Measures for Population Management

The population of any place or a country is not constant. There is always change in population. This note has information about the various measures of management of population.

Summary

The population of any place or a country is not constant. There is always change in population. This note has information about the various measures of management of population.

Things to Remember

  • The population of any place or a country is not constant.
  • There is an interrelationship between population growth and available means and resources.
  • The population growth should be balanced according to the available means and resources.
  • The essential factors such as food, education, clothes, shelter, health services, employment, etc. are to be provided to maintain the quality of life.

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Subjective Questions

Q1:

In the given right angled triangle, identify h, p and b.

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Type: Short Difficulty: Easy

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Answer: <p>Solution:</p> <p>Here,</p> <p>\(\angle\)ABC = \(\theta\), angle of reference.</p> <p>\(\angle\)ABC = 90&deg;</p> <p>Hence,</p> <p>AC = h, AB = p and BC = b</p>

Q2:

In the given right angled triangle, identify h, p and b.

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Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>Here,</p> <p>\(\angle\)QPR = &alpha;, angle of reference.</p> <p>\(\angle\)PQR = 90&deg;</p> <p>Hence,</p> <p>RP = h, RQ = p and PQ = b</p>

Q3:

In the given right angle triangled, identify h, p and b.

z


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>Here,</p> <p>\(\angle\)EFG = &beta;, angle of reference.</p> <p>\(\angle\)GEF 90&deg;</p> <p>Hence,</p> <p>FG = h, EG = p and EF = b.</p>

Q4:

In the given right angled triangle, find the value of unknown angles.

s


Type: Short Difficulty: Easy

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Answer: <p>Solution:</p> <p>Here,</p> <p>\(\angle\)ABC = 90&deg;</p> <p>Hence,</p> <p>or, \(\theta\) + 35&deg; = 90&deg;</p> <p>or, \(\theta\) = 90&deg;&minus; 35&deg;</p> <p>&there4; \(\theta\) = 55&deg;</p>

Q5:

In the given right angled triangle, find the value of unknown angle.

az


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>\(\angle\)XYZ = 90&deg;</p> <p>Here,</p> <p>or, &beta; + 56&deg; = 90&deg;</p> <p>or, &beta; = 90&deg;&minus; 56&deg;</p> <p>&there4; &beta; = 34&deg;</p>

Q6:

In the given right angled triangle, find the value of unknown angle.

zasx


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>\(\angle\)QPR = 90&deg;</p> <p>Here,</p> <p>or, &alpha; + 28 = 90&deg;</p> <p>or, &alpha; = 90&deg;&minus; 28&deg;</p> <p>&there4; &alpha; = 52&deg;</p>

Q7:

In the given right angled triangle, find the length of unknown sides.

z


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution,</p> <p>Here,</p> <p>AB = P = 3cm</p> <p>BC = b = 4cm</p> <p>AC = h = ?</p> <p>From the right angled &Delta;ABC we have,</p> <p>or, h = \(\sqrt{p^2+b^2}\)</p> <p>or, h = \(\sqrt{3^2+4^2}\)</p> <p>or, h = \(\sqrt{9+16}\)</p> <p>or, h = \(\sqrt{25}\)</p> <p>&there4; h = 5cm</p>

Q8:

In the given angled triangle, find the length of unknown sides.

za


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>Here,</p> <p>PQ = h = 13cm</p> <p>RQ = p = 12cm</p> <p>PR = b = ?</p> <p>From right angled &Delta;PQR we have,</p> <p>or, b = \(\sqrt{h^2&minus;p^2}\)</p> <p>or, b = \(\sqrt{13^2&minus;12^2}\)</p> <p>or, b = \(\sqrt{169&minus;12}\)</p> <p>or, b = \(\sqrt{25}\)</p> <p>&there4; b = 5cm</p>

Q9:

In the given right angled, find the length of unknown sides.

d


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>Here,</p> <p>YZ = h = 10cm</p> <p>XY = b = 6cm</p> <p>XZ = p = ?</p> <p>From right angled &Delta;XYZ we have,</p> <p>or, p = \(\sqrt{h^2&minus;b^2}\)</p> <p>or, p = \(\sqrt{10^2&minus;6^2}\)</p> <p>or, p = \(\sqrt{100&minus;36}\)</p> <p>or, p = \(\sqrt{64}\)</p> <p>&there4; p = 8cm</p>

Q10:

Show that a triangle having the sides 5cm, 12cm and 13cm is a right angled triangle.

a


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>The three sides of a triangle are 5cm, 12cm and 13cm</p> <p>Here,</p> <p>or, 13<sup>2</sup> = 5<sup>2</sup> + 12<sup>2</sup></p> <p>or, 169 = 25 + 144</p> <p>&there4; 169 = 169</p> <p>i.e. h<sup>2</sup> = p<sup>2</sup>+ b<sup>2</sup></p> <p>Hence, the triangle is right angle triangle.</p> <p></p>

Q11:

\(\sqrt{\frac{secβ + 1}{secβ − 1}}\) = \(\frac{1 + cosβ}{sinβ}\)


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>L.H.S = \(\sqrt{\frac{sec&beta; + 1}{sec&beta;&minus; 1}}\) <br>= \(\sqrt{\frac{sec&beta; + 1}{sec&beta;&minus; 1}&times; \frac{sec&beta; + 1}{sec&beta; + 1}}\)<br>= \(\sqrt{\frac{(sec&beta; + 1)^2}{sec&beta;^2 &minus; 1}}\)<br>= \(\sqrt{\frac{(sec&beta; + 1)^2}{tan^2&beta;}}\)<br>= \(\sqrt{(\frac{sectan&beta; + 1}{tan&beta;})^2}\)<br>= \(\frac{sec&beta; + 1}{tan&beta;}\)<br>= \(\frac{\frac{1}{cos&beta;} + 1}{\frac{sin&beta;}{cos&beta;}}\)<br>= \(\frac{\frac{1 + cos&beta;}{cos&beta;}}{\frac{sin&beta;}{cos&beta;}}\)<br>= \(\frac{1 + cos&beta;}{cos&beta;}\) &times; \(\frac{cos&beta;}{sin&beta;}\)<br>= \(\frac{1 + cos&beta;}{sin&beta;}\)<br>= R.H.S proved</p> <p></p>

Q12:

\(\frac{tanθ + tanβ}{cotθ + cotβ}\) = tanθ . tanβ


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>L.H.S = \(\frac{tan&theta; + tan&beta;}{cot&theta; + cot&beta;}\)<br>= \(\frac{\frac{sin&theta;}{cos&theta;} + \frac{sin&beta;}{cos&beta;}}{\frac{cos&theta;}{sin&theta;} + \frac{cos&beta;}{sin&beta;}}\)<br>= \(\frac{\frac{sin&theta; . cos&beta; + sin&beta; . cos&theta;}{cos&theta; . cos&beta;}}{\frac{cos&theta; . sin&beta; + cos&beta; . sin&theta;}{sin&theta; . sin&beta;}}\)<br>=\(\frac{sin&theta; . cos&beta; + sin&beta; . cos&theta;}{cos&theta; . cos&beta;}\) &times; \(\frac{sin&theta; . sin&beta; }{cos&theta; . sin&beta; + cos&beta; . sin&theta; }\)<br>= \(\frac{sin&theta; . sin&beta;}{cos&theta; . cos&beta;}\)<br>= \(\frac{sin&theta;}{cos&theta;}\) .\(\frac{sin&beta;}{cos&beta;}\)<br>= tan&theta; . tan&beta;<br>= R.H.S proved</p>

Q13:

\(\frac{cosα + sinα}{cosα − sinα}\) = \(\frac{1 + tanα}{1 − tanα}\) = \(\frac{cotα + 1}{cotα − 1}\)


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>l.H.S = \(\frac{cos&alpha; + sin&alpha;}{cos&alpha;&minus; sin&alpha;}\) [Dividing numerator and denominator by cos&alpha;]<br>= \(\frac{\frac{cos&alpha; + sin&alpha;}{cos&alpha;}}{\frac{cos&alpha;&minus; sin&alpha;}{cos&alpha;}}\)<br>= \(\frac{\frac{cos&alpha;}{cos&alpha;} + \frac{sin&alpha;}{cos&alpha;}}{\frac{cos&alpha;}{cos&alpha;} &minus; \frac{sin&alpha;}{cos&alpha;}}\)<br>= \(\frac{1 + tan&alpha;}{1&minus; tan&alpha;}\) (Middle Term)</p> <p>Middle Term = \(\frac{1 + tan&alpha;}{1&minus; tan&alpha;}\)<br>= \(\frac{1 + \frac{1}{cot&alpha;}}{1 &minus;\frac{1}{cot&alpha;}}\)<br>= \(\frac{cot&alpha; + 1}{cot&alpha;}\)&times; \(\frac{cot&alpha; }{cot&alpha;&minus; 1}\)<br>= \(\frac{cot&alpha; + 1}{cot&alpha;&minus; 1}\) (R.H.S)<br>\(\therefore\) L.H.S = M.T = R.H.S proved</p>

Q14:

\(\frac{1}{1 + cosθ}\) + \(\frac{1}{1 − cosθ}\) = 2cosec2θ


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>L.H.S = \(\frac{1}{1 + cos&theta;}\) + \(\frac{1}{1&minus; cos&theta;}\)<br>= \(\frac{1(1&minus; cos&theta;) + 1(1 + cos&theta;)}{(1 + cos&theta;)(1&minus; cos&theta;)}\)<br>= \(\frac{2}{1&minus; cos^2&theta;}\)<br>= \(\frac{2}{sin^2&theta;}\)<br>= \(\frac{2}{\frac{1}{cosec^2&theta;}}\)<br>= 2cosec<sup>2</sup>&theta;<br>= R.H.S proved</p>

Q15:

\(\frac{tanθ}{1 − cotθ}\) + \(\frac{cotθ}{1 − tanθ}\) = secθ . cosecθ + 1


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>L.H.S = \(\frac{tan&theta;}{1&minus; cot&theta;}\) + \(\frac{cot&theta;}{1&minus; tan&theta;}\)<br>= \(\frac{\frac{sin&theta;}{cos&theta;}}{1&minus; \frac{cos&theta;}{sin&theta;}}\) +\(\frac{\frac{cos&theta;}{sin&theta;}}{1&minus; \frac{sin&theta;}{cos&theta;}}\)<br>= \(\frac{\frac{sin&theta;}{cos&theta;}}{\frac{sin&theta;&minus;cos&theta;}{sin&theta;}}\)+\(\frac{\frac{cos&theta;}{sin&theta;}}{\frac{cos&theta;&minus;sin&theta;}{cos&theta;}}\)<br>= \(\frac{sin&theta;}{cos&theta;}\)&times; \(\frac{sin&theta;}{(sin&theta;&minus;cos&theta;)}\) + \(\frac{cos&theta;}{sin&theta;}\)&times;\(\frac{cos&theta;}{(cos&theta;&minus;sin&theta;)}\)<br>= \(\frac{sin^2&theta;}{cos&theta;(sin&theta;&minus;cos&theta;)}\) + \(\frac{cos^2&theta;}{sin&theta;(cos&theta;&minus;sin&theta;)}\)<br>= \(\frac{sin^2&theta;}{cos&theta;(sin&theta;&minus;cos&theta;)}\)&minus;\(\frac{cos^2&theta;}{sin&theta;(sin&theta;&minus;cos&theta;)}\)<br>= \(\frac{sin^2&theta;&times; sin&theta;&minus;cos^2&theta;&times;cos&theta;}{sin&theta; . cos&theta;(sin&theta;&minus;cos&theta;)}\)<br>= \(\frac{sin^3&theta;&minus;cos^3&theta;}{sin&theta; . cos&theta;(sin&theta;&minus;cos&theta;)}\)<br>= \(\frac{(sin&theta;&minus;cos&theta;)(sin^2&theta; +sin&theta; . cos&theta; +cos^2&theta;)}{sin&theta; . cos&theta;(sin&theta;&minus;cos&theta;)}\)<br>= \(\frac{(sin^2&theta; +sin&theta; . cos&theta; +cos^2&theta;)}{sin&theta; . cos&theta;}\)<br>= \(\frac{sin^2&theta; +cos^2&theta; +sin&theta; . cos&theta;}{sin&theta; . cos&theta;}\)<br>= \(\frac{1 +sin&theta; . cos&theta;}{sin&theta; . cos&theta;}\)<br>= \(\frac{1}{sin&theta; . cos&theta;}\) +\(\frac{sin&theta; . cos&theta;}{sin&theta; . cos&theta;}\)<br>= \(\frac{1}{sin&theta; . cos&theta;}\) + 1<br>= \(\frac{1}{sin&theta;}\)&times; \(\frac{1}{cos&theta;}\) + 1<br>= sec&theta; . cosec&theta; + 1<br>= R.H.S proved</p>

Q16:

What is the length of the hypotenuse?


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Use the Pythagorean theorem, with a = 12 and b = 5.</p> <p>or, c<sup>2</sup>= a<sup>2</sup>+ b<sup>2</sup>[Pythagorean theorem]</p> <p>or, c<sup>2</sup>= 12<sup>2</sup>+ 5<sup>2</sup> [Putting value of a &amp; b]</p> <p>or, c<sup>2</sup>= 144 + 25</p> <p>or, c<sup>2</sup>= 169</p> <p>or, \(\sqrt{c^2}\) = \(\sqrt{169}\) [Squaring on both sides]</p> <p>\(\therefore\) c =13</p> <p>Hence, The length of the hypotenuse is 13 metres.</p> <p></p> <p></p>

Q17:

What is the length of the hypotenuse?


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>Use the Pythagorean theorem, with a = 3 and b = 4.</p> <p>or, c<sup>2</sup>= a<sup>2</sup>+ b<sup>2</sup>[Pythagorean theorem]</p> <p>or, c<sup>2</sup>= 3<sup>2</sup>+ 4<sup>2</sup>[Putting value of a &amp; b]</p> <p>or, c<sup>2</sup>= 9 + 16</p> <p>or, c<sup>2</sup>= 25</p> <p>or, \(\sqrt{c^2}\) = \(\sqrt{25}\) [Squaring on both side]</p> <p>\(\therefore\) c = 5</p> <p>Hence, The length of the hypotenuse is 5 metres.</p> <p></p>

Q18: The perimeter of a right angled triangle is 30 cm. if the hypotenuse is 1cm more than the perpendicular and 8cm more than the base ,find the sides of the right angled triangle.
Type: Long Difficulty: Easy

Q19: 1+sinA/1-sinA
Type: Short Difficulty: Easy

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Measures for Population Management

Measures for Population Management

The population of any place or a country is not constant. There is always change in population. We usually find the population being increased in most of the places but in some of the places we can find the population being decreased. The countries of north-west Europe such as Norway, Denmark, etc. can be taken as examples of it.

People usually takes population management as related only to the control of the population but its meaning is little different from that. Population management is a bit different from population control. Population management is the practice of keeping the human population at a sustainable population size. It includes the size of population, structure, distribution, population growth etc. Regarding the different factors related to birth, death and migration and keeping the balance on means and resources to control the population is called population management. The essential factors such as food, education, clothes, shelter, health services, employment, etc. are to be provided to maintain the quality of life. All these works are called population management. The change in population does not occur in the same way at all the places, it differs.

There is an interrelationship between population growth and available means and resources. There must be a balance between these resources and population growth otherwise there is danger of the lives of the human. The population growth should be balanced according to the available means and resources. The factors such as quality education, small family, proper age of marriage, birth spacing, breastfeeding, women participation in the decision taking on the important matters of the family, etc. should be considered by developing countries for the management of population. It is essential to bring about changes in the area of mother and child health care in order to decrease infant mortality rate. Involving women in income-generating activities, women empowerment, population education, etc. should be promoted with regard to all these activities.

Lesson

Population Growth and Its Management

Subject

Social Studies and Population Education

Grade

Grade 8

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