Zones and Districts of Nepal

Nepal is a landlocked Himalayan country bordered by China to the north and India to the south, east, and west. This note has information about zones and districts of Nepal.

Summary

Nepal is a landlocked Himalayan country bordered by China to the north and India to the south, east, and west. This note has information about zones and districts of Nepal.

Things to Remember

  • Nepal is a landlocked Himalayan country bordered by China to the north and India to the south, east and west. 
  • Nepal has an area of 1,47,181 sq.km which is only 0.3% of the total landmass of Asia and 0.03% of the world. 
  • Nepal is divided into 5 development regions, 14 zones and 75 districts.
  • Nepal, though small, is one of the country of the world with most varied land topography.

MCQs

No MCQs found.

Subjective Questions

Q1:

If the A(2, 3), B(2, 0) and C(−1, 0) are the vertices of ΔABC, find the length of sides AB, AC , and CA.


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>For, AB<br>A(2, 3) = (x, y)<br>B(2, 0) = (x<sub>2</sub>, y<sub>2</sub>)<br>AB = ?<br>We know that,<br>or, AB = \(\sqrt{(x_2&minus; x_1)^2 + (y_2&minus; y_1)^2}\)<br>or, AB =\(\sqrt{(2&minus; 2)^2 + (0 &minus; 3)^2}\)<br>or, AB =\(\sqrt{(0)^2 + (3)^2}\)<br>or, AB =\(\sqrt{3^2}\)<br>&there4; AB = 3 units.</p> <p>For BC<br>B(2, 0) = (x<sub>1</sub>, y<sub>1</sub>)<br>C(&minus;1, 0) = (x<sub>2</sub>, y<sub>2</sub>)<br>BC = ?<br>We know that,<br>or, BC = \(\sqrt{(x_2&minus; x_1)^2 + (y_2&minus; y_1)^2}\)<br>or, BC = \(\sqrt{(&minus;1 &minus; 2)^2 + 0 &minus; 0)^2}\)<br>or, BC = \(\sqrt{(3)^2}\)<br>&there4; BC = 3 units<br><br>For CA<br>C(&minus; 1, 0) = (x<sub>1</sub>, y<sub>1</sub>)<br>A(2, 3) =(x<sub>2</sub>, y<sub>2</sub>)<br>CA = ?<br>We know that,<br>or, CA =\(\sqrt{(x_2&minus; x_1)^2 + (y_2&minus; y_1)^2}\)<br>or, CA =\(\sqrt{(2 + 1)^2 + (3 &minus; 0)^2}\)<br>or, CA =\(\sqrt{(3)^2 + (3)^2}\)<br>or, CA =\(\sqrt{9 + 9}\)<br>or, CA =\(\sqrt{18}\)<br>&there4; 3\(\sqrt{2}\) units</p> <p></p>

Q2:

A(−3, −4), B(−1, −2), C(8, 6) and D(10, 8) are four points. Show that AB = CD


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>For AB<br>A(&minus;3,&minus;4) = (x<sub>1</sub>, y<sub>1</sub>)<br>B(&minus;1,&minus;2) =(x<sub>2</sub>, y<sub>2</sub>)<br>AB = ?<br>We know that,<br>or, AB = \(\sqrt{(x_2&minus;x_1)^2 + (y_2&minus; y_1)^2}\)<br>or, AB = \(\sqrt{(&minus;1 + 3)^2 + (&minus;2 + 4)^2}\)<br>or, AB = \(\sqrt{(2)^2 + (2)^2}\)<br>or, AB = \(\sqrt{4 +4}\)<br>or, AB = \(\sqrt{8}\)<br>or, AB = 2\(\sqrt{2}\) units.<br><br></p> <p>For CD<br>C(8, 6) = (x<sub>1</sub>, y<sub>1</sub>)<br>D(10, 8) =(x<sub>2</sub>, y<sub>2</sub>)<br>CD = ?<br>We know that,<br>or, CD = \(\sqrt{(x_2&minus;x_1)^2 + (y_2&minus; y_1)^2}\)<br>or, CD = \(\sqrt{(10 &minus; 8)^2 + (8 &minus; 6)^2}\)<br>or, CD = \(\sqrt{(2)^2 + (2)^2}\)<br>or, CD = \(\sqrt{4+4}\)<br>or, CD =\(\sqrt{8}\)<br>or, CD = 2\(\sqrt{2}\) units.<br><br>&there4; AB = CD =2\(\sqrt{2}\) units. proved.<br><br><br><br></p>

Q3:

If A is a point on x-axis whose abscissa is −4 and B is point on y-axis whose ordinate is −3, find the distance between A and B.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>A(&minus;4, 0) = (x1, y1)<br>B(0,&minus;3) = (x2, y2)<br>AB = ?<br>We know that,<br>or, AB = \(\sqrt{(x_2&minus; x_1)^2 + (y_2&minus;y_1)^2}\)<br>or, AB = \(\sqrt{(0 + 4)^2 + (&minus; 3 &minus; 0)^2}\)<br>or, AB = \(\sqrt{4)^2 + (3)^2}\)<br>or, AB = \(\sqrt{16 + 9}\)<br>or, AB = \(\sqrt{25}\)<br>&there4; AB = 5 units.<br><br><br><br></p>

Q4:

If P is a point on x-axis whose abscissa  is −8 and B is point on y-axis whose ordinate is 6. Find the length of PB.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>Here, <br>P(&minus;8, 0) = (x<sub>1</sub>, y<sub>1</sub>)<br>B(0, 6) =(x<sub>2</sub>, y<sub>2</sub>)<br>PB = ?<br>We know that,<br>or, PB = \(\sqrt{(x_2&minus; x_1)^2 + (y_2&minus; y_1)^2}\)<br>or, PB = \(\sqrt{(0 + 8)^2 + (6 &minus; 0)^2}\)<br>or, PB = \(\sqrt{(8)^2 + (6)^2}\)<br>or, PB = \(\sqrt{64 + 36}\)<br>or, PB = \(\sqrt{100}\)<br>&there4; PB = 10 units.</p>

Q5:

Find the slope of line joining the points if A(0, −5) and B(4 , 8).


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>Here,<br>A(0,&minus;5) = (x<sub>1</sub>, y<sub>1</sub>)<br>B(4 , 8).= (x<sub>2</sub>, y<sub>2</sub>)<br>Slope (m) = ?</p> <p>We know that,<br>Slope (m) = \(\frac{y_2&minus; y_1}{x_2&minus; x_1}\)<br>=\(\frac{8 + 5}{4 &minus; 0}\)<br>= \(\frac{13}{4}\)</p>

Q6:

Find the slope of the point if E(4, −7) and F(−3, 4).


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>Here,<br>E(4,&minus;7) = (x<sub>2</sub>, y<sub>1</sub>)<br>F(&minus;3, 4) = (x<sub>2</sub>, y<sub>2</sub>)<br>Slope (m) = ?</p> <p>we know that,<br>Slope (m) = \(\frac{y_2&minus; y_1}{x_2&minus; x_1}\)<br>=\(\frac{4 + 7}{&minus;3 &minus; 4}\)<br>= \(\frac{11}{&minus;7}\)</p>

Q7:

Find the distance between the given points (−1, 7) and (3, 10).


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>Let,<br>A (&minus;1, 7) = (x<sub>1</sub>, y<sub>1</sub>)<br>B (3, 10) = (x<sub>2</sub>, y<sub>2</sub>)<br>Distance of AB = ?</p> <p>We know that,<br>or, AB = \(\sqrt{(x_2&minus; x_1)^2 + (y_2&minus; y_1)^2}\)<br>or, AB = \(\sqrt{(3 + 1)^2 + (10 &minus; 7)^2}\)<br>or, AB = \(\sqrt{(4)^2 + (3)^2}\)<br>or, AB = \(\sqrt{16+ 9}\)<br>or, AB = \(\sqrt{25}\)<br>&there4; AB = 5 units</p> <p></p>

Q8:

P(2, −5), Q(4, 1)and R(10, 3) are three points. Show that PQ = QR.


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>For PQ<br>P(2,&minus;5) = (x<sub>1</sub>, y<sub>1</sub>)<br>Q(4, 1) = (x<sub>2</sub>, y<sub>2</sub>)<br>PQ = ?<br>We know that,<br>or, PQ = \(\sqrt{(x_2&minus; x_1)^2 + (y_2&minus; y_1)^2}\)<br>or, PQ = \(\sqrt{(4 &minus; 2)^2 + (1 +5)^2}\)<br>or, PQ = \(\sqrt{(2)^2 + (6)^2}\)<br>or, PQ = \(\sqrt{4 + 36}\)<br>or, PQ = \(\sqrt{40}\)<br>&there4; PQ = 2\(\sqrt{10}\) units.</p> <p>For QR<br>Q(4, 1) =(x<sub>1</sub>, y<sub>1</sub>)<br>R(10, 3) =(x<sub>2</sub>, y<sub>2</sub>)<br>QR = ?<br>We know that,<br>or, QR = \(\sqrt{(x_2&minus; x_1)^2 + (y_2&minus; y_1)^2}\)<br>or, QR = \(\sqrt{(10&minus; 4)^2 + (3 &minus; 1)^2}\)<br>or, QR = \(\sqrt{(6)^2 + (2)^2}\)<br>or, QR = \(\sqrt{36 + 4}\)<br>or, QR = \(\sqrt{40}\)<br>&there4; QR = 2\(\sqrt{10}\)<br>&there4; PQ = QR proved.</p>

Q9:

Show that (1, 2), (4, 5) and (7, 2) are the vertices of the right angle isosceles triangle.


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>Let, A(1, 2), B(4, 5) and C(7, 2) be the three points<br>For AB<br>A(1, 2) = (x<sub>1</sub>, y<sub>1</sub>)<br>B(4, 5) = (x<sub>2</sub>, y<sub>2</sub>)<br>Distance of AB = ?<br>We knoe that,<br>or, AB = \(\sqrt{(x_2&minus; x_1)^2 + (y_2&minus; y_1)^2}\)<br>or, AB = \(\sqrt{(4 &minus; 1)^2 + (5 &minus; 2)^2}\)<br>or, AB = \(\sqrt{(3)^2 + (3)^2}\)<br>or, AB = \(\sqrt{18}\)<br>or, AB = 3\(\sqrt{2}\) units</p> <p>For BC<br>B(4, 5) = (x<sub>1</sub>, y<sub>1</sub>)<br>C(7, 2) = (x<sub>2</sub>, y<sub>2</sub>)<br>BC = ?<br>we know that,<br>or, BC = \(\sqrt{(x_2&minus; x_1)^2 + (y_2&minus; y_1)^2}\)<br>or, BC = \(\sqrt{(7&minus; 4)^2 + (2 &minus; 5)^2}\)<br>or,BC = \(\sqrt{(3)^2 + (&minus; 3)^2}\)<br>or, BC = \(\sqrt{9 + 9}\)<br>or, BC = 3\(\sqrt{2}\) units</p> <p>For CA<br>C(7, 2) = (x<sub>1</sub>, y<sub>1</sub>)<br>A(1, 2) = (x<sub>2</sub>, y<sub>2</sub>)<br>CA = ?<br>We know that,<br>or, CA = \(\sqrt{(x_2&minus; x_1)^2 + (y_2&minus; y_1)^2}\)<br>or, CA = \(\sqrt{(1 &minus; 7)^2 + (2&minus; 2)^2}\)<br>or, CA = \(\sqrt{( &minus; 6)^2 + (0)^2}\)<br>or, CA = \(\sqrt{36}\)<br>or, CA = 6 units</p> <p>Now,<br>CA has the longest side<br>or, 6<sup>2</sup> = (3\(\sqrt{2}\))<sup>2</sup>+(3\(\sqrt{2}\))<sup>2</sup><br>or, 36 = 18 + 18<br>or, 36 = 36</p> <p>AB = BC = 3\(\sqrt{2}\) units and is satisfied the pythogorus theorem (i.e. h<sup>2</sup> = p<sup>2</sup> + b<sup>2</sup>). So, the given points are the vertices of isosceles right angled triangle</p>

Q10:

Find the distance between point (0, 0) and (−6, −4)


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>Let, Q(0, 0) and R(&minus;6,&minus;4)<br>Distance of QR = ?<br>We know that,<br>or, QR = \(\sqrt{(x_2&minus; x_1)^2 + (y_2&minus; y_1)^2}\)<br>or, QR = \(\sqrt{(&minus; 6 &minus; 0)^2 + (&minus; 4 &minus; 0)^2}\)<br>or, QR = \(\sqrt{(6)^2 + (4)^2}\)<br>or, QR = \(\sqrt{36 + 16}\)<br>or, QR = \(\sqrt{52}\)<br>&there4; QR = 2\(\sqrt{3}\)<br><br></p>

Q11:

Show that (1, −1), (−1, 1) and (\(\sqrt{3}\), \(\sqrt{3}\)) are the vertices of an equilateral triangle.


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>For AB<br>A(1,&minus;1) = (x<sub>1</sub>, y<sub>1</sub>)<br>B(&minus;1, 1) = (x<sub>2</sub>, y<sub>2</sub>)<br>AB = ?<br>We know that,<br>or, AB = \(\sqrt{(x_2&minus; x_1)^2 + (y_2&minus; y_1)^2}\)<br>or, AB = \(\sqrt{(&minus; 1&minus; 1)^2 + (1 +1)^2}\)<br>or, AB = \(\sqrt{(&minus;2)^2 + 2^2}\)<br>or, AB = \(\sqrt{4 + 4}\)<br>or, AB = \(\sqrt{8}\)<br>or, AB = 2\(\sqrt{2}\) units.</p> <p>For BC<br>B(&minus;1, 1) = (x<sub>1</sub>, y<sub>1</sub>)<br>C(\(\sqrt{3}\), \(\sqrt{3}\)) = (x<sub>2</sub>, y<sub>2</sub>)<br>BC = ?<br>We know that,<br>or, BC = \(\sqrt{(x_2&minus; x_1)^2 + (y_2&minus; y_1)^2}\)<br>or, BC = \(\sqrt{(\sqrt{3}+1)^2 + (\sqrt{3} &minus; 1)^2}\)<br>or, BC = \(\sqrt{(\sqrt{3})^2 + 2 . \sqrt{3} .1 + 1^2 + (\sqrt{3})^2 &minus; 2 . \sqrt{3} . 1 + 1^2}\)<br>or, BC = \(\sqrt{3 +2 . \sqrt{3} + 1 + 3&minus; 2 . \sqrt{3} + 1}\)<br>or, BC = \(\sqrt{8}\)<br>or, BC = 2\(\sqrt{2}\)</p> <p>For AC<br>A(1,&minus;1) = (x<sub>1</sub>, y<sub>1</sub>)<br>C(\(\sqrt{3}\),\(\sqrt{3}\)) = (x<sub>2</sub>, y<sub>2</sub>)<br>AC = ?<br>We know that,<br>or, AC = \(\sqrt{(x_2&minus; x_1)^2 + (y_2&minus; y_1)^2}\)<br>or, AC = \(\sqrt{(\sqrt{3} &minus; 1)^2 + (\sqrt{3}+1)^2}\)<br>or, AC = \(\sqrt{(\sqrt{3})^2 &minus; 2 . \sqrt{3} .1 + 1^2 + (\sqrt{3})^2 +2 . \sqrt{3} . 1 + 1^2}\)<br>or, AC = \(\sqrt{3 &minus; 2 . \sqrt{3} + 1 + 3 +2 . \sqrt{3} + 1}\)<br>or, AC = \(\sqrt{8}\)<br>or, AC = 2\(\sqrt{2}\)</p> <p>Here, AB = BC = AC. So, the given points are the vertices of an equilateral triangle.</p>

Q12:

Show the points A(1, 1), B(4, 4), C(4, 8) and D(1, 5) are the vertices of a parallelogram.


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>For AB<br>A(1, 1) = (x<sub>1</sub>, y<sub>1</sub>)<br> B(4, 4) = (x<sub>2</sub>, y<sub>2</sub>)<br>AB = ?<br>We know that,<br>or, AB = \(\sqrt{(x_2&minus; x_1)^2 + (y_2&minus; y_1)^2}\)<br>or,AB = \(\sqrt{(4 &minus; 1)^2 + (4 &minus; 1)^2}\)<br>or,AB = \(\sqrt{(3)^2 + (3)^2}\)<br>or, AB = \(\sqrt{9 + 9}\)<br>or, AB = \(\sqrt{18}\)<br>or, AB = 3\(\sqrt{2}\) units</p> <p>For BC<br>B(4, 4) =(x<sub>1</sub>, y<sub>1</sub>)<br> C(4, 8) = (x<sub>2</sub>, y<sub>2</sub>)<br>BC = ?<br>We know that,<br>or, BC =\(\sqrt{(x_2&minus; x_1)^2 + (y_2&minus; y_1)^2}\)<br>or, BC =\(\sqrt{(4&minus; 4)^2 + (8&minus; 4)^2}\)<br>or, BC = \(\sqrt{4^2}\)<br>or, BC = 4 units.</p> <p>For CD<br>C(4, 8) =(x<sub>1</sub>, y<sub>1</sub>)<br>D(1, 5) =(x<sub>2</sub>, y<sub>2</sub>)<br>CD = ?<br>We know that,<br>or, CD =\(\sqrt{(x_2&minus; x_1)^2 + (y_2&minus; y_1)^2}\)<br>or, CD =\(\sqrt{(1 &minus; 4)^2 + (5 &minus; 8)^2}\)<br>or, CD =\(\sqrt{(&minus; 3)^2 + (&minus; 3)^2}\)<br>or, CD = \(\sqrt{9 + 9}\)<br>or, CD = 3\(\sqrt{2}\) units</p> <p>For AD<br> A(1, 1) =(x<sub>1</sub>, y<sub>1</sub>)<br>D(1, 5) =(x<sub>2</sub>, y<sub>2</sub>)<br>AD = ?<br>We know that,<br>or, AD =\(\sqrt{(x_2&minus; x_1)^2 + (y_2&minus; y_1)^2}\)<br>or, AD =\(\sqrt{(1&minus; 1)^2 + (5 &minus; 1)^2}\)<br>or, AD = \(\sqrt{4^2}\)<br>or, AD = 4 units</p> <p>Here, AB = CD = 3\(\sqrt{2}\) units<br> BC = AB = 4 units<br>So, the given points are the vertices of a parallelogram.</p>

Q13:

Show that the points A(0, −1), B(2, 1), C(0, 3) and D(−2, 1) are the vertices of a square.


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>For AB<br>A(0,&minus;1) = (x<sub>1</sub>, y<sub>1</sub>)<br> B(2, 1) = (x<sub>2</sub>, y<sub>2</sub>)<br>AB = ?<br>We know that,<br>or, AB = \(\sqrt{(x_2&minus; x_1)^2 + (y_2&minus; y_1)^2}\)<br>or, AB = \(\sqrt{2&minus; 0)^2 + (1+1)^2}\)<br>or, AB = \(\sqrt{2^2 + 2^2}\)<br>or, AB = \(\sqrt{4 + 4}\)<br>or, AB = \(\sqrt{8}\)<br>or, AB = 2\(\sqrt{2}\) units</p> <p>For BC<br>B(2, 1) = (x<sub>1</sub>, y<sub>1</sub>)<br> C(0, 3) = (x<sub>2</sub>, y<sub>2</sub>)<br>BC = ?<br>We know that,<br>or, BC = \(\sqrt{(x_2&minus; x_1)^2 + (y_2&minus; y_1)^2}\)<br>or, BC = \(\sqrt{(0&minus; 2)^2 + (3 &minus; 1)^2}\)<br>or, BC = \(\sqrt{(&minus; 2)^2 + (2)^2}\)<br>or, BC = \(\sqrt{4 + 4}\)<br>or, BC = \(\sqrt{8}\)<br>or, BC = 2\(\sqrt{2}\) units</p> <p>For CD<br>C(0, 3) = (x<sub>1</sub>, y<sub>1</sub>)<br>D(&minus;2, 1) = (x<sub>2</sub>, y<sub>2</sub>)<br>CD = ?<br>We know that,<br>or, CD = \(\sqrt{(x_2&minus; x_1)^2 + (y_2&minus; y_1)^2}\)<br>or, CD = \(\sqrt{(&minus; 2&minus; 0)^2 + (1 &minus; 3)^2}\)<br>or, CD = \(\sqrt{(&minus; 2)^2 + 2^2}\)<br>or, CD = \(\sqrt{4 + 4}\)<br>or, CD = \(\sqrt{8}\)<br>or, CD = 2\(\sqrt{2}\) units</p> <p>For AD<br> A(0,&minus;1) = (x<sub>1</sub>, y<sub>1</sub>)<br>D(&minus;2, 1) = (x<sub>2</sub>, y<sub>2</sub>)<br>AD = ?<br>We know that,<br>or, AD = \(\sqrt{(x_2&minus; x_1)^2 + (y_2&minus; y_1)^2}\)<br>or, AD = \(\sqrt{(&minus; 2&minus; 0)^2 + (1 &minus; 1)^2}\)<br>or, AD = \(\sqrt{(&minus; 2)^2 + (2)^2}\)<br>or, AD = \(\sqrt{4 + 4}\)<br>or, AD = \(\sqrt{8}\)<br>or, AD = 2\(\sqrt{2}\) units</p> <p>For AC<br>A(0,&minus;1) = (x<sub>1</sub>, y<sub>1</sub>)<br>C(0, 3) = (x<sub>2</sub>, y<sub>2</sub>)<br>or, AC = \(\sqrt{(x_2&minus; x_1)^2 + (y_2&minus; y_1)^2}\)<br>or, AC = \(\sqrt{(0 &minus; 0)^2 + (3+1)^2}\) <br>or, AC = \(\sqrt{4^2}\)<br>or, AC = 4 units.</p> <p>For DB<br>D(&minus;2, 1) = (x<sub>1</sub>, y<sub>1</sub>)<br>B(2, 1) = (x<sub>2</sub>, y<sub>2</sub>)<br>DB = ?<br>We know that,<br>or, DB = \(\sqrt{(x_2&minus; x_1)^2 + (y_2&minus; y_1)^2}\)<br>or, DB = \(\sqrt{(2+2)^2 + (1 &minus; 1)^2}\)<br>or, DB = \(\sqrt{4^2}\)<br>or, DB = 4 units</p> <p>Here, AB = BC = CD = AD = 2\(\sqrt{[2}\) units<br> AC = BD = 4 units<br>Since the given points are the vertices of a square.</p>

Q14:

Show that (2, −1), (7, 4) and (8, 11) are the vertices of an isoscles trinagle.


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>Let, A(2,&minus;1), B(7, 4) and C(8, 11) be the vertices of an isoscles trinagle.<br>For AB<br>A(2,&minus;1) = (x<sub>1</sub>, y<sub>1</sub>)<br> B(7, 4) = (x<sub>2</sub>, y<sub>2</sub>)<br>AB = ?<br>We know that,<br>or, AB = \(\sqrt{(x_2&minus; x_1)^2 + (y_2&minus; y_1)^2}\)<br>or, AB =\(\sqrt{(7 &minus; 2)^2 + (4 +1)^2}\)<br>or, AB =\(\sqrt{(5)^2 + (5)^2}\)<br>or, AB = \(\sqrt{25 + 25}\)<br>or, AB = \(\sqrt{50}\)<br>or, AB = 5\(\sqrt{2}\) units</p> <p>For BC<br>B(7, 4) =(x<sub>1</sub>, y<sub>1</sub>)<br>C(8, 11) =(x<sub>2</sub>, y<sub>2</sub>)<br>BC = ?<br>we know that,<br>or, BC =\(\sqrt{(x_2&minus; x_1)^2 + (y_2&minus; y_1)^2}\)<br>or, BC =\(\sqrt{(8 &minus; 7)^2 + (11 &minus; 4)^2}\)<br>or, BC = \(\sqrt{1 + 7^2 }\)<br>or, BC = \(\sqrt{50}\)<br>or, BC = 5\(\sqrt{2}\) units</p> <p>For AC<br>A(2,&minus;1) = (x<sub>1</sub>, y<sub>1</sub>)<br>C(8, 11) =(x<sub>2</sub>, y<sub>2</sub>)<br>AC = ?<br>We know that,<br>or, AC =\(\sqrt{(x_2&minus; x_1)^2 + (y_2&minus; y_1)^2}\)<br>or, AC =\(\sqrt{(8 &minus; 2)^2 + (11+1)^2}\)<br>or, AC =\(\sqrt{(6)^2 + (12)^2}\)<br>or, AC = \(\sqrt{36 + 144}\)<br>or, AC = \(\sqrt{180}\)<br>or, AC = 6\(\sqrt{5}\) units</p> <p>Here, AB = BC = 5\(\sqrt{2}\) units<br>So, the given points are the vertices of an isosceles triangle</p>

Q15:

Find the slope of the line A(4, 2) and B(6, 8)

 


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>Here,<br>A(4, 2) = (x<sub>1</sub>, y<sub>1</sub>)<br> B(6, 8) = (x<sub>2</sub>, y<sub>2</sub>)<br>Slope (m) = ?<br>We know that,<br>Slope (m) = \(\frac{y_2&minus; y_1}{x_2&minus; x_1}\)<br>=\(\frac{8 &minus; 2}{6 &minus; 4}\)<br>=\(\frac{6}{2}\)<br>= 3<br><br></p>

Q16:

Show that the points (1, −1), (3, 2), (1, 4) and (−2, 2) are the vertices of a rectangular


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>For AB<br>A((1,?1) = (x<sub>1</sub>, y<sub>1</sub>)<br>B(3, 2) = (x<sub>2</sub>, y<sub>2</sub>)<br>AB = ?<br>We know that,<br>or, AB = \(\sqrt{(x_2? x_1)^2 + (y_2? y_1)^2}\)<br>or, AB = \(\sqrt{(3 ? 1)^2 + (2 ? 1)^2}\)<br>or, AB = \(\sqrt{(2)^2 + (3)^2}\)<br>or, AB = \(\sqrt{4 + 9}\)<br>or, AB = \(\sqrt{13}\) units</p> <p>For BC<br>B(3, 2) = (x<sub>1</sub>, y<sub>1</sub>)<br>C(1, 4) = (x<sub>2</sub>, y<sub>2</sub>)<br>BC = ?<br>We know that,<br>or, BC = \(\sqrt{(x_2? x_1)^2 + (y_2? y_1)^2}\)<br>or, BC = \(\sqrt{(1 ? 3)^2 + (4 ? 2)^2}\)<br>or, BC = \(\sqrt{(? 2)^2 + (2)^2}\)<br>or, BC = \(\sqrt{4 + 4}\)<br>or, BC = \(\sqrt{8}\) units</p> <p>For CD<br>C(1, 4) = (x<sub>1</sub>, y<sub>1</sub>)<br>D(?2, 2) = (x<sub>2</sub>, y<sub>2</sub>)<br>CD = ?<br>We know that,<br>or, CD = \(\sqrt{(x_2? x_1)^2 + (y_2? y_1)^2}\)<br>or, CD = \(\sqrt{(? 2 ? 1)^2 + (2 ? 4)^2}\)<br>or, CD = \(\sqrt{(? 3)^2 + (? 2)^2}\)<br>or, CD = \(\sqrt{9 + 4}\)<br>or, CD = \(\sqrt{13}\) units</p> <p>For AD<br>A(1,?1) = (x<sub>1</sub>, y<sub>1</sub>)<br>D(?2, 2) = (x<sub>2</sub>, y<sub>2</sub>) <br>AD = ?<br>We know that,<br>or, AD = \(\sqrt{(x_2? x_1)^2 + (y_2? y_1)^2}\)<br>or, AD = \(\sqrt{(1 ? 1)^2 + (4+1)^2}\)<br>or, AD = \(\sqrt{(0)^2 + (5)^2}\)<br>or, AD = 5 units</p> <p>For BC<br>B(3, 2) = (x<sub>1</sub>, y<sub>1</sub>)<br>D(?2, 2) = (x<sub>2</sub>, y<sub>2</sub>)<br>BD = ?<br>We know thats,<br>or, BD = \(\sqrt{(x_2? x_1)^2 + (y_2? y_1)^2}\)<br>or, BD = \(\sqrt{(? 2 ? 3)^2 + (2? 2)^2}\)<br>or, BD = \(\sqrt{(?5)^2 + (0)^2}\)<br>or, BD = 5 units</p> <p>Here,<br>AC = BD = 5 units<br>But other points cannot be proved. So it is not rectangle.<br><br></p>

Videos

No videos found.

Zones and Districts of Nepal

Zones and Districts of Nepal

Nepal is a landlocked Himalayan country bordered by China to the north and India to the south, east and west. It has an area of 1,47,181 sq.km which is only 0.3% of the total landmass of Asia and 0.03% of the world. Despite its vast resources, Nepal has remained a backward country for years. Among several causes, one is the country’s difficult physical structure. Nepal, though small, is one of the few countries of the world with most varied land topography.

Based on its landform and climate, Nepal is divided into three regions: the Himalayan Region, the Hilly Region and the Terai Region. These are distinct ecological regions running west-east horizontally. The high mountains in the north, the moderate Mahabharat hills in the centre and undulating chure hills with their sheltered lowland in the south are the predominant landscapes of the country.

As we know government alone cannot take care of the whole country, so to carry out uniform development to increase people's participation, to use resources, to decentralize the power our country is divided into 5 development regions, 14 zones and 75 districts. Such regions are known as Administrative Regions. Each district is headed by a chief district officer (CDO) and is responsible for maintaining law and order and coordinating the work of field agencies of the various government ministries.

The division of districts of Nepal according to the zones is shown in the following table:

Development Region Zones Districts
Eastern Development Region Mechi

Koshi

Sagarmatha

 Taplejung, Panchthar, Ilam, Jhapa

Morang, Sunsari, Dhankuta, Terhathum, Bhojpur, Sankhuwasabha

Saptari, Siraha, Udaypur, Khotang, Okhaldhunga, Solukhumbu
Central Development Region Janakpur

Bagmati


Narayani

 Mahottari, Dhanusha, Sarlahi, Sindhuli, Ramechhap, Dolakha

Kavre Palanchok, Kathmandu, Lalitpur, Bhaktapur, Sindhupalchok, Rasuwa, Dhading, Nuwakot

Rautahat, Bara, Parsa, Chitwan, Makwanpur
Western Development Region Gandaki

Dhaulagiri

Lumbini

 Kaski, Tanahun, Lamjung, Gorkha, Syanjhya, Manang

Parbat, Myagdi, Baglung, Mustang

Rupandehi, Nawalparasi, Kapilbastu, Arghakhanchi, Gulmi, Palpa
Mid-Western Development Region Karnali

Rapti

Bheri

 Dolpa, Jumla, Mugu, Humla, Kalikot

Dang, Salyan, Pyuthan, Rolpa, Rukum

Banke, Bardiya, Surkhet, Dailekh, Jajarkot
Far-Western Development Region Seti

Mahakali

 Kailali, Doti, Accham, Bajura, Bajhang

Kanchanpur, Dadeldhura, Baitadi, Darchula

Lesson

Our Earth

Subject

Social Studies and Population Education

Grade

Grade 8

Recent Notes

No recent notes.

Related Notes

No related notes.