Equations in One Variable InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable InText Questions and Answers.

8th Class Maths 2nd Lesson Linear Equations in One Variable InText Questions and Answers

Do this

Question 1.
Which of the following are linear equations:        [Page No. 35]
i) 4x + 6 = 8
ii) 4x – 5y = 9
iii) 5x² + 6xy – 4y² = 16
iv) xy + yz + zx = 11
v) 3x + 2y – 6 = 0
vi) 3 = 2x + y
vii) 7p + 6q + 13s = 11
Answer:
(i), (ii), (v), (vi), (vii) are the linear equations.

Question 2.
Which of the following are simple equations?        [Page No. 36]
i) 3x + 5 = 14
ii) 3x – 6 = x + 2
iii) 3 = 2x + y
iv) \(\frac{x}{3}\) + 5 = 0
v) x² + 5x + 3 = 0
vi) 5m – 6n = 0
vii) 7p + 6q + 13s = 11
viii) 13t – 26 = 39
Answer:
(i), (ii), (iv), (viii) are the simple equations.
Since these are all in the form of ax + b = 0.