Mahesh

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 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 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 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.

AP State Syllabus 8th Class Maths Solutions 1st Lesson Rational Numbers InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 1 Rational Numbers InText Questions and Answers.

8th Class Maths 1st Lesson Rational Numbers InText Questions and Answers

Do this

Question 1.
Consider the following collection of numbers 1, \(\frac{1}{2}\), -2, 0.5, 4\(\frac{1}{2}\), \(\frac{-33}{7}\), 0, \(\frac{4}{7}\), \(0 . \overline{3}\), 22, -5, \(\frac{2}{19}\), 0.125. Write these numbers under the appropriate category. [A number can be written in more than one group]  (Page No. 2)
Answer:
i) Natural numbers 1, 22
ii) Whole numbers 0, 1, 22
iii) Integers 0, 1, 22, -5, -2
iv) Rational numbers 1, \(\frac{1}{2}\), -2, 0.5, 4\(\frac{1}{2}\), \(\frac{-33}{7}\), 0, \(\frac{4}{7}\), \(0 . \overline{3}\), 22, -5, \(\frac{2}{19}\), 0.125 etc.
Would you leave out any of the given numbers from rational numbers? No
Is every natural number, whole number and integer is a rational number? Yes

Question 2.
Fill the blanks in the table.     (Page No. 6)
Answer:

Question 3.
Complete the following table.     (Page No. 9)
Answer:

Question 4.
Complete the following table.      (Page No. 13)
Answer:

Question 5.
Complete the following table.      (Page No. 16)
Answer:

Question 6.
Complete the following table.      (Page No. 17)
Answer:

Question 7.
Represent – \(\frac{13}{5}\) on the number line.     (Page No. 22)
Answer:
Representing – \(\frac{13}{5}\) on the number line.

Try These

Question 1.
Hamid says \(\frac{5}{3}\) is a rational number and 5 is only a natural number. Shikha says both are rational numbers. With whom do you agree?       (Page No. 3)
Answer:
I would not agree with Hamid’s argument. Since \(\frac{5}{3}\) is a rational number. But ‘5’ is not only
a natural number, it is also a rational number.
Since every natural number is a rational number,
According to Shikha’s opinion \(\frac{5}{3}\), 5 are rational numbers.
∴ I agree with Shikha’s opinion.

Question 2.
Give an example to satisfy the following statements.        (Page No.3)
i) All natural numbers are whole numbers but all whole numbers need not be natural numbers.
ii) All whole numbers are integers but all integers are not whole numbers.
iii) All integers are rational numbers but all rational numbers need not be integers.
Answer:
i) ‘0’ is not a natural number.
∴ Every whole number is not a natural number. (∵ N ⊂ W)
ii) -2, -3, -4 are not whole numbers.
∴ All integers are not whole numbers. (∵ W ⊂ Z)
iii) \(\frac{2}{3}\), \(\frac{7}{4}\) are not integers.
∴ Every rational number is not an integer. (∵ Z ⊂ Q)

Question 3.
If we exclude zero from the set of integers is it closed under division? Check the same for natural numbers.    (Page No. 6)
Answer:
If ‘0’ is subtracted from the set of integers then it becomes Z – {0}.
Closure property under division on integers.
Ex: -4 ÷ 2 = -2 is an integer.
3 ÷ 5 = \(\frac{3}{5}\) is not an integer.
∴ Set of integers doesn’t satisfy closure property under division.
Closure property under division on natural numbers.
Ex: 2 ÷ 4 = \(\frac{1}{2}\) is not a natural number.
∴ Set of natural numbers doesn’t satisfy closure property under division.

Question 4.
Find using distributivity.     (Page No. 16)
A) \(\left\{\frac{7}{5} \times\left(\frac{-3}{10}\right)\right\}+\left\{\frac{7}{5} \times\left(\frac{9}{10}\right)\right\}\)
B) \(\left\{\frac{9}{16} \times 3\right\}+\left\{\frac{9}{16} \times-19\right\}\)
Answer:
Distributive law: a × (b + c) = ab + ac
A)

B)

Question 5.
Write the rational number for the points labelled with letters, on the number line.       (Page No. 22)
i)

ii)

Answer:
i) A = \(\frac{1}{5}\), B = \(\frac{4}{5}\), C = \(\frac{5}{5}\) = 1, D = \(\frac{7}{5}\), E = \(\frac{8}{5}\), F = \(\frac{10}{5}\) = 2.
ii) S = \(\frac{-6}{4}\), R = \(\frac{-6}{4}\), Q = \(\frac{-3}{4}\), P = \(\frac{-1}{4}\)

Think, discuss and write

Question 1.
If a property holds good with respect to addition for rational numbers, whether it holds good for integers? And for whole numbers? Which one holds good and which doesn’t hold good?     (Page No. 15)
Answer:
Under addition the properties which are followed by set of rational numbers are also followed by integers.

Question 2.
Write the numbers whose multiplicative inverses are the numbers themselves.      (Page No. 15)
Answer:
The number T is multiplicative inverse of itself.
∵ 1 × \(\frac{1}{1}\) = 1 ⇒ 1 × 1 = 1
∴ The multiplicative inverse of 1 is 1.

Question 3.
Can you find the reciprocal of ‘0’ (zero)? Is there any rational number such that when it is multiplied by ‘0’ gives ‘1’?
          (Page No. 15)
Answer:
The reciprocal of ‘0’ is \(\frac{1}{0}\).
But the value of \(\frac{1}{0}\) is not defined.
∴ There is no number is found when it is multiplied ‘0’ gives 1.
∵ 0 × (Any number) = 0

∴ No, there is no number is found in place of ‘A’.

Question 4.
Express the following in decimal form.     (Page No. 28)
i) \(\frac{7}{5}\), \(\frac{3}{4}\), \(\frac{23}{10}\), \(\frac{5}{3}\),\(\frac{17}{6}\),\(\frac{22}{7}\)
ii) Which of the above are terminating and which are non-terminating decimals?
iii) Write the denominators of above rational numbers as the product of primes.
iv) If the denominators of the above simplest rational numbers has no prime divisors other than 2 and 5 what do you observe?
Answer:
i) \(\frac{7}{5}\) = 0.4,
\(\frac{3}{4}\) = 0.75,
\(\frac{23}{10}\) = 2.3,
\(\frac{5}{3}\) = 1.66… = \(1 . \overline{6}\),
\(\frac{17}{6}\) = 2.833… = \(2.8 \overline{3}\),
\(\frac{22}{7}\) = 3.142
ii) From the above decimals \(\frac{7}{5}\), \(\frac{3}{4}\), \(\frac{23}{10}\) are terminating decimals.
While \(\frac{5}{3}\),\(\frac{17}{6}\),\(\frac{22}{7}\) are non-terminating decimals
iii) By writing the denominators of above decimals as a product of primes is

iv) If the denominators of integers doesn’t have factors other than 2 or 5 and both are called terminating decimals.

Question 5.
Convert the decimals \(0 . \overline{9}\), \(14 . \overline{5}\) and \(1.2 \overline{4}\) to rational form. Can you find any easy method other than formal method?     (Page No. 31)
Answer:
Let x = \(0 . \overline{9}\)
⇒ x = 0.999 ……. (1)
The periodicity of the above equation is ‘1’. So it is to be multiplied by 10 on both sides.
⇒ 10 × x = 10 × 0.999
10x = 9.999 …….. (2)
From (1) & (2)

∴ x = 1 or \(0 . \overline{9}\) = 1
Second Method:
\(0 . \overline{9}\) = 0 + \(0 . \overline{9}\)
= 0 + \(\frac{9}{9}\)
= 0 + 1
= 1

Let x = \(14 . \overline{5}\)
⇒ x = 14.55 …….. (1)
The periodicity of the equation (1) is 1.
So it should be multiplied by 10 on both sides.
⇒ 10 × x = 10 × 14.55
10x = 145.55 …….. (2)

Second Method:

Let x = \(1.2 \overline{4}\)
⇒ x= 1.244 …….. (1)
Here periodicity of equation (1) is 1. So it should be multiplied by 10 on both sides.
⇒ 10 × x = 10 × 1.244
10 x = 12.44 …….. (2)

Second Method:

AP State Syllabus SSC 10th Class Maths Solutions 14th Lesson Statistics InText Questions

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 14 Statistics InText Questions and Answers.

10th Class Maths 14th Lesson Statistics InText Questions and Answers

Think & Discuss

(Page No. 327)

Question 1.
The mean value can be calculated from both ungrouped and grouped data. Which one do you think is more accurate? Why?
Answer:
Mean calculated from ungrouped data is more accurate than, mean calculated from the grouped data. Since its calculation takes all the observations in the data into consideration.

Question 2.
When it is more convenient to use grouped data for analysis?
Answer:
Grouped data is convenient when the values f_i and x_i are low.

(Page No. 331)

Question 3.
Is the result obtained by all the three methods the same?
Answer:
Yes. Mean obtained by all the three methods are equal.

Question 4.
If x_i and f_i are sufficiently small, then which method is an appropriate choice?
Answer:
Direct method.

Question 5.
If x_i and f_i are numerically large numbers, then which methods are appropriate to use?
Answer:
Assumed – Mean method and Step – Deviation method.

Do these

(Page No. 334)

Question 1.
Find the mode of the following data.
a) 5, 6, 9, 10, 6, 12, 3, 6, 11, 10, 4, 6, 7.
Answer:
Mode = 6 (most repeated value of the data).

b) 20, 3, 7, 13, 3, 4, 6, 7, 19, 15, 7, 18, 3.
Answer:
Mode = 3,7.

c) 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6.
Answer:
Mode = 2, 3, 4, 5, 6.

Question 2.
Is the mode always at the centre of the data?
Answer:
No. Mode may not beat the centre always.

Question 3.
Does the mode change, if another observation is added to the data in Example? Comment.
Answer:
In the example the observations are 0, 1, 2, 2, 2, 3, 3, 4, 5, 6.
Here mode = 2.
If we add ‘2’ then mode doesn’t change.
It we add ‘3’ then the mode will be ‘2’ and ‘3’.
Even if we add other then 3 the mode will not be changed.
So we cannot decide whether the mode changes or not. It depends on the situation.

Question 4.
If the maximum value of an observation in the data in Example 4 is changed to 8, would the mode of the data be affected? Comment.
Answer:
If the maximum value is altered to 8, the mode remains the same. Mode doesn’t consider the values but consider their frequencies only.

Think & Discuss

(Page No. 336)

Question 1.
It depends upon the demand of the situation whether we are interested in finding the average marks obtained by the students or the marks obtained by most of the students.
a) What do we find in the first situation?
Answer:
We find A.M.

b) What do we find in the second situation?
Answer:
We find the mode.

Question 2.
Can mode be calculated for grouped data with unequal class sizes?
Answer:
Yes. Mode can be calculated for grouped data with unequal class sizes.

AP State Syllabus SSC 10th Class Maths Solutions 13th Lesson Probability InText Questions

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 13 Probability InText Questions and Answers.

10th Class Maths 13th Lesson Probability InText Questions and Answers

Do This

(Page No. 307)

Outcomes of which of the following experiments are equally likely ?
Question 1.
Getting a digit 1, 2, 3, 4, 5 or 6 when a die is rolled.
Answer:
Equally likely.

Question 2.
Picking a different colour ball from a bag of 5 red balls, 4 blue balls and 1 black ball.
Note: Picking two different colour balls …………..
i.e., picking a red or blue or black ball from a …………
Answer:
Not equally likely.

Question 3.
Winning in a game of carrom.
Answer:
Equally likely.

Question 4.
Units place of a two digit number selected may be 0, 1, 2, 3, 4, 5, 6, 7, 8 or 9.
Answer:
Equally likely.

Question 5.
Picking a different colour ball from a bag of 10 red balls, 10 blue balls and 10 black balls.
Answer:
Equally likely.

Question 6.
a) Raining on a particular day of July.
Answer:
Not equally likely.

b) Are the outcomes of every experiment equally likely?
Answer:
Outcomes of all experiments need not necessarily be equally likely.

c) Give examples of 5 experiments that have equally likely outcomes and five more examples that do not have equally likely outcomes.
Answer:
Equally likely events:

1. Getting an even or odd number when a die is rolled.
2. Getting tail or head when a coin is tossed.
3. Getting an even or odd number when a card is drawn at random from a pack of cards numbered from 1 to 10.
4. Drawing a green or black ball from a bag containing 8 green balls and 8 black balls.
5. Selecting a boy or girl from a class of 20 boys and 20 girls.
6. Drawing a red or black card from a deck of cards.

Events which are not equally likely:

1. Getting a prime or composite number when a die is thrown.
2. Getting an even or odd number when a card is drawn at random from a pack of cards numbered from 1 to 5.
3. Getting a number which is a multiple of 3 or not a multiple of 3 from numbers 1, 2, …… 10.
4. Getting a number less than 5 or greater than 5.
5. Drawing a white ball or green ball from a bag containing 5 green balls and 8 white balls.

Question 7.
Think of 5 situations with equally likely events and find the sample space.    (Page No. 309)
Answer:
a) Tossing a coin: Getting a tail or head when a coin is tossed.
Sample space = {T, H}.
b) Getting an even or odd number when a die is rolled.
Sample space = (1, 2, 3, 4, 5, 6}.
c) Winning a game of shuttle.
Sample space = (win, loss}.
d) Picking a black or blue ball from a bag containing 3 blue balls and 3 blackballs = {blue, black}.
e) Drawing a blue coloured card or black coloured card from a deck of cards = {black, red}.

Question 8.
i) Is getting a head complementary to getting a tail? Give reasons.   (Page No. 311)
Answer:
Number of outcomes favourable to head = 1
Probability of getting a head = \(\frac{1}{2}\) [P(E)]
Number of outcomes not favourable to head = 1
Probability of not getting a head = \(\frac{1}{2}\) [P(\(\overline{\mathrm{E}}\))]
Now P(E) + P(\(\overline{\mathrm{E}}\)) = \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1
∴ Getting a head is complementary to getting a tail.

ii) In case of a die is getting a 1 comple-mentary to events getting 2, 3, 4, 5, 6? Give reasons for your answer.
Answer:
Yes. Complementary events.
∵ Probability of getting 1 = \(\frac{1}{6}\) [P(E)]
Probability of getting 2, 3, 4, 5, 6 = P(E) = P(\(\overline{\mathrm{E}}\)) = \(\frac{5}{6}\)
P(E) + P(\(\overline{\mathrm{E}}\)) = \(\frac{1}{6}\) + \(\frac{5}{6}\) = \(\frac{6}{6}\) = 1

iii) Write of five new pair of events that are complementary.
Answer:

1. When a dice is thrown, getting an even number is complementary to getting an odd number.
2. Drawing a red card from a deck of cards is complementary to getting a black card.
3. Getting an even number is complementary to getting an odd number from numbers 1, 2, ….. 8.
4. Getting a Sunday is complementary to getting any day other than Sunday in a week.
5. Winning a running race is complementary to loosing it.

Try This

Question 1.
A child has a dice whose six faces show the letters A, B, C, D, E and F. The dice is thrown once. What is the probability of getting (i) A? (ii) D?     (Page No. 312)
Answer:
Total number of outcomes (A, B, C, D, E and F) = 6.
i) Number of favourable outcomes to A = 1
Probability of getting A =
P(A) = \(\frac{\text { No.of favourable outcomesto } \mathrm{A}}{\text { No.of all possible outcomes }}\) = \(\frac{1}{6}\)

ii) No. of outcomes favourable to D = 1
Probability of getting D
= \(\frac{\text { No.of outcomes favourble to } \mathrm{D}}{\text { All possible outcomes }}\) = \(\frac{1}{6}\)

Question 2.
Which of the following cannot be the probability of an event?     (Page No. 312)
(a) 2.3
(b) -1.5
(c) 15%
(d) 0.7
Answer:
a) 2.3 – Not possible
b) -1.5 – Not possible
c) 15% – May be the probability
d) 0.7 – May be the probability

Question 3.
You have a single deck of well shuffled cards. Then, what is the probability that the card drawn will be a queen?     (Page No. 313)
Answer:
Number of all possible outcomes = 4 × 13 = 1 × 52 = 52
Number of outcomes favourable to Queen = 4 [♥ Q, ♦ Q, ♠ Q, ♣ Q]
∴ Probability P(E) = \(\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}\)
= \(\frac{4}{52}\) = \(\frac{1}{13}\)

Question 4.
What is the probability that it is a face card?     (Page No. 314)
Answer:
Face cards are J, Q, K.
∴ Number of outcomes favourable to face card = 4 × 3 = 12
No. of all possible outcomes = 52
P(E) = \(\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}\)
= \(\frac{12}{52}\) = \(\frac{3}{13}\)

Question 5.
What is the probability that it is a spade?       (Page No. 314)
Answer:
Number of spade cards = 13
Total number of cards = 52
Probability
= \(\frac{\text { Number of outcomes favourable to spades }}{\text { Number of all outcomes }}\)
= \(\frac{13}{52}\) = \(\frac{1}{4}\)

Question 6.
What is the probability that is the face card of spades?       (Page No. 314)
Answer:
Number of outcomes favourable to face cards of spades = (K, Q, J) = 3
Number of all outcomes = 52
P(E) = \(\frac{3}{52}\)

Question 7.
What is the probability it is not a face card?       (Page No. 314)
Answer:
Probability of a face card = \(\frac{12}{52}\) from (1)
∴ Probability that the card is not a face card

(or)
Number of favourable outcomes = 4 × 10 = 40
Number of all outcomes = 52
∴ Probability
= \(\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}\)
= \(\frac{40}{52}\) = \(\frac{10}{13}\)

Think & Discuss

(Page No. 312)

Question 1.
Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of any game?
Answer:
Probability of getting a head is \(\frac{1}{2}\) and of a tail is \(\frac{1}{2}\) are equal.
Hence tossing a coin is a fair way.

Question 2.
Can \(\frac{7}{2}\) be the probability of an event? Explain.
Answer:
\(\frac{7}{2}\) can’t be the probability of any event.
Since probability of any event should lie between 0 and 1.

Question 3.
Which of the following arguments are correct and which are not correct? Give reasons.
i) If two coins are tossed simultaneously, there are three possible outcomes – two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is \(\frac{1}{3}\).
Answer:
False.
Reason:
All possible outcomes are 4
HH, HT, TH, TT
Thus, probability of two heads = \(\frac{1}{4}\)
Probability of two tails = \(\frac{1}{4}\)
Probability of one each = \(\frac{2}{4}\) = \(\frac{1}{2}\).

ii) If a dice is thrown, there are two possible outcomes – an odd number or an even number. Therefore, the probability of getting an odd number is \(\frac{1}{2}\).
Answer:
True.
Reason:
All possible outcomes = (1, 2, 3, 4, 5, 6) = 6
Outcomes favourable to an odd number (1, 3, 5) = 3
Outcomes favourable to an even number = (2, 4, 6) = 3
∴ Probability (odd number)
= \(\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}\)
= \(\frac{3}{6}\) = \(\frac{1}{2}\).

AP State Syllabus SSC 10th Class Maths Solutions 12th Lesson Applications of Trigonometry InText Questions

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 12 Applications of Trigonometry InText Questions and Answers.

10th Class Maths 12th Lesson Applications of Trigonometry InText Questions and Answers

Do This

(Page No. 297)

Question 1.
Draw diagram for the following situations:
i) A person is flying a kite at an angle of elevation a and the length of thread from his hand to kite is ‘l’.
ii) A person observes two banks of a river at angles of depression θ₁ and θ₂ (θ₁ < θ₂) from the top of a tree of height ‘h’ which is at a side of the river. The width of the river is ‘d’.
Answer:

In the figure
‘D’ is the position of person
CD is the height of the tree
AB is the width of the river
Angles of depression are θ₁ and θ₂.

Think & Discuss

(Page No. 297)

Question 1.
You are observing top of your school building at an angle of elevation a from a point which is at d meter distance from foot of the building. Which trigonometric ratio would you like to consider to find the height of the building?
Answer:

The trigonometric ratio which connects d and h is tan α.

Question 2.
A ladder of length x meter is leaning against a wall making angle θ with the ground. Which trigonometric ratio would you like to consider to find the height of the point on the wall at which the ladder is touching?
Answer:

We use ‘sin θ’ as it is the ratio between the side opp. to θ and hypotenuse.

AP State Syllabus SSC 10th Class Maths Solutions 11th Lesson Trigonometry InText Questions

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 11 Trigonometry InText Questions and Answers.

10th Class Maths 11th Lesson Trigonometry InText Questions and Answers

Do This

Identify “Hypotenuse”, “Opposite side” and “Adjacent side” for the given angles in the given triangles.  (Page No. 271)

Question 1.
For angle R
Answer:

In the △PQR
Opposite side = PQ
Adjacent side = QR
Hypotenuse = PR

Question 2.
i) For angle X
ii) For angle Y        (Page No. 271)
Answer:

In the △XYZ,
i) For angle X
Opposite side = YZ
Adjacent side = XZ
Hypotenuse = XY
ii) For angle Y
Opposite side = XZ
Adjacent side = YZ
Hypotenuse = XY

Question 3.
Find (i) sin C (ii) cos C and (iii) tan C in the given triangle.    (Page No. 274)

Answer:
By Pythagoras theorem
AC² = AB² + BC²
13² = AB² + 5²
AB² = 169 – 25
AB² = √144
⇒ AB = √144 = 12

Question 4.
In a triangle XYZ, ∠Y is right angle, XZ = 17 cm and YZ = 15 cm, then find (i) sin X (ii) cos Z (iii) tan X.    (Page No. 274)
Answer:
Given △XYZ, ∠Y is right angle.

By Pythagoras theorem
XZ² = YZ² + XY²
17² = 15² + XY²
XY² = 17² – 15² = 289 – 225
XY² = 64
XY = √64 = 8

Question 5.
In a triangle PQR with right angle at Q, the value of ∠P is x, PQ = 7 cm and QR = 24 cm, then find sin x and cos x.    (Page No. 274)
Answer:
Given right angled triangle is PQR with right angle at Q. The value of ∠P is x.

By Pythagoras theorem
PR² = PQ² + QR²
PR² = 7² + 24²
PR² = 49 + 576
PR² = 625
PR² = √625 = 25
sin x = \(\frac{QR}{PR}\) = \(\frac{24}{25}\)
cos x = \(\frac{PQ}{PR}\) = \(\frac{7}{25}\)

Try This

Question 1.
Write lengths of “Hypotenuse”, “Opposite side” and “Adjacent side” for the given angles in the given triangles.
1. For angle C
2. For angle A          (Page No. 271)
Answer:

By Pythagoras theorem
AC² = AB² + BC²
(5)² = AB² + 4²
25 = AB² + 16
AB² = 25 – 16
AB² = 9
AB = √9 = 3
For angle C:
Opposite side = AB = 3 cm
Adjacent side = BC = 4 cm
Hypotenuse = AC = 5 cm
For angle A:
Opposite side = BC = 4 cm
Adjacent side = AB = 3 cm
Hypotenuse = AC = 5 cm

Question 2.
In a right angle triangle ABC, right angle is at C. BC + CA = 23 cm and BC – CA = 7 cm, then find sin A and tan B.    (Page No. 274)
Answer:
In a right angle triangle ABC, right angle is at C.

BC = \(\frac{30}{2}\) = 15
BC = 15
Substituting BC = 15 in equation (1)
BC + CA = 23
CA = 23 – BC = 23 – 15
CA = 8
By Pythagoras theorem
AB² = AC² + BC²
= 8² + 15²
= 64 + 225 = 289
= √289 = 17
sin A = \(\frac{BC}{AB}\) = \(\frac{15}{17}\)
cos B = \(\frac{AC}{BC}\) = \(\frac{8}{15}\)

Question 3.
What will be the ratios of sides for sec A and cot A?    (Page No. 275)
Answer:
sec A = \(\frac{\text { Hypotenuse }}{\text { Adjacent side of the angle } \mathrm{A}}\)
cot A = \(\frac{\text { Adjacent side of the angle } A}{\text { Opposite side of the angle } A}\)

Think & Discuss

Question 1.
Discuss with your friends
i) sin x = \(\frac{4}{3}\) does exist for some value of angle x?
ii) The value of sin A and cos A is always less than 1. Why?
iii) tan A is product of tan and A.        (Page No. 274)
Answer:
i) The value of sin 0 always lies between 0 and 1. Here sin x = \(\frac{4}{3}\) which is greater than 1. So, it does not exist.
ii)

We observe in above sin A, cos A, hypotenuse is in denominator which is greater than other two sides
∴ sin A = \(\frac{\text { Opposite side }}{\text { Hypotenuse }}\) = \(\frac{Nr}{Dr}\)
here denominator is more than numerator. Hence its value will be less than 1.
cos A = \(\frac{\text { Adjacent side }}{\text { Hypotenuse }}\)
here also adjacent side is always less than hypotenuse. Hence its value is also less than or equal to 1.

iii) The symbol tan A is used as an abbreviation for “the tan of the angle A”.
tan A is not the product of “tan” and A. “tan” separated from A’ has no meaning.

Question 2.
Is \(\frac{\sin \mathrm{A}}{\cos \mathrm{A}}\) equal to tan A?      (Page No. 275)
Answer:
Yes, \(\frac{\sin \mathrm{A}}{\cos \mathrm{A}}\) = tan A
Proof:

Question 3.
Is \(\frac{\cos \mathrm{A}}{\sin \mathrm{A}}\) equal to cot A?     (Page No. 275)
Answer:
Yes, \(\frac{\cos \mathrm{A}}{\sin \mathrm{A}}\) = tan A
Proof:

cot A = \(\frac{\text { Adjacent side }}{\text { Opposite side }}\)

Do this

Question 1.
Find cosec 60°, sec 60° and cot 60°.       (Page No. 279)
Answer:

Consider an equilateral triangle ABC. Since each angle is 60° in an equilateral triangle, we have ∠A = ∠B = ∠C = 60° and the sides of equilateral triangle is AB = BC = CA = 2a units.
Draw the perpendicular line AD from vertex A to BC as shown in the given figure.
Perpendicular AD acts as “angle bisector of angle A” and “bisector of the side BC” in the equilateral triangle ABC.
Therefore, ∠BAD = ∠CAD = 30°. Since point D divides the side BC into equal halves.
BD = \(\frac{1}{2}\) BC = \(\frac{2a}{2}\) = a units.
Consider right angle triangle ABD in the above given figure.
We have AB = 2a and BD = a
Then AD² = AB² – BD²
(By Pythagoras theorem)
= (2a)² – (a)² = 3a²
Therefore, AD = a√3
From definitions of trigonometric ratios,

Try this

Question 1.
Find sin 30°, cos 30°, tan 30°, cosec 30°, sec 30° and cot 30° by using the ratio concepts.     (Page No. 279)
Answer:

Consider an equilateral triangle ABC. Since each angle is 60° in an equilateral triangle, we have ∠A = ∠B = ∠C = 60° and the sides of equilateral triangle is AB = BC = CA = 2a units.
Draw the perpendicular line AD from vertex A to BC as shown in the given figure.
Perpendicular AD acts as “angle bisector of angle A” and “bisector of the side BC” in the equilateral triangle ABC.
Therefore, ∠BAD = ∠CAD = 30°. Since point D divides the side BC into equal halves.
BD = \(\frac{1}{2}\) BC = \(\frac{2a}{2}\) = a units.
Consider right angle triangle ABD in the above given figure.
We have AB = 2a and BD = a
Then AD² = AB² – BD²
(By Pythagoras theorem)
= (2a)² – (a)² = 3a²
Therefore, AD = \(\sqrt{3 a^{2}}\) = √3a
BD = a, AD = √3a and hypotenuse = AB = 2a and ∠DAB = 30°.
sin 30° = \(\frac{BD}{AB}\) = \(\frac{a}{2a}\) = \(\frac{1}{2}\)

Question 2.
Find the values for tan 90°, cosec 90°, sec 90° and cot 90°.     (Page No. 281)
Answer:
From the adjacent figure, the trigonometric ratios of ∠A, gets larger and larger in △ABC till it becomes 90°.

As ∠A get larger and larger, ∠C gets smaller and smaller. Therefore, the length of the side AB goes on decreasing. The point A gets closer to point B. Finally when ∠A is very close to 90°, ∠C becomes very close to 0° and the side AC almost coincides with side BC.

∴ AB = 0 and AC = BC = r
From trigonometric ratios
sin A = \(\frac{BC}{AC}\)
sin A = \(\frac{AB}{AC}\)
If A = 90°, then AB = 0 and AC = BC = r,

Think & Discuss

Question 1.
Discuss with your friend about the following conditions:
What can you say about cosec 0° = \(\frac{1}{\sin 0^{\circ}}\)? Is it defined? Why?    (Page No. 280)
Answer:
sin 0° = 0
cosec 0° = \(\frac{1}{\sin 0^{\circ}}\) = \(\frac{1}{0}\) = not defined.
It is not defined.
Reason:
Division by ‘0’ is not allowed, hence \(\frac{1}{0}\) is indeterminate.

Question 2.
What can you say about cot 0° = \(\frac{1}{\tan 0^{\circ}}\). Is it defined? Why?    (Page No. 281)
Is it defined? Why?
Answer:
tan 0° = 0
cot 0° = \(\frac{1}{\tan 0^{\circ}}\) = \(\frac{1}{0}\) = undefined.
Reason:
Division by ‘0’ is not allowed, hence \(\frac{1}{0}\) is indeterminate.

Question 3.
sec 0° = 1. Why?      (Page No. 281)
Answer:
sec 0° = \(\frac{1}{\cos 0^{\circ}}\) [∵ cos 0° = 1]
= \(\frac{1}{1}\) = 1

Question 4.
What can you say about the values of sin A and cos A, as the value of angle A increases from 0° to 90°? (Observe the above table)
i) If A ≥ B, then sin A ≥ sin B. Is it true?
ii) If A ≥ B, then cos A ≥ cos B. Is it true? Discuss.      (Page No. 282)
Answer:
i) Given statement
“If A ≥ B, then sin A ≥ sin B”
Yes, this statement is true.
Because, it is clear from the table below that the sin A increases as A increases.

ii) Given statement
“If A ≥ B, then cos A ≥ cos B”
No, this statement is not true.
Because, it is clear from the table below that cos A decreases as A increases.

Think & Discuss

Question 1.
For which value of acute angle
(i) \(\frac{\cos \theta}{1-\sin \theta}\) + \(\frac{\cos \theta}{1+\sin \theta}\) = 4 is true?
For which value of 0° ≤ θ ≤ 90°, above equation is not defined?    (Page No. 285)
Answer:


⇒ cos θ = cos 60° [from trigonometric ratios table]
⇒ θ = 60°
Given statement is true for the acute angle i.e., θ = 60°.

Question 2.
Check and discuss the above relations in the case of angles between 0° and 90°, whether they hold for these angles or not? So,          (Page No. 286)
i) sin (90° – A) = cos A
ii) cos (90° – A) = sin A
iii) tan (90° – A) = cot A and
iv) cot (90° – A) = tan A
v) sec (90° – A) = cosc A
vi) cosec (90° – A) = sec A
Answer:
Let A = 30°
i) sin (90° – A) = cos A
⇒ sin (90° – 30°) = cos 30°
⇒ sin 60° = cos 30° = \(\frac{\sqrt{3}}{2}\)

ii) cos (90° – A) = sin A
⇒ cos (90° – 30°) = sin 30°
⇒ cos 60° = sin 30° = \(\frac{1}{2}\)

iii) tan (90° – A) = cot A
⇒ tan (90° – 30°) = cot 30°
⇒ tan 60° = cot 30° = √3

iv) cot (90° – A) = tan A
⇒ cot (90° – 30°) = tan 30°
⇒ cot 60° = tan 30° = \(\frac{1}{\sqrt{3}}\)

v) sec (90° – A) = cosec A
⇒ sec (90° – 30°) = cosec 30°
⇒ sec 60° = cosec 30° = 2

vi) cosec (90° – A) = sec A
⇒ cosec (90° – 30°) = sec 30°
⇒ cosec 60° = sec 30° = \(\frac{2}{\sqrt{3}}\)

So, the above relations hold for all the angles between 0° and 90°.

Do this

(Page No. 290)

Question 1.
i) If sin A = \(\frac{15}{17}\) then find cos A.
Answer:
Given sin A = \(\frac{15}{17}\)
cos A = \(\sqrt{1-\sin ^{2} A}\) [From Identity -I]

ii) If tan x = \(\frac{5}{12}\), then find sec x. (AS j)
Answer:
Given tan x = \(\frac{5}{12}\)
We know that sec² x – tan² x = 1
sec² x = 1 + tan² x

iii) If cosec θ = \(\frac{25}{7}\), then find cot θ.
Answer:
Given cosec θ = \(\frac{25}{7}\)
We know that cosec² θ – cot² θ = 1
cot² θ = cosec² θ – 1

Try This

(Page No. 290)

Question 1.
Evaluate the following and justify your answer.

i) \(\frac{\sin ^{2} 15^{\circ}+\sin ^{2} 75^{\circ}}{\cos ^{2} 36^{\circ}+\cos ^{2} 54^{\circ}}\)
Answer:
Given \(\frac{\sin ^{2} 15^{\circ}+\sin ^{2} 75^{\circ}}{\cos ^{2} 36^{\circ}+\cos ^{2} 54^{\circ}}\)

[∵ sin (90° – θ) = cos θ; cos (90° – θ) = sin θ]
= \(\frac{1}{1}\) = 1 [By sin² θ + cos² θ = 1]

ii) sin 5° cos 85° + cos 5° sin 85°
Answer:
Given sin 5° cos 85° + cos 5° sin 85°
= sin 5° . cos (90° – 5°) + cos 5° . sin (90° – 5°)
= sin 5° . sin 5° + cos 5° . cos 5°
[∵ sin (90° – θ) = cos θ; cos (90° – θ) = sin θ]
= sin² 5° + cos² 5°
= 1 [∵ sin² θ + cos² θ = 1]

iii) sec 16° cosec 74° – cot 74° tan 16°.
Answer:
Given sec 16° cosec 74° – cot 74° tan 16°
= sec 16° . cosec (90° – 16°) – cot (90° – 16°) . tan 16°
= sec 16°. sec 16° – tan 16° . tan 16° [∵ cosec (90° – θ) = sec θ; cot (90° – θ) — tan θ]
= sec² 16° – tan² 16°
= 1 [∵ sec² θ – tan² θ = 1]

Think & Discuss

(Page No. 290)

Question 1.
Are these identities true for 0° ≤ A ≤ 90°? If not for which values of A they are true?
i) sec² A – tan² A = 1
Answer:
Given identity is sec² A – tan² A = 1
Let A = 0°
L.H.S. = sec² 0° – tan² 0°
= 1 – 0 = 1 = R.H.S.
Let A = 90°
tan A and sec A are not defined.
So it is true.
∴ For all given values of ‘A’ such that 0° ≤ A ≤ 90° this trigonometric identity is true.

ii) cosec² A – cot² A = 1
Answer:
Given identity is cosec² A – cot² A = 1
Let A = 0°
cosec A and cot A are not defined for A = 0°.
Therefore identity is true for A = 0°.
Let A = 90°
cosec A = cosec 90° = 1
cot A = cot 90° = 0
L.H.S. = l² – 0² = 1 – 0 = 1 = R.H.S.
∴ This identity is true for all values of A, such that 0° ≤ A ≤ 90°.

AP State Syllabus SSC 10th Class Maths Solutions 10th Lesson Mensuration InText Questions

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 10 Mensuration InText Questions and Answers.

10th Class Maths 10th Lesson Mensuration InText Questions and Answers

Try this

(Page No. 245)

Question 1.
Consider the following situations. In each find out whether you need volume or area and why?
i) Quantity of water inside a bottle.
ii) Canvas needed for making a tent.
iii) Number of bags inside the lorry.
iv) Gas filled in a cylinder.
v) Number of match sticks that can be put in the match box.
Answer:
i) Volume: 3-d shape
ii) Area: L.S.A. / T.S.A.
iii) Volume: 3-d shape
iv) Volume: 3-d shape
v) Volume: 3-d shape

Question 2.
Write 5 more such examples and ask your friends to choose what they need?
Answer:
Student’s Activity.

Question 3.
Break the pictures in the previous figure into solids of known shapes.    (Page No. 246)
Answer:

Question 4.
Think of 5 more objects around you that can be seen as a combination of shapes. Name the shapes that combined to make them.    (Page No. 246)
Answer:
Student’s Activity.

Try this

Question 1.
Use known solid shapes and make as many objects (by combining more than two) as possible that you come across in your daily life.
[Hint: Use clay, or balls, pipes, paper cones, boxes like cube, cuboid etc]      (Page No. 252)
Answer:
Student’s Activity.

Think & Discuss

Question 1.
A sphere is inscribed in a cylinder. Is the surface of the sphere equal to the curved surface of the cylinder? If yes, explain how.      (Page No. 252)

Answer:
Yes, the surface area of the sphere is equal to the curved surface area of the cylinder.
Let the radius of the “cylinder be ‘r’ and its height ‘h’.
Then radius of sphere = r.
Then curved surface area of cylinder = 2πrh = 2πr (r + r)
[∵ height = diameter of the sphere = diameter of the cylinder = 2r]
= 2πr (2r) = 4πr²
And surface area of the sphere = 4πr²
∴ C.S.A. of cylinder = Surface area of sphere.

Try This

(Page No. 257)

Question 1.
If the diameter of the cross – section of a wire is decreased by 5%, by what percentage should the length be increased so that the volume remains the same ?
Answer:
Let the radius of wire = r
(if diameter = 2r)
and length of wire = h = l (say)
then volume of wire = πr²l₁ = v₁ ….(1)
Now the diameter after decreasing 5%

then radius of wire after decreasing
= \(\frac{190r}{100}\) × \(\frac{1}{2}\) = \(\frac{95r}{100}\)
and let the length of wire = (l₂)

So it length is increased by 24.22%. Its volume remains same.

Question 2.
Surface area of a sphere and cube are equal. Then find the ratio of their volumes.
Answer:
Let a cube with side ‘a’.
Then its surface area = 6a²
By problem, surface area of the sphere = 4πr²
Surface area of the cube = 6a²

Do This

(Page No. 263)

Question 1.
A copper rod of diameter 1 cm. and length 8 cm; is drawn into a wire of length 18 m of uniform thickness. Find the thickness of the wire.
Answer:
Volume of the copper rod (cylinder)
= πr²h
= \(\frac{22}{7}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) × 8
= \(\frac{44}{7}\) m²
If V is the radius of the wire, then its volume = πr²h
∴ The volume of rod is equal to the volume of the wire. We have

∴ Thickness = d = 2 × 0.03 ≃ 0.06 cm.

Question 2.
Pravali house has a water tank in the shape of a cylinder on the roof. This is filled by pumping water from a sump (an underground tank) which is in the shape of a cuboid. The sump has dimensions 1.57 m × 1.44 m × 9.5 cm. The water tank has radius 60 cm. and height 95 cm. Find the height of the water left in the sump after the water tank has been completely filled with water from the sump which had been full of water. Compare the capacity of the tank with that of the sump, (π = 3.14)
Answer:
Volume of the water in the sump = 1.57 × 1.44 × 0.95 [∵ V = lbh]
= 2.14776 m³ = 2147760 cm³
Volume of the tank on the roof = πr²h
= 3.14 × 60 × 60 × 95
= 1073880 cm³
∴ Volume of the water left in the sump after filling the tank
= 2147760 – 1073880 = 1073880 cm³
Let the height of the water in the tank be h.
∴ 157 × l44h = 1073880
h = \(\frac{1073880}{157 \times 144}\) = 47.5 cm.
∴ Ratio of the volume of the sump and tank = 2147760 : 1073880 = 2 : 1
∴ Sump can hold two times the water that can be hold in the tank.

Think & Discuss

(Page No. 262)

Question 1.
Which barrel shown in the below figure can hold more water? Discuss with your friends.

Answer:
r₁ = \(\frac{1}{2}\) = 0.5 cm; h₁ = 4 cm
Volume of the 1st barrel = πr²h
= \(\frac{22}{7}\) × 0.5 × 0.5 × 4 = 3.142 cm³
r₂ = \(\frac{4}{2}\) = 2 cm
∴ h = 1 cm
Volume of the 2nd barrel
V = πr²h = \(\frac{22}{7}\) × 2 × 2 × 1 = 12.57 cm³
Hence the volume of the 2nd barrel is more than the first barrel.

AP State Syllabus SSC 10th Class Maths Solutions 9th Lesson Tangents and Secants to a Circle InText Questions

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 9 Tangents and Secants to a Circle InText Questions and Answers.

10th Class Maths 9th Lesson Tangents and Secants to a Circle InText Questions and Answers

Do this

Question 1.
Draw a circle with any radius. Draw four tangents at different points. How many tangents can you draw to this circle?    (Page No. 226)
Answer:

Let ‘O’ be the centre of the circle with radius OA.
l, m, n, p and q be the tangents to the circles at A, B, C, D and E. We can draw a tangent at each point on the circle, i.e., infinitely many tangents can be drawn to a circle.

Question 2.
How many tangents you can draw to circle from a point away from it?    (Page No. 226)
Answer:

We can draw only two tangents from an exterior point.

Question 3.
In the below figure which are tangents to the given circles?      (Page No. 226)

Answer:
P and M are the tangents to the given circles.

Question 4.
Draw a circle and a secant PQ of the circle on a paper as shown below. Draw various lines parallel to the secant on both sides of it. What happens to the length of chord coming closer and closer to the centre of the circle?      (Page No. 227)

Answer:

The length of the chord increases as it comes closer to the centre of the circle.

Question 5.
What is the longest chord?        (Page No. 227)
Answer:
Diameter is the longest of all chords.

Question 6.
How many tangents can you draw to a circle, which are parallel to each other?    (Page No. 227)
Answer:
Only one tangent can be drawn parallel to a given tangent.
To a circle, we can draw infinitely many pairs of parallel tangents.

Try this

(Page No. 228)

How can you prove the converse of the above theorem.
Question 1.
“If a line in the plane of a circle is perpendicular to the radius at its end point on the circle, then the line is tangent to the circle”.
Answer:
Given: Circle with centre ‘O’, a point A on the circle and the line AT perpendicular to OA.
R.T.P: AT is a tangent to the circle at A.
Construction:

Suppose AT is not a tangent then AT produced either way if necessary, will meet the circle again. Let it do so at P, join OP.
Proof: Since OA = OP (radii)
∴ ∠OAP = ∠OPA But ∠OPA = 90°
∴ Two angles of a triangle are right angles which is impossible.
∴ Our supposition is false.
∴ Hence AT is a tangent.

We can find some more results using the above theorem.
i) Since there can be only one perpendicular OP at the point P, it follows that one and only one tangent can be drawn to a circle at a given point on the circumference.
ii) Since there can be only one perpendicular to XY at the point P, it follows that the perpendicular to a tangent at its point of contact passes through the centre.

Question 2.
How can you draw the tangent to a circle at a given point when the centre of the circle is not known?    (Page No. 229)
Answer:

Steps of Construction:

1. Take a point P and draw a chord PR through P.
2. Construct ∠PRQ and measure it.
3. Construct ∠QPX at P equal to ∠PRQ.
4. Extend PX on other side. XY is the required tangent at P.

Note: Angle between a tangent and chord is equal to angle in the alternate segment.

Hint: Draw equal angles ∠QPX and ∠PRQ. Explain the construction.
Answer:

Steps of construction:

1. Draw any two chords AB and AC in the given circle.
2. Draw the perpendicular bisectors to AB and AC, they meet at the centre of the circle.
3. Tet O be the centre, join OP.
4. Draw a perpendicular to OP at P and extend it on either sides which forms a tangent to the circle at ‘P’.

Try this

Question 1.
Use Pythagoras theorem and write proof of above theorem “the lengths of tangents drawn from an external point to a circle are equal.”      (Page No. 231)
Answer:
Given: Two tangents PA and PB to a circle with centre O, from an exterior point P.
R.T.P: PA = PB

Proof: In △OAP; ∠OAP = 90°
∴ AP² = OP² – OA²
[∵ Square of the hypotenuse is equal to the sum of squares on the other two sides – Pythagoras theorem]
[∵ OA = OB, radii of the same circle]
= BP² [∵ In AOBP; OB² + BP² = OP²
⇒ BP² – OP² – OB²]
⇒ AP² – BP²
⇒ PA – PB Hence proved.

Question 2.
Draw a pair of radii OA and OB such that ∠BOA = 120°. Draw the bisector of ∠BOA and draw lines perpendiculars to OA and OB at A and B. These lines meet on the bisector of ∠BOA at a point which is the external point and the perpendicular lines are the required tangents. Construct and justify.    (Page No. 235)
Answer:

Justification:
OA ⊥ PA
OB ⊥ PB
Also in △OAP, △OBP
OA = OB
∠OAP = ∠OBP
OP – OP
∴ △OAP ≅ △OBP
∴ PA = PB. [Q.E.D.]

Do this

Question 1.
Shankar made the following pictures also with washbasin.

What shapes can they be broken into that we can find area easily?    (Page No. 237)
Answer:

Question 2.
Make some more pictures and think of the shapes they can be divided into different parts.      (Page No. 237)
Answer:

Question 3.
Find the area of sector, whose radius is 7 cm. with the given angles.    (Page No. 239)
i) 60° ii) 30° iii) 72° iv) 90° v) 120°
Answer:

Question 4.
The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 10 minutes.    (Page No. 239)
Answer:
Angle made by minute hand in 1 m = \(\frac{360^{\circ}}{60}\) = 6°
Angle made by minute hand in 10m = 10 × 6 = 60°
The area swept by minute hand is in the shape of a sector with radius r = 14 cm and angle x = 60°

Area swept by the minute hand in 10 minutes = 102.66 cm².

Try this

Question 1.
How can you find the area of major segment using area of minor segment?    (Page No. 239)
Answer:

Area of the major segment = Area of the circle – Area of the minor segment.

AP State Syllabus SSC 10th Class Maths Solutions 8th Lesson Similar Triangles InText Questions

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 8 Similar Triangles InText Questions and Answers.

10th Class Maths 8th Lesson Similar Triangles InText Questions and Answers

Do these

(Page No. 194)

Question 1.
Fill in the blanks with similar/not similar.
i) All squares are ………. (similar)
ii) All equilateral triangles are ………. (similar)
iii) All isosceles triangles are ………. (similar)
iv) Two polygons with same number of sides are ………, if their corresponding angles are equal and corresponding sides are equal. (similar)
v) Reduced and enlarged photographs of an object are ………. (similar)
vi) Rhombus and squares are ……… to each other. (not similar)

Question 2.
Write True / False for the following statements.
i) Any two similar figures are congruent.
Answer:
False
ii) Any two congruent figures are similar.
Answer:
True
iii) Two polygons are similar if their corresponding angles are equal.
Answer:
False

Question 3.
Give two different examples of pair of
i) Similar figures
ii) Non-similar figures
Answer:
i) Similar figures:
a) Any two circles
b) Any two squares
c) Any two equilateral triangles
ii) Non-similar figures:
a) A square and a rhombus
b) A square and a rectangle

Question 4.
What value(s) of x will make DE || AB, in the given figure?  (Page No. 200)
AD = 8x + 9, CD = x + 3,
BE = 3x + 4, CE = x.

Answer:
Given : In AABC, DE // AB AD = 8x + 9, CD = x + 3,
BE = 3x + 4, CE = x
By Basic proportionality theorem,
If DE // AB then we should have
\(\frac{\mathrm{CD}}{\mathrm{DA}}\) = \(\frac{\mathrm{CE}}{\mathrm{EB}}\)
\(\frac{x+3}{8x+9}\) = \(\frac{x}{3x+4}\)
⇒ (x + 3) (3x + 4) = x (8x + 9)
⇒ x (3x + 4) + 3 (3x + 4) – 8x² + 9x
⇒ 3x² + 4x + 9x + 12 = 8x² + 9x
⇒ 8x² + 9x – 3x² – 4x – 9x -12 = 0
⇒ 5x² – 4x – 12 = 0
⇒ 5x² – 10x + 6x – 12 = 0
⇒ 5x (x – 2) + 6 (x – 2) = 0
⇒ (5x + 6) (x – 2) = 0
⇒ 5x + 6 = 0 or x – 2 = 0
⇒ x = \(\frac{-6}{5}\) or x = 2;
x cannot be negative.
∴ The value x = 2 will make DE // AB.

Question 5.
In △ABC, DE || BC. AD = x, DB = x – 2, AE = x + 2 and EC = x – 1. Find the value of x.    (Page No. 200)

Answer:
Given: In △ABC, DE // BC
∴ By Basic proportionality theorem, we have
\(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{\mathrm{AE}}{\mathrm{EC}}\)
\(\frac{\mathrm{x}}{\mathrm{x}-2}\) = \(\frac{x+2}{x-1}\)
⇒ x (x – 1) = (x + 2) (x – 2)
⇒ x² – x = x² – 4
⇒ -x = -4
∴ x = 4

Try This

Question 1.
E and F are points on the sides PQ and PR respectively of △PQR. For each of the following, state whether EF || QR or not?      (Page No. 197)
i) PE = 3.9 cm, EQ = 3 cm,
PF = 3.6 cm and FR = 2.4 cm.
Answer:

Here
\(\frac{\mathrm{PE}}{\mathrm{EQ}}\) = \(\frac{3.9}{3}\) = \(\frac{1.3}{1}\)
\(\frac{\mathrm{PF}}{\mathrm{FR}}\) = \(\frac{3.6}{2.4}\) = \(\frac{3}{2}\)
\(\frac{\mathrm{PE}}{\mathrm{EQ}}\) ≠ \(\frac{\mathrm{PF}}{\mathrm{FR}}\)
Hence, EF is not parallel to QR.

ii) PE = 4 cm, QE = 4.5 cm,
Answer:
Here
\(\frac{\mathrm{PE}}{\mathrm{EQ}}\) = \(\frac{4}{4.5}\) = \(\frac{8}{9}\)
\(\frac{\mathrm{PF}}{\mathrm{RF}}\) = \(\frac{8}{9}\)
\(\frac{\mathrm{PE}}{\mathrm{EQ}}\) = \(\frac{\mathrm{PF}}{\mathrm{RF}}\)
∴ EF // QR
Hence, EF is parallel to QR.

iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 1.8 cm and PF = 3.6 cm.
Answer:

Given: PQ = 1.28 cm, PE = 1.8 cm
⇒ EQ = PE – PQ = 1.8 – 1.28
⇒ EQ = 0.52 cm
Also, PR = 2.56 cm, PE = 3.6 cm, FR = PF – PR = 3.6 cm – 2.56 cm
FR = 1.04 cm

∴ EF // QR (By converse of Basic proportionality theorem)
Hence, EF is parallel to QR.

Question 2.
In the following figures DE || BC.    (Page No. 198)
i) Find EC.

Answer:
\(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{AE}{EC}\)
⇒ \(\frac{1.5}{3}\) = \(\frac{1}{EC}\)
∴ EC = \(\frac{3}{1.5}\) = 2 cm

ii) Find AD.

Answer:

Think & Discuss

Question 1.
Can you give some more examples from your daily life where scale factor is used?    (Page No. 192)
Answer:
Scale factor is used in drawing maps, designing machines and in sculpture, etc.

Question 2.
Can you say that a square and a rhombus are similar? Discuss with your friends.Write why the conditions are not sufficient.       (Page No. 193)
Answer:
A square □ ABCD and a rhombus ▱ PQRS are not similar.
Though the ratio of their corresponding sides are equal, the corresponding angles are not equal.

but ∠A ≠ ∠P; ∠B ≠ ∠Q
∠C ≠ ∠R; ∠D ≠ ∠S

Try This

(Page No. 207)

Question 1.
Are the triangles similar ? If so, name the criterion of similarity. Write the similarity relation in symbolic form.
i)

!! ∠G = ∠I alt. int. angles for the ∠F = ∠K parallel lines GF // KI
Answer:
∠FHG = ∠IHK (Vertically opp. angle)
∴ ∠GHF and ∠IKH are similar by AAA similarity rule.
△GHF ~ △IKH.

ii)

Answer:
\(\frac{PQ}{LM}\) = \(\frac{6}{3}\) = 2;
\(\frac{QR}{MN}\) = \(\frac{10}{4}\) = 2.5;
\(\frac{PQ}{LM}\) ≠ \(\frac{QR}{MN}\)
△PQR and △LMN are not similar.
△PQR ~ △LMN

iii)

Answer:
∠A = ∠A (Common)
\(\frac{AB}{AX}\) = \(\frac{5}{3}\); \(\frac{AC}{AY}\) = \(\frac{5}{3}\)
∴ △ABC and △AXY are similar by SAS similarity condition.
△ABC ~ △AXY.

iv)

Answer:

∴ △ABC and △APJ are not similar.

v)

Answer:
∠A = ∠B = 90°
∠AOQ = ∠POB (∵ Vertically opposite angles)
∠Q = ∠P (alternate interior angles)
∴ △AOQ and △BOP are similar by AAA criterion.
△AOQ ~ △BOP.

vi)

Answer:
△ABC and △QPR are similar by AAA similarity condition.
△ABC ~ △QPR.

vii)

Answer:
∠A = ∠P
\(\frac{AB}{PQ}\) = \(\frac{2}{5}\); \(\frac{AC}{PR}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
∴ \(\frac{AB}{PQ}\) ≠ \(\frac{AC}{PR}\)
Hence not similar.

viii)

Answer:
\(\frac{AB}{PQ}\) = \(\frac{6}{2.5}\); \(\frac{AC}{PR}\) = \(\frac{10}{5}\)
∴ △ABC and △PQR are not similar.

Question 2.
Explain why the triangles are similar and then find the value of x.
i)

Answer:
Given: In △PQR and △LTS
∠Q = ∠T; ∠R = ∠S = 90°
∴ ∠P = ∠T
(by angle sum property of triangles)
Hence, △PQR ~ △LTS [∵ AAA]

ii)

Answer:
Given: In △ABC and △PQC
∠B = ∠Q
[∵ ∠PQC = 180°- 110° = 70° – linear pair of angles]
∠C = ∠C [∵ Common]
∠A = ∠P [∵ Angle Sum property of triangles]
△ABC ~ △PQC by AAA similarity condition.
Then the ratio of their corresponding sides are equal.

iii)

Answer:
Given: In △ABC and △ECD
∠A = ∠E
∠ACB = ∠ECD [∵ Vertically opposite]
∴ ∠B = ∠D [∵ angle sum property]
∴ △ABC ~ △EDC

iv)

Answer:
Given: In △RAB and △RST
∠R = ∠R (common)
∠A = ∠S and ∠B = ∠T
[∵ Pair of corresponding angles for AB // ST]
∴ △RAB ~ △RST [∵ AAA similarity]

v)

Answer:
Given: In △PQR and △PMN
∠P = ∠P [∵ Common]
∠Q = ∠M [∵ Pair of corresponding angles for MN // QR]
∠R = ∠N
△PQR ~ △PMN [∵ AAA similarity]

[From the figure, PR = 4 + x]
⇒ 3 × 4 = 4 + x
⇒ x = 12 – 4 = 8

vi)

Answer:
Given: In △XYZ and △XBA,
∠X = ∠X [∵ Common]
∠B = ∠Y [∵ Pair of corresponding ∠A = ∠Z angles for AB // ZY]
∴ △XYZ ~ △XBA [∵ AAA similarity]

[From the figure, XZ = 7.5 + x]
\(\frac{3}{2}\) = \(\frac{7.5+\mathrm{x}}{\mathrm{x}}\)
3x = 15 + 2x;
3x – 2x = 15
x = 15

vii)

Answer:
Given: With the given conditions, we can’t find the value of x.
Note: If it is given that ∠A = ∠E then
we can say that △ABC ~ △EDC by AAA rule

viii)

Answer:
In △ABC and △BEC
∠ABC = ∠CEB (given)
∠C = ∠C (Common angle)
∴ △ABC ~ △BEC
(A.A. Criterion similarity)

Think & Discuss

(Page No. 203)

Question 1.
Discuss with your friends that in what way similarity of triangles is dif¬ferent from similarity of other polygons?
Answer:
In two triangles if the corresponding angles are equal then they are similar, whereas in two polygons if the corre-sponding angles are equal, they may not be similar, i.e., In triangles,
(Pairs of corresponding angles are equal) ⇔ (Ratio of corresponding sides are equal).
But this is not so with respect to polygons.

Do This

Question 1.
In △ACB, ∠C = 90° and CD ⊥ AB. Prove that \(\frac{\mathrm{BC}^{2}}{\mathrm{AC}^{2}}\) = \(\frac{BD}{AD}\).    (Page No. 218)

Answer:
Proof: △ADC and △CDB are similar.

[Ratio of areas of similar triangles is equal to the ratio of squares of their corresponding sides.]
From (1) and (2),
\(\frac{BD}{AD}\) = \(\frac{\mathrm{BC}^{2}}{\mathrm{AC}^{2}}\) (Q.E.D.)

Question 2.
A ladder 15 m long reaches a window which is 9 m above the ground on one side of a street. Keeping its foot at the same point, the ladder is turned to other side of the street to reach a window 12 m high. Find the width of the street.    (Page No. 218)
Answer:

Let A and D be the windows on the either sides of the street.
From Pythagoras theorem,
AC² = AB² + BC²
15² = 9² + BC²
BC² = 225 – 81
BC = √144 = 12 ….. (1)
Also, CD² = DE² + CE²
15² = 12² + CE²
CE² = 225 – 144
CE = √81 = 9
∴ BE = BC + CE = 12 + 9 = 21
Width of the street = 21 m.

Question 3.
In the given figure if AD ⊥ BC, prove that AB² + CD² = BD² + AC².    (Page No. 219)

Answer:
Given: In △ABC, AD ⊥ BC.
R.T.P: AB² + CD² = BD² + AC²
Proof: △ABD is a right angled triangle
AB² – BD² = AD² ……. (1)
△ACD is a right angle triangle
AC² – CD² = AD²
From (1) and (2)
AB² – BD² = AC² – CD²
AB² + CD² = BD² + AC²

Think & Discuss

Question 1.
For a right angled triangle with integer sides atleast one of its measurements must be an even number. Why? Discuss this with your friends and teachers.    (Page No. 215)
Answer:

Let l, m, n are integer sides of a right
angled triangle.
then l² – m² + n²
⇒ n = l² – m² = (l + m) (l – m)
Now
Case I: Both l, m are even the (l + m) is even then (l + m) (l – m) is also even. So ‘n’ is even. Here all are even.
Case II: Both l, m are odd then (l + m) and (l – m) become even. Then the product of even numbers is even so ‘n’ is even.
Here only ‘n’ is even.
Case III: If we consider l is even, m is’ odd then ‘n’ will be odd. So here T is even. We observe in all above three cases at least one of l, m, n is even; Hence proved.