1. What drawback exists in comparing line segments purely through observation?
1. What drawback exists in comparing line segments purely through observation?
Explanation:
We cannot compare the line segments with a tiny length difference by simple observation. We are unable to determine which line segment is longer. As a result, the likelihood of errors brought on by inappropriate sight is higher.
2. Why is it better to use a divider than a ruler, while measuring the length of a line segment?
2. Why is it better to use a divider than a ruler, while measuring the length of a line segment?
Explanation:
Errors can happen when using a ruler because of its thickness and angular viewing. Hence, precise measuring is feasible when employing dividers.
3. Draw whatever line segment you like, such as a (AB). Take a look at any location C between A and B. AB, BC, and AC are measured in length. Is AB equal to AC + CB?
3. Draw whatever line segment you like, such as a (AB). Take a look at any location C between A and B. AB, BC, and AC are measured in length. Is AB equal to AC + CB?
Explanation:
Considering 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡 𝑐 is located halfway between the letters a and b. As a result, all points are located on the same line segment of the →→. Hence, we may state that in every instance AB
where point C is located between A and B. A = CB + AC
For instance:
C is a position between 𝐴 𝑎𝑛𝑑 𝐵 such that 𝐴𝐶 = 3cm and 𝐶𝐵 = 4 cm, and AB is a line segment of length AB of 7 cm.
Therefore, 𝐴𝐶 + 𝐶𝐵 = 7
𝑆𝑖𝑛𝑐𝑒 𝐴𝐵 = 7 𝑐𝑚
𝐴 = 𝐵𝐶 + 𝐴𝐵 is confirmed.
4. Which point is between 𝑨 𝒂𝒏𝒅 𝑩 if 𝑨, 𝑩, 𝒂𝒏𝒅 𝑪 are three points on a line where 𝑨𝑩 =𝟓 𝒄𝒎, 𝑩𝑪 = 𝟑𝒄𝒎, and 𝑨𝑪 = 𝟖 𝒄𝒎?
4. Which point is between 𝑨 𝒂𝒏𝒅 𝑩 if 𝑨, 𝑩, 𝒂𝒏𝒅 𝑪 are three points on a line where 𝑨𝑩 =𝟓 𝒄𝒎, 𝑩𝑪 = 𝟑𝒄𝒎, and 𝑨𝑪 = 𝟖 𝒄𝒎?
Explanation:
assuming 𝐴𝐵 = 5 𝑐𝑚
𝐵𝐶 = 3 𝑐𝑚
𝐴𝐶 = 8 𝑐𝑚
now evident that 𝐴𝐶 = 𝐴𝐵 + 𝐵𝐶
𝑃𝑜𝑖𝑛𝑡 𝐵 is thus situated between 𝐴 𝑎𝑛𝑑 𝐶.
5. Confirm that 𝑫 lies at 𝒕𝒉𝒆 → midpoint.
𝑨𝑮
5. Confirm that 𝑫 lies at 𝒕𝒉𝒆 → midpoint.
𝑨𝑮
Explanation:
Therefore AD = DG = 3 units is evident from the figure. D is therefore the middle of the →→
AG
6. Explain why AB = CD 𝒊𝒇 𝑩 𝒊𝒔 𝒕𝒉𝒆 𝒎𝒊𝒅𝒑𝒐𝒊𝒏𝒕 of the → and 𝑪 𝒊𝒔 𝒕𝒉𝒆 𝒎𝒊𝒅𝒑𝒐𝒊𝒏𝒕 𝒐𝒇 the
𝑨𝑪
→→, where 𝑨, 𝑩, 𝑪, and 𝑫 𝒍𝒊𝒆 𝒐𝒏 𝒕𝒉𝒆 𝒔𝒂𝒎𝒆 𝒔𝒕𝒓𝒂𝒊𝒈𝒉𝒕 𝒍𝒊𝒏𝒆.
𝑩𝑫
6. Explain why AB = CD 𝒊𝒇 𝑩 𝒊𝒔 𝒕𝒉𝒆 𝒎𝒊𝒅𝒑𝒐𝒊𝒏𝒕 of the → and 𝑪 𝒊𝒔 𝒕𝒉𝒆 𝒎𝒊𝒅𝒑𝒐𝒊𝒏𝒕 𝒐𝒇 the
𝑨𝑪
→→, where 𝑨, 𝑩, 𝑪, and 𝑫 𝒍𝒊𝒆 𝒐𝒏 𝒕𝒉𝒆 𝒔𝒂𝒎𝒆 𝒔𝒕𝒓𝒂𝒊𝒈𝒉𝒕 𝒍𝒊𝒏𝒆.
𝑩𝑫
Explanation:
Given
The middle of AC is B. Hence, 𝐴𝐵 Equals 𝐵𝐶 (1)
𝑇ℎ𝑒 𝑚𝑖𝑑𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝐵𝐷 is C. Thus, 𝐵𝐶 = 𝐶𝐷 (2)
Using (1)𝑎𝑛𝑑 (2)
CD = 𝐴𝐵 is confirmed.
7. Draw five triangles, then count their sides. In each instance, verify that the third side is always smaller than the sum of the lengths of any two sides.
7. Draw five triangles, then count their sides. In each instance, verify that the third side is always smaller than the sum of the lengths of any two sides.
Explanation:
Case 1: In the triangle ABC,
AB equals 2.5 𝑐𝑚
𝐵𝐶 = 4.8 𝑐𝑚
𝐴𝐶 = 5.2 𝑐𝑚
A + B = 2.5 𝑐𝑚 plus 4.8 𝑐𝑚 = 7.3 𝑐𝑚
𝐴𝑠 7.3 > 5.2
𝐵𝐶 = 𝐴𝐵 > 𝐴𝐶
As a result, the third side Of a triangle is bigger than the total of any two of its sides.
8. Draw five triangles, then count their sides. In each instance, verify that the third side is always smaller than the sum of the lengths of any two sides.
8. Draw five triangles, then count their sides. In each instance, verify that the third side is always smaller than the sum of the lengths of any two sides.
Explanation:
Case 2. 𝑃𝑄𝑅 , 𝑃𝑄 = 2 𝑐𝑚 in the triangle.
𝑄𝑅 = 2.5 𝑐𝑚
𝑃𝑅 = 3.5 𝑐𝑚
𝑃𝑄 + 𝑄𝑅 equals 2 𝑐𝑚 + 2.5 𝑐𝑚 = 4.5 𝑐𝑚
𝐴𝑠 4.5 > 3.5
𝑃𝑄 + 𝑄𝑅 = 𝑃𝑅
𝐻𝑒𝑛𝑐𝑒 , 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑎𝑛𝑦 𝑡𝑤𝑜 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝑠𝑖𝑑𝑒𝑠 𝑖𝑠 𝑔𝑟𝑒𝑎𝑡𝑒𝑟 𝑡ℎ𝑎𝑛 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑡ℎ𝑖𝑟𝑑 𝑠𝑖𝑑𝑒
9. Draw five triangles, then count their sides. In each instance, verify that the third side is always smaller than the sum of the lengths of any two sides.
9. Draw five triangles, then count their sides. In each instance, verify that the third side is always smaller than the sum of the lengths of any two sides.
Explanation:
Case 3. 𝑋𝑌𝑍 , 𝑋𝑌 = 5 𝑐𝑚 in the triangle
𝑌𝑍 = 3 𝑐𝑚
𝑍𝑋 = 6.8 𝑐𝑚
𝑋𝑌 + 𝑌𝑍 = 5 𝑐𝑚 + 3 𝑐𝑚 = 8 𝑐𝑚
As 8 > 6.8 𝑐𝑚
𝑍𝑋 = 𝑋𝑌 + 𝑌𝑍
𝐻𝑒𝑛𝑐𝑒 , 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑎𝑛𝑦 𝑡𝑤𝑜 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝑠𝑖𝑑𝑒 𝑖𝑠 𝑔𝑟𝑒𝑎𝑡𝑒𝑟 𝑡ℎ𝑎𝑛 𝑡ℎ𝑒 𝑠𝑢𝑚𝑜𝑓 𝑡ℎ𝑒 𝑡ℎ𝑖𝑟𝑑 𝑠𝑖𝑧𝑒
10. Draw five triangles, then count their sides. In each instance, verify that the third side is always smaller than the sum of the lengths of any two sides.
10. Draw five triangles, then count their sides. In each instance, verify that the third side is always smaller than the sum of the lengths of any two sides.
Explanation:
Case 4. 𝑀𝑁𝑆 , 𝑀𝑁 = 2.7 𝑐𝑚 in the triangle
𝑁𝑆 = 4 𝑐𝑚
𝑀𝑆 = 4.7 𝑐𝑚
𝑁𝑆 + 𝑀𝑁 = 2.7 𝑐𝑚 + 4 𝑐𝑚 = 6.7 𝑐𝑚
As 6.7 𝑐𝑚 > 4.7 𝑐𝑚
𝑀𝑆 = 𝑀𝑁 + 𝑁𝑆
𝐻𝑒𝑛𝑐𝑒 , 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑎𝑛𝑦 𝑡𝑤𝑜 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝑠𝑖𝑑𝑒 𝑖𝑠 𝑔𝑟𝑒𝑎𝑡𝑒𝑟 𝑡ℎ𝑎𝑛 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑡ℎ𝑖𝑟𝑑 𝑠𝑖𝑑𝑒
11. 𝑫𝒓𝒂𝒘 𝒇𝒊𝒗𝒆 𝒕𝒓𝒊𝒂𝒏𝒈𝒍𝒆𝒔, then count 𝒕𝒉𝒆𝒊𝒓 𝒔𝒊𝒅𝒆𝒔. 𝑰𝒏 𝒆𝒂𝒄𝒉 instance, verify that the third side is always smaller than 𝒕𝒉𝒆 𝒔𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒍𝒆𝒏𝒈𝒕𝒉𝒔 𝒐𝒇 𝒂𝒏𝒚 𝒕𝒘𝒐 𝒔𝒊𝒅𝒆𝒔.
11. 𝑫𝒓𝒂𝒘 𝒇𝒊𝒗𝒆 𝒕𝒓𝒊𝒂𝒏𝒈𝒍𝒆𝒔, then count 𝒕𝒉𝒆𝒊𝒓 𝒔𝒊𝒅𝒆𝒔. 𝑰𝒏 𝒆𝒂𝒄𝒉 instance, verify that the third side is always smaller than 𝒕𝒉𝒆 𝒔𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒍𝒆𝒏𝒈𝒕𝒉𝒔 𝒐𝒇 𝒂𝒏𝒚 𝒕𝒘𝒐 𝒔𝒊𝒅𝒆𝒔.
Explanation:
𝐶𝑎𝑠𝑒 5 . 𝑖𝑛 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝐾𝐿𝑀, 𝐾𝐿 = 3.5 𝑐𝑚
𝐿𝑀 = 3.5 𝑐𝑚
𝐾𝑀 = 3.5 𝑐𝑚
𝐾𝐿 + 𝐿𝑀 = 3.5 𝑐𝑚 + 3.5 𝑐𝑚 = 7 𝑐𝑚
7 cm are greater than 3.5 𝑐𝑚.
∴ 𝐾𝐿 + 𝐿𝑀 = 𝐾𝑀
𝐻𝑒𝑛𝑐𝑒, 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑎𝑛𝑦 𝑡𝑤𝑜 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝑠𝑖𝑑𝑒𝑠 𝑖𝑠 𝑔𝑟𝑒𝑎𝑡𝑒𝑟 𝑡ℎ𝑎𝑛 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑡ℎ𝑖𝑟𝑑 𝑠𝑖𝑧𝑒
12. How many fractions of a clockwise revolution does the hour hand of a clock make when it moves from one hour to the next?
(𝒂) 𝟑 𝒕𝒐 𝟗
12. How many fractions of a clockwise revolution does the hour hand of a clock make when it moves from one hour to the next?
(𝒂) 𝟑 𝒕𝒐 𝟗
Explanation:
We know that the hour hand will revolve 3600 times in one complete clockwise rotation.
(𝒂) Whenever 𝑡ℎ𝑒 ℎ𝑜𝑢𝑟 ℎ𝑎𝑛𝑑 travels clockwise 𝑓𝑟𝑜𝑚 3 𝑡𝑜 9 , it revolves by
2 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒𝑠 𝑜𝑟 1800
𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛 = 1800 = 1/2
3600
13.Where will the clock hand stop if it
(a) starts at 12 and makes 1/2 of a clockwise revolution?
13.Where will the clock hand stop if it
(a) starts at 12 and makes 1/2 of a clockwise revolution?
Explanation:
We knew that hour hand will spin 3600 times 𝑖𝑛 𝑜𝑛𝑒 𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑒 clockwise rotation.
(𝒂) The ℎ𝑜𝑢𝑟 ℎ𝑎𝑛𝑑 𝑜𝑓 𝑎 𝑐𝑙𝑜𝑐𝑘 will spin by 1800 if it starts at 12 and makes 1rotation
2
clockwise.
As a result, 𝑡ℎ𝑒 ℎ𝑜𝑢𝑟 ℎ𝑎𝑛𝑑 on 𝑎 𝑐𝑙𝑜𝑐𝑘 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑎𝑡 at 6.
14. Which side will you be faced if you started facing?
east and make 1/2 of a clockwise revolution?
14. Which side will you be faced if you started facing?
east and make 1/2 of a clockwise revolution?
Explanation:
We shall revolve by 3600 after
𝑜𝑛𝑒 𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑒 𝑟𝑜𝑢𝑛𝑑 𝑖𝑛 𝑒𝑖𝑡ℎ𝑒𝑟 𝑐𝑙𝑜𝑠𝑒𝑤𝑖𝑠𝑒 𝑜𝑟 𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛, and two nearby directions are 900 or one-fourth of a complete rotation apart.
(a) 𝑖𝑓 𝑤𝑒 begin 𝑓𝑎𝑐𝑖𝑛𝑔 𝑒𝑎𝑠𝑡 𝑎𝑛𝑑 rotate one-half 𝑜𝑓 𝑎 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑐𝑙𝑜𝑐𝑘 𝑤𝑖𝑠𝑒,
𝑤𝑒 𝑤𝑖𝑙𝑙 𝑓𝑎𝑐𝑒 𝑤𝑒𝑠𝑡
15. Which side will you be faced if you started facing?
(1) east and make 𝟏⁄𝟏/𝟐 of a clockwise revolution?
15. Which side will you be faced if you started facing?
(1) east and make 𝟏⁄𝟏/𝟐 of a clockwise revolution?
Explanation:
𝑖𝑓 𝑤𝑒 begin 𝑓𝑎𝑐𝑖𝑛𝑔 𝑒𝑎𝑠𝑡 𝑎𝑛𝑑 rotate 𝑜𝑛𝑒 − ℎ𝑎𝑙𝑓 𝑜𝑓 𝑎 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑐𝑙𝑜𝑐𝑘 𝑤𝑖𝑠𝑒,
𝑤𝑒 𝑤𝑖𝑙𝑙 𝑓𝑎𝑐𝑒 𝑤𝑒𝑠𝑡
16. If you stand facing (a) east and turn clockwise to face north, what portion of a revolution have you completed?
16. If you stand facing (a) east and turn clockwise to face north, what portion of a revolution have you completed?
Explanation:
By rotating one full revolution either clockwise or anti-clockwise, we shall rotate by 3600, and two nearby directions are 900 or one-fourth of a complete revolution apart (a) If we start facing East and turn clockwise to face North, we must make three-quarters of a revolution.
17. What is the shape?
Is that your instrument case?
Is it a brick?
Is it a matchbox?
Is it a road roller?
A tasty laddu?
17. What is the shape?
Is that your instrument case?
Is it a brick?
Is it a matchbox?
Is it a road roller?
A tasty laddu?
Explanation:
An instrument box has a cuboid form.
A brick has a cuboid form.
A match box has a cuboid form.
A road roller has a cylindrical form.
A delicious laddu has a spherical form.
18. Complete the following:
Provide two additional instances of each shape.
18. Complete the following:
Provide two additional instances of each shape.
Explanation:
Examples of cones are an ice cream cone and a birthday hat.
Spheres include cricket balls and tennis balls.
Examples of cylinders include a road roller and a lawn roller.
A book and a brick are both cuboids.
Examples of pyramids include a diamond and Egypt's pyramids.
19. A diagonal is a line segment that connects any two polygon vertices but is not a side of the polygon. Sketch a rough outline of a pentagon and its diagonals.
19. A diagonal is a line segment that connects any two polygon vertices but is not a side of the polygon. Sketch a rough outline of a pentagon and its diagonals.
Explanation:
The diagonals in the illustration are AC, AD, BD, BE, and CE.
20. Create a rough drawing of an octagon. (If desired, use squared paper). Create a rectangle by connecting exactly four of the octagon's vertices.
20. Create a rough drawing of an octagon. (If desired, use squared paper). Create a rectangle by connecting exactly four of the octagon's vertices.
Explanation:
The image below shows a regular octagon with a rectangle formed by connecting four of its vertices.
21. Create a rough drawing of a regular hexagon. Draw a triangle by connecting any three of its vertices. Determine the sort of triangle you drew.
21. Create a rough drawing of a regular hexagon. Draw a triangle by connecting any three of its vertices. Determine the sort of triangle you drew.
Explanation:
We may make an isosceles triangle by connecting three vertices of a hexagon, as seen in the image below.
Also Read: Chapter 6: Integers
