Category: Class 8

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Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 4 Statistics Ex 4.3

Miscellaneous and Practice Problems

Question 1.
Draw a pie chart for the given table.

Solution:
Converting the area in percentage into components parts of 360°, we have.

Question 2.
The data on modes of transport used by the students to come to school are given below. Draw a pie chart for the data.

Solution:
Converting the percentage into components parts of 360°, we have

Mode of Transport by students.

Question 3.
Draw a histogram for the given frequency distribution.

Solution:
The given distribution is discontinuous.
Lower boundary = lower limit — \(\frac { 1 }{ 2 } \) (gap between the adjacent class interval)
= 41 – \(\frac { 1 }{ 2 } \) (1) = 40.5
Upper boundary = Upper limit + \(\frac { 1 }{ 2 } \) (gap between the adjacent class interval)
= 45 + \(\frac { 1 }{ 2 } \) (1) = 45.5
Now continuous frequency table is as below

Question 4.
Draw a histogram and the frequency polygon in the same diagram to represent the following data.

Solution:
The given distribution is discontinuous.
Lower boundary = lower limit – \(\frac { 1 }{ 2 } \) (gap between the adjacent class interval)
= 50 – \(\frac { 1 }{ 2 } \) (1) = 49.5
Upper boundary = Upper limit + \(\frac { 1 }{ 2 } \) (gap between the adjacent class interval)
= 55 + \(\frac { 1 }{ 2 } \) (1) = 55.5
∴ The continuous frequency table is as below.

Question 5.
The daily income of men and women is given below, draw a separate histogram for men and women.

Solution:
The given data is continuous frequency distribution. So we take Income in X axis and No. of men in Y axis.

Now we consider the number of women and their income we have.

Challenging Problems

Question 1.
Form a continuous frequency distribution table and draw histogram from the following data.

Solution:
Converting into continuous distribution we have

Question 2.
A rupee spent in a cloth manufacturing company is distributed as follows. Represent this in a pie chart.

Solution:
1 Rupee = 100 paise.

Expenditure of a cloth manufacturing company.

Question 3.
Draw a histogram for the following data.

Solution:
Since mid values are given, the given distributors is discontinuous.
Lower boundary = lower limit — \(\frac { 1 }{ 2 } \) (gap between the adjacent class interval)
= 15 – \(\frac { 1 }{ 2 } \) (10) = 10
Upper boundary = Upper limit + \(\frac { 1 }{ 2 } \) (gap between the adjacent class interval)
15 + \(\frac { 1 }{ 2 } \) (10) = 20
The continuous distribution will be as follows.

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Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 4 Statistics Ex 4.2

Question 1.
Which of the following data can be represented in a histogram?

1. The number of mountain climbers in the age group 20 to 60.
2. Production of cycles in different years.
3. The number of students in each class of a school.
4. The number votes polled from 7 am to 6 pm in an election.
5. The wickets fallen from 1 over to 50th over in a one day cricket match.

Answers:

1. Yes
2. No
3. No
4. Yes
5. Yes

Question 2.
Fill in the blanks

1. The area of the rectangles are proportional to the _______ given.
2. The total area of the histogram is _______ to the total frequency of the given data.
3. _______ is a graphical representation of continuous frequency distribution with rectangles.
4. Histogram is a graphical representation of _______ data.

Answers:

1. frequency
2. proportional
3. Histogram
4. grouped

Question 3.
In a village, there are 570 people who have cell phones. An NGO survey their cell phone usage. Based on this survey a histogram is drawn. Answer the following questions.

(i) How many people use the cell phone for less than 3 hours?
Solution:
330 people (110+ 220)

(ii) How many of them use the cell phone for more than 5 hours?
Solution:
150 of them (100 + 50)

(iii) Are people using cell phone for less than 1 hour?
Solution:
No.

(iv) Give your suggestions on the data.
Solution:
People should minimise the usage of cell phones.

Question 4.
Draw a histogram for the following data.

Solution:
The given data is continuous frequency distribution taking class intervals on X axis and No. of students on Y-axis, the histogram is given below.

Question 5.
Construct a histogram from the following distribution of total marks of 40 students in a class.

Solution:
The given distribution is continuous taking marks on X axis and No. of students on Y-axis the histogram is constructed.

Question 6.
The distribution of heights (in cm) of 100 people is given below. Construct a histogram and the frequency polygon imposed on it.

Solution:
The given distribution is discontinuous.
Converting into continuous distribution we have
Lower boundary = lower limit – \(\frac { 1 }{ 2 } \) (gap between the adjacent class interval)
= 125 – \(\frac { 1 }{ 2 } \) (1) = 124.5
Upper boundary = Upper limit + \(\frac { 1 }{ 2 } \) (gap between adjacent class interval)
= 135 + \(\frac { 1 }{ 2 } \) = 135.5
∴ The new frequency table is

+

Question 7.
In a study of dental problem, the following data were obtained.

Represent the above data by a frequency polygon.
Solution:
Finding the midpoints of the class interval we get.

The points to be plotted are A (5,5), B (15, 13), C (25, 25), D (35, 14), E (45, 30), G (65, 43.), H (75, 50) to obtain the frequency polygon ZABCDEFGHI.
Where I imagined class between 80 and 90.

Question 8.
The marks obtained by 50 students in Mathematics are given below (i) Make a frequency distribution table taking a class size of 10 marks (ii) Draw a histogram and a frequency polygon.

Solution:
Maximum marks obtained = 89
Minimum marks obtained = 08
Range = Maximum marks – Minimum marks = 89 – 08 = 81
Taking the class size = 10, then

Now we have the continuous frequency table.

We will draw the histogram taking class interval in x axis and frequency in y axis as follows.

Objective Type Questions

Question 1.
Data is a collection of _______
(a) numbers
(b) words
(c) measurements
(d) all the three
Answer:
(d) all the three

Question 2.
The number of times an observation occurs in the given data is called _______
(a) tally marks
(b) data
(c) frequency
(d) none of these
Answer:
(c) frequency

Question 3.
The difference between the largest value and the smallest value of the given data is _______
(a) range
(b) frequency
(c) variable
(d) none of these
Answer:
(a) range

Question 4.
The data that can take values between a certain range is called _______
(a) ungrouped
(b) grouped
(c) frequency
(d) none of these
Answer:
(b) grouped

Question 5.
Inclusive series is a _______ series
(a) continuous
(b) discontinuous
(c) both
(d) none of these
Answer:
(b) discontinuous

Question 6.
In a class interval the upper limit of one class is the lower limit of the other class. This is _______ series.
(a) Inclusive
(b) exclusive
(c) ungrouped
(d) none of these
Answer:
(b) exclusive

Question 7.
The graphical representation of un grouped data is _______
(a) histogram
(b) frequency polygon
(c) pie chart
(d) all the three

Question 8.
Histogram is a graph of a _______ frequency distribution.
(a) continuous
(b) discontinuous
(c) discrete
(d) none of these
Answer:
(a) continuous

Question 9.
A _______ is a line graph for the graphical representation of the continuous frequency distribution.
(a) frequency polygon
(b) histogram
(c) pie chart
(d) bar graph
Answer:
(a) frequency polygon

Question 10.
The graphical representation of grouped data is _______
(a) bar graph
(b) pictograph
(c) pie chart
(d) histogram
Answer:
(d) histogram

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Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 5 Information Processing Intext Questions

Exercise 5.1
Activity 1 (Text book Page No. 104)

Question 1.
Consider that you are going to a store with your total budget of ₹ 220 to buy things without changing the quantity of the items given in the list below with the following conditions.

Conditions :
(i) First you have to complete the price list given.
(ii) You have to buy three items as per the given price list but within your budget ₹ 220.
(iii) You won’t carry exceeding 5kg because you have to walk home carrying them, so they cannot be bulky.

Now, answer the following questions:
1. In how many ways can you buy your items? Complete the price lists given below. One is done for you.
2. Which one is the best purchase price list and why?

Answer:

Try this (Text book Page No. 106)

The teacher divides the class into four groups and setup a mock market in the class room and ask the students to involve in role play as two groups of businessmen and two groups of consumers. Consumers have to buy products at different shops and prepare a price list.

The two supermarkets in which the two groups buy are Star Food Mart and Super Provisions. Th is week they each have got a special deal on some products. At Star Food Mart, you can buy items at discount prices. At Super Provisions, there are some “BUY ONE GET ONE” deals. Flave a look at their deal:

Now, answer the following questions.

I. Here is your shopping list:
4 bottles of Protein Milk (200 ml size), 2 packets of Peanut candies(200 gm), 1 packet of Chocolate biscuits and 1 packet of Badam nuts (500 gm)
(i) If you buy all the items in one shop, where will you get the best price?
(ii) If you buy the items from different shops, how will you do it to spend the least amount of money?

II. You have ₹ 1000/- to spend to buy the following shopping list:
6 bottles of Protein Milk (200 ml size), 3 packets of Peanut candies (200 gm), 3 packets of Chocolate biscuits and 1 packet of Badam nuts (250 gm).

(i) How can you do this so that you don’t go over your budget amount ₹ 1000?
(ii) Which shop offers you the best value for money on each item?
(iii) Is the “BUY ONE GET ONE” deal at Super Provisions the same as “50% off’ deal?
Solution:
Prices of Star Food Mart.


(i) In star Food Mart
(ii) Badam nuts from super provisions and other items from Star Food Mart.

II. Comparing two stores.

(i) We can buy them from any one shop.
(ii) Star Food Mart.
(iii) Yes

Exercise 5.2
Try this (Text book Page No. 117)

Question 1.
If Kumaran cut the woods using first fit method then find the wastage pieces.
Solution:

Waste pieces = 2 feet + 4 feet + 2 feet = 8 feet

Activity 1 (Text book Page No. 117)

Question 1.
Seva Sangam has decided to deliver some aids to flood victims via lorries with a maximum capacity of 5000kg. All of these items that are given below are to be packed and sent in the lorry.

Solution:

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Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 4 Statistics Intext Questions

Exercise 4.1
Try These (Text book Page no. 77)

Question 1.
Arrange the given data in ascending and descending order:
9, 34, 4, 13, 42, 10, 25, 7, 31, 4, 40
Solution:
Ascending order: 4, 4, 7, 9, 10, 13, 25, 31, 34, 40, 42.
Descending order : 42, 40, 34, 31, 25, 13, 10, 9, 7, 4, 4

Question 2.
Find the range of the given data : 53, 42, 61, 9, 39, 63, 14, 20, 06, 26, 31, 4, 57
Solution:
Ascending order of the given data:
4, 6, 9, 14, 20, 26, 31, 39, 42, 53, 57, 61, 63
Here largest value = 63
Smallest value = 4
∴ Range = Largest value – smallest value = 63 – 4 = 59

Think (Text book Page no. 79)

How will you change the given series as continuous series
15 – 25
28 – 38
41 – 51
54 – 64
Solution:
Given series
15 – 25
28 – 38
41 – 51
54 – 64
Difference in the gap = 28 – 25 = 3
Here half of the gap = \(\frac{1}{2}\)(3) = 1.5
∴ 1.5 is the adjustment factor. So we subtract 1.5 from the lower limit and add 1.5 to the upper limit to make it as a continuous series.

Think (Text book Page no. 80)

If we want to represent the given data by 5 classes, then how shall we find the interval?
Solution:
We can find the class size by the formula
Number of class intervals = \(\frac{Range}{Class size}\)

Try These (Text book Page no. 82)

Question 1.
Prepare a frequency table for the data : 3, 4, 2, 4, 5, 6, 1, 3, 2, 1, 5, 3, 6, 2, 1, 3, 2, 4
Solution:
Ascending order of the given data.
1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 5, 5, 6, 6
The distribution table:

∴ Frequency Table:

Question 2.
Prepare a grouped frequency table for the data :
10, 9, 3, 29, 17, 34, 23, 20, 39, 42, 5, 12, 19, 47, 18, 19, 27, 7, 13, 40, 38, 24, 34, 15, 40
Largest value = 47
Smallest value = 3
Range = Largest value – Smallest value = 47 – 3 = 44
Suppose we take class size as 10, then Number of class intervals possible
= \(\frac{Range}{Class size}\) = \(\frac{44}{10}\) = 4.4
\(\tilde { – } \) 5

Exercise 4.2
Think (Text book Page no. 94)

When joining two adjacent midpoints w ithout using a ruler, can you get a polygon?
Solution:
No, because it may be curved lines and they are not considered as polygons.

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Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 3 Geometry Additional Questions

Question 1.
In ΔPQR, PS is a median. If QS = 7 cm find the length of QR?
Solution:
PS is the median ⇒ S is the midpoint of QR

Given
QS = 7 cm
∴ SR = 7 cm
QR = QS + SR
= 7 + 7 = 14cm

Question 2.
In ΔABC, G is the centroid. If AD = 6 cm, BC = 4 cm and BE = 9 cm find the perimeter of ΔBDG.
Solution:
In ΔABC, G is the centroid.

If AD = 6 cm
⇒ GD = \(\frac{1}{3}\) of AD = \(\frac{1}{3}\) (6)
BE = 9 cm
⇒ BG = \(\frac{2}{3}\) of BE = \(\frac{2}{3}\) (9) = 6cm
Also D is the midpoint of BC ⇒ BD
= \(\frac{1}{1}\) of BC = \(\frac{1}{2}\) (4) = 2cm
∴ Perimeter of ΔBDG = BD + GD + BG = 2 + 2 + 6 = 10 cm

Question 3.
Construct a rhombus FISH with FS = 8 cm and ∠F = 80°
Solution:
Given FS = 8 cm and ∠F = 80°

Steps :
(i) Drawn a line segment FS = 7 cm.
(ii) At F, made ∠SFX = ∠SFY = 40° on either side of FS.
(iii) At S, made ∠FSP = ∠FSQ = 40° on either side of FS
(iv) Let FX and SP cut at H and FY and SQ cut at I.
(v) FISH is the required rhombus

Calculation of Area:
Area of the rhombus FISH = \(\frac{1}{2}\) × d₁ × d₂ sq.units = \(\frac{1}{2}\) × 5.9 × 7 cm² = 20.65 cm²

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Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 2 Life Mathematics Additional Questions

Question 1.
120 men had food for 200 days. After 5 days 30 men left the camp. How long will the remaining food last.
Solution:
Since 30 men left after 5 days, the remaining food is sufficient for 120 men for 195 days. Suppose the remaining food lasts for x days for the remaining 90 men.
We have

More men means less days the food lasts
∴ It is inverse proportion
120 : 90 = x : 195
Product of extremes = Product of means
120 × 195 = 90 × x
x = \(\frac{120×195}{90}\)
x = 90
x = 260.
∴ Remaining food last for 260 days.

Question 2.
15 men earn Rs 900 in 5 days, how much will 20 men earn in 7 days?
Solution:
In one day 15 men earn Rs 900
In one day 15 men earn \(\frac{900}{5}\) = Rs 180
In one day 1 men earn \(\frac{180}{5}\) = Rs 12
∴ 1 men earn in 7 days = 12 × 7 = Rs 84
∴ 20 men earn in 7 days = 84 × 20 = 1680

Question 3.
A and B together can do a piece of work in 10 days, B and C can do the same work together in 12 days, A and C can do together in 15 days. How long will it take to complete the work working three of them altogether?
Solution:
(A + B)’s 1 day’s work = \(\frac{1}{10}\)……….(1)
(B + C)’s 1 day’s work = \(\frac{1}{12}\)……….(2)
(A + C)’s 1 day’s work = \(\frac{1}{15}\)……….(3)
(l) + (2) + (3) ⇒
[A + B + B + C + A + C]’s 1 day work = \(\frac{1}{10}\) + \(\frac{1}{12}\) + \(\frac{1}{15}\)
(2A + 2B + 2C)’s 1 day work = \(\frac{6 + 5 + 4}{60}\)
2(A + B + C)’s 1 day work = \(\frac{15}{60}\)
(A + B + C)’s 1 day’s work = \(\frac{1}{4 × 2}\) = \(\frac{1}{8}\)
∴ A + B + C work together to finish the work in 8 days.

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Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 3 Geometry Ex 3.3

Question 1.
Construct the following rhombuses with the given measurements and also find their area.
(i) FACE, FA = 6 cm and FC = 8 cm
Solution:
Given FA = 6 cm and FC = 8cm

Steps :
(i) Drawn a line segment FA = 6 cm.
(ii) With F and A as centres, drawn arcs of radii 8 cm and 6 cm respectively and let them cut at C.
(iii) Joined FC and AC.
(iv) With F and C as centres, drawn arcs of radius 6 cm each and let them cut at E. Joined FE and EC.
(v) FACE is the required rhombus.

Calculation of Area :
Area of the rhombus = \(\frac{1}{2}\) × d₁ × d₂ sq.units = \(\frac{1}{2}\) × 8 × 9 sq.units = 36 cm²

(ii) RACE, RA = 5.5 cm and AE = 7 cm
Solution:
Given RA = 5.5 cm and AE = 7 cm

Steps :
(i) Drawn a line segment RA = 5.5 cm.
(ii) With R and A as centres, drawn arcs of radii 5.5 cm and 7 cm respectively and let them cut at E.
(iii) Joined RE and AE.
(iv) With E and A as centres, drawn arcs of radius 5.5 cm each and let them cut at C.
(v) Joined AC and EC.
(vi) RACE is the required rhombus.

Calculation of Area :
Area of the rhombus = \(\frac{1}{2}\) × d₁ × d₂ sq.units = \(\frac{1}{2}\) × 7 × 8.5 cm² = 29.75 cm²

(iii) CAKE, CA = 5 cm and ∠A = 65°
Solution:
Given CA = 5 cm and ∠A = 65°

(i) Drawn a line segment CA = 5 cm.
(ii) At A on AC, made ∠CAX = 65°
(iii) With A as centre, drawn arc of radius 5 cm. Let it cut AX at K.
(iv) With K and C as centres, drawn arcs of radius 5 cm each and let them cut at E. Joined KE and CE.
(v) CAKE is the required rhombus.

Calculation of Area :
Area of the rhombus = \(\frac{1}{2}\) × d₁ × d₂ sq.units = \(\frac{1}{2}\) × 5.4 × 8.5 cm² = 22.95 cm²

(iv) MAKE, MA= 6.4 cm and ∠M = 80°
Solution:
Given MA = 6.4 cm and ∠M = 80°

Steps :
(i) Drawn a line segment MA = 6.4 cm.
(ii) At M on MA, made ∠AMX = 80°
(iii) With M as centres, drawn arc of radius 6.4 cm. Let it cut MX at E.
(iv) With E and A as centres, drawn arcs of radius 6.4 cm each and let them cut at K.
(v) Joined EK and AK.
(vi) MAKE is the required rhombus.

Calculation of Area :
Area of the rhombus = \(\frac{1}{2}\) × d₁ × d₂ sq.units = \(\frac{1}{2}\) × 8.2 × 9.8 cm² = 40.18 cm²

(v) LUCK, LC = 7.8 cm and UK = 6 cm
Solution:
Given LC = 7.8 cm and UK = 6 cm

Steps :
(i) Drawn a line segment LC = 7.8 cm.
(ii) Drawn the perpendicular bisector XY to LC. Let it cut LC at ‘O’
(iii) With O as centres, drawn arc of radius 3 cm on either side of O which cut OX at K and OY at U.
(iv) Joined LU, UC, CK and LK.
(v) LUCK is the required rhombus.

Calculation of Area :
Area of the rhombus = \(\frac{1}{2}\) × d₁ × d₂ sq.units = \(\frac{1}{2}\) × 7.8 × 6 cm² = 23.4 cm²

(vi) DUCK, DC = 8 cm and UK = 6 cm
Solution:
Given DC = 8 cm and UK = 6 cm

Steps :
(i) Drawn a line segment DC = 8 cm.
(ii) Drawn the perpendicular bisector XY to DC. Let it cut DC at ‘O’
(iii) With O as centres, drawn arc of radius 3 cm on either side of O which cut OX at U and OYat K.
(iv) Joined DK, KC, CU and DU.
(v) DUCK is the required rhombus.

Calculation of Area :
Area of the rhombus = \(\frac{1}{2}\) × d₁ × d₂ sq.units = \(\frac{1}{2}\) × 8 × 6 cm² = 24 cm²

(vii) PARK, PR = 9 cm and ∠P = 70°
Solution:
Given PR = 9 cm and ∠P = 70°

Steps :
(i) Drawn a line segment PR = 9 cm.
(ii) At P, made ∠RPX = ∠RPY = 35° on either side of PR.
(iii) At R, made ∠PRQ = ∠PRS = 35° on either side of PR
(iv) Let PX and RQ cut at A and PY and RS at K.
(v) PARK is the required rhombus

Calculation of Area :
Area of the rhombus = \(\frac{1}{2}\) × d₁ × d₂ sq.units = \(\frac{1}{2}\) × 9 × 6.2 cm² = 27.9 cm²

(viii) MARK, AK =7.5 cm and ∠A = 80°
Solution:
Given AK = 7.5 cm and ∠A = 80°

(i) Drawn a line segment AK = 7.5 cm.
(ii) At A, made ∠KAX = ∠KAY = 40° on either side of AK.
(iii) At K, made ∠AKP = ∠AKQ = 40° on either side of AK
(iv) Let AX and KP cut at M and AY and KQ at R.
(v) MARK is the required rhombus

Calculation of Area :
Area of the rhombus = \(\frac{1}{2}\) × d₁ × d₂ sq.units = \(\frac{1}{2}\) × 7.5 × 6.4 cm² = 24 cm²

Question 2.
(i) Construct the following rectangles with the given measurements and also find their area.
(i) HAND, HA = 7 cm and AN = 4 cm
Solution:
Given HA = 7 cm and AN = 4 cm

Steps :
(i) Drawn a line segment HA = 7 cm.
(ii) At H, constructed HX ⊥ HA.
(iii) With H as centre, drawn an arc of radius 4 cm and let it cut at HX at D.
(iv) With A and D as centres, drawn arcs of radii 4 cm and 7 cm respectively and let them cut at N.
(v) Joined AN and DN.
(vi) HAND is the required rectangle.

Calculation of Area :
Area of the rectangle HAND = l × b sq.units = 7 × 4 cm² = 28 cm²

(ii) SAND, SA = 5.6 cm and SN = 4.4 cm
Solution:
Given SA = 5.6 cm and SN = 4.4 cm

Steps :
(i) Drawn a line segment SA = 5.6 cm.
(ii) At S, constructed SX ⊥ SA.
(iii) With S as centre, drawn an arc of radius 4.4 cm and let it cut at SX at D.
(iv) With A and D as centres, drawn arcs of radii 4.4 cm and 5.6 cm respectively and let them cut at N.
(v) Joined DN and AN.
(vi) SAND is the required rectangle.

Calculation of Area :
Area of the rectangle SAND = l × b sq.units = 5.6 × 4.4 cm² = 26.64 cm²

(iii) LAND, LA = 8 cm and AD = 10 cm
Solution:
Given LA = 8 cm and AD = 10 cm

Steps :
(i) Drawn a line segment LA = 8 cm.
(ii) At L, constructed LX ⊥ LA.
(iii) With A as centre, drawn an arc of radius 10 cm and let it cut at LX at D.
(iv) With A as centre and LD as radius drawn an arc. Also with D as centre and LA as radius drawn another arc. Let then cut at N.
(v) Joined DN and AN.
(vi) LAND is the required rectangle.

Calculation of Area :
Area of the rectangle LAND = l × b sq.units = 8 × 5.8 cm² = 46.4 cm²

(iv) BAND, BA = 7.2 cm and BN = 9.7 cm
Solution:
Given = 7.2 cm and BN = 9.7 cm

Steps :
(i) Drawn a line segment BA = 7.2 cm.
(ii) At A, constructed NA ⊥ AB.
(iii) With B as centre, drawn an arc of radius 9.7 cm and let it cut at AX at N.
(iv) With B as centre and AN as radius drawn an arc. Also with N as centre and BA as radius drawn another arc. Let then cut at D.
(v) Joined ND and BD.
(vi) BAND is the required rectangle.

Calculation of Area :
Area of the rectangle BAND = l × b sq.units = 7.2 × 6.7 cm² = 48.24 cm²

Question 3.
Construct the following squares with the given measurements and also find their area.
(i) EAST, EA = 6.5 cm
Solution:
Given side = 6.5 cm

Steps :
(i) Drawn a line segment EA = 6.5 cm.
(ii) At E, constructed EX ⊥ EA.
(iii) With E as centre, drawn an arc of radius 6.5 cm and let it cut EX at T.
(iv) With A and T as centre drawn an arc of radius 6.5 cm each and let them cut at S.
(v) Joined TS and AS.
(vi) EAST is the required square.

Calculation of Area :
Area of the square EAST = a² sq.units = 6.5 × 6.5 cm² = 42.25 cm²

(ii) WEST, ST = 6 cm
Solution:
Given side of the square = 6 cm

Steps :
(i) Drawn a line segment ST = 6 cm.
(ii) At S, constructed SX ⊥ ST.
(iii) With S as centre, drawn an arc of radius 6 cm and let it cut SX at E.
(iv) With E and T as centre drawn an arc of radius 6 cm each and let them cut at W.
(v) Joined TW and EW.
(vi) WEST is the required square.

Calculation of Area :
Area of the square WEST = a² sq.units = 6 × 6 cm² = 36 cm²

(iii) BEST, BS = 7.5 cm
Solution:
Given diagonal = 7.5 cm

Steps :
(i) Drawn a line segment BS = 7.5 cm.
(ii) Drawn the perpendicular bisector XY to BS. Let it bisect BS at O.
(iii) With O as centre, drawn an arc of radius 3.7 cm on either side of O which cut OX at T and OY at E
(iv) Joined BE, ES, ST and BT.
(v) BEST is the required square.

Calculation of Area :
Area of the square BEST = a² sq.units = 5.3 × 5.3 cm² = 28.09 cm²

(iv) REST, ET = 8 cm
Solution:
Given diagonal = 8 cm

Steps:
(i) Drawn a line segment ET = 8 cm.
(ii) Drawn the perpendicular bisector XY to ET. Let it bisect ET at O.
(iii) With O as centre, drawn an arc of radius 4 cm on either side of O which cut OX at R and OY at S
(iv) Joined ES, ST, TR and ER.
(v) REST is the required square.

Calculation of Area :
Area of the square REST = a² sq.units = 5.7 × 5.7 cm² = 32.49 cm²

Students can Download Maths Chapter 3 Geometry Ex 3.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 3 Geometry Ex 3.2

Miscellaneous and Practice Problems

Question 1.
Identify the centroid of ΔPQR.

Solution:
In ΔPQR, PT = TR ⇒ QT is a median from vertex Q.
QS = SR ⇒ PS is a median from vertex P.
QT and PS meet at W and therefore W is the centroid of ΔPQR.

Question 2.
Name the orthocentre of ΔPQR.

Solution:
∠P = 90°
This is a right triangle
∴ orthocentre = P [∴ In right triangle orthocentre is the vertex containing 90°]

Question 3.
In the given figure, A is the midpoint of YZ and G is the centroid of the triangle XYZ. If the length of GA is 3 cm, find XA.

Solution:
Given A is the midpoint of YZ.
∴ ZA = AY
G is the centroid of ΔXYZ centroid divides each median in a ratio 2 : 1 ⇒ XG : GA = 2 : 1
\(\frac{XG}{GA}\) = \(\frac{2}{1}\)
\(\frac{XG}{3}\) = \(\frac{2}{1}\)
XG = 2 × 3
XG = 6 cm
XA = XG + GA
= 6 + 3 ⇒ XA = 9 cm

Challenging Problems

Question 4.
Find the length of an altitude on the hypotenuse of a right angled triangle of legs of length 15 feet and 20 feet.
Solution:

Since ∠B = 90°
Using pythagoras theorem
AC² = AB² + BC² = 20² + 15²
= 400 + 225
AC² = 62⁵
AC² = 25⁵
AC = 25
Area of a triangle ΔABC = \(\frac{1}{2}\) × 15 × 20 = 150 feet²
Again Area of ΔABC = \(\frac{1}{2}\) × AC × BD
150 = \(\frac{1}{2}\) × 25 × BD
BD = \(\frac{2 × 150}{25}\) = \(\frac{300}{25}\)
BD = 12 feet
∴ Length of the altitude on the hypotenuse of the right angled triangle is 12 feet.

Question 5.
If I is the incentre of ΔXYZ, ∠IYZ = 30° and ∠IZY = 40°, find ∠YXZ.

Solution:
Since I is the incentre of ΔXYZ
∠IYZ = 30° ⇒ ∠IYX = 30°
∠IZY = 40° ⇒ ∠IZX = 40°
∴ ∠XYZ = ∠XYI + ∠IYZ = 30° + 30°
∠XYZ = 60°
lll ly ∠XYZ = ∠XZI + ∠IZY = 40° + 40°
∠XYZ = 80°
By angle sum property of a triangle
∠XZY + ∠XYZ + ∠YXZ = 180°
80° + 60° + ∠YXZ = 180°
140° + ∠YXZ = 180°
∠YXZ = 180° – 140°
∠YXZ = 40°

Question 6.
In ΔDEF, DN, EO, FM are medians and point P is the centroid. Find the following.
(i) If DE = 44, then DM = ?
(ii) If PD = 12, then PN = ?
(iii) If DO = 8, then FD = ?
(iv) If OE = 36, then EP = ?

Solution:
Given DN, EO, FM are medians.
∴ FN = EN
DO = FO
EM = DM
(i) If DE = 44, then
DM = \(\frac{44}{2}\) = 22
DM = 22

(ii) If PD = 12, PN = ?
\(\frac{PD}{PN}\) = \(\frac{2}{1}\)
\(\frac{12 }{PN}\) = \(\frac{2}{1}\) ⇒ PN = \(\frac{12}{2}\) = 6
PN = 6

(iii) If DO = 8, then
FD = DO + OF = 8 + 8
FD = 16

(iv) If OE = 36,
then \(\frac{EP}{PO}\) = \(\frac{2}{1}\)
\(\frac{EP}{2}\) = PO
OE = OP + PE
36 = \(\frac{PE}{2}\) + PE
36 = \(\frac{PE}{2}\) + \(\frac{2PE}{2}\)
36 = \(\frac{3PE}{2}\)
PE = \(\frac{36 × 2}{3}\)
PE = 24

Students can Download Maths Chapter 3 Geometry Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 3 Geometry Intext Questions

Exercise 3.1
Think (Text book Page no. 53)

Question 1.
In any acute angled triangle, all three altitudes are inside the triangle. Where will be the orthocentre? In the interior of the triangle or in its exterior?

Solution:
Interior of the triangle.

Question 2.
In any right angled triangle, the altitude perpendicular to the hypotenuse is inside the triangle; the other two altitudes are the legs of the triangle. Can you identify the orthocentre in this case?

Solution:
Vertex containing 90°

Question 3.
In any obtuse angled triangle, the altitude connected to the obtuse vertex is inside the triangle, and the two altitudes connected to the acute vertices are outside the triangle. Can you identify the orthocentre in this case?

Solution:
Exterior of the triangle.

Try These (Text book Page no. 56)

Identify the type of segment required in each triangle:
(median, altitude, perpendicular bisector, angle bisector)

(i) AD = ……….
(ii) l1 = ………..
(iii) BD = …………
(iv) CD = …………
Solution:
(i) AD = Altitude
(ii) l₁ = Perpendicular bisector
(iii) BD = Median
(iv) CD = Angular bisector

Exercise 3.3
Activity 1. (Text book Page no. 60)

Question 1.
A pair of identical 30°-60°-90° set-squares are needed for this activity. Place them as shown in the figure.

1. What is the shape we get? It is a parallelogram.
2. Are the opposite sides parallel?
3. Are the opposite sides equal?
4. Are the diagonals equal?
5. Can you get this shape by using any other pair of identical set-squares?

Solution:

1. It is a parallelogram.
2. Yes
3. Yes
4. no
5. yes

Question 2.
We need a pair of 30°-60°-90° set- squares for this activity. Place them as shown in the figure.

(i) What is the shape we get?
(ii) Is it a parallelogram?
It is a quadrilateral; infact it is a rectangle. (How?)
(iii) What can We say about its lengths of sides, angles and diagonals? Discuss and list them out.
Solution:
(i) Rectangle
(ii) Yes, Opposite sides are equal. All angles = 90°
(iii) Opposite sides are equal.
All angles are equal and are = 90°.
Diagonals are equal

Question 3.
Repeat the above activity, this time with a pair of 45°-450-90° set-squares.

(i) How does the figure change now? Is it a parallelogram? It becomes a square! (How did it happen?)
(ii) What can we say about its lengths of sides, angles and diagonals? Discuss and list them out.
(iii) How does it differ from the list we prepared for the rectangle?
Solution:
(i) All sides are equal
(ii) All sides are equal
All angles = 90°
Diagonals equal
(iii) All sides are equal.
Diagonals bisects each other.

Question 4.
We again use four identical 30°-60°-90° set- squares for this activity.
Note carefully how they are placed touching one another.

(i) Do we get a parallelogram now?
(ii) What can we say about its lengths of sides, angles and diagonals?
(iii) What is special about their diagonals?
Solution:
(i) Yes
(ii) All sides equal.
(iii) Diagonals bisects perpendicularly.

Try These (Text book Page no. 62)

Question 1.
Say True or False:
(a) A square is a special rectangle.
(b) A square is a parallelogram.
(c) A square is a special rhombus.
(d) A rectangle is a parallelogram
Solution:
(a) True
(b) True
(c) True
(d) True

Question 2.
Name the quadrilaterals
(a) Which have diagonals bisecting each other.
(b) In which the diagonals are perpendicular bisectors of each other.
(c) Which have diagonals of different lengths.
(d) Which have equal diagonals.
(e) Which have parallel opposite sides.
(f) In which opposite angles are equal.
Solution:
(a) Square, rectangle, parallelogram, rhombus.
(b) Rhombus and square.
(c) Parallelogram and Rhombus
(d) Rectangle, square.
(e) Square, Rectangle, Rhombus, parallelogram.
(f) Square, rectangle, rhombus, parallelogram

Question 3.
Two sticks are placed on a ruled sheet as shown. What figure is formed if the four corners of the sticks are joined?
(a)
Two unequal sticks. Placed such that their midpoints coincide.
Solution:
Parallelogram

(b)
Two equal sticks. Placed such that their midpoints coincide.
Solution:
Rectangle

(c)
Two unequal sticks. Placed intersecting at mid points perpendicularly.
Solution:
Rhombus

(d)
Two equal sticks. Placed intersecting at mid points perpendicularly.
Solution:
Square

(e)
Two unequal sticks. Tops are not on the same ruling. Bottoms on the same ruling. Not cutting at the mid point of either.
Solution:
Quadrilateral

(f)
Two unequal sticks. Tops on the same ruling. Bottoms on the same ruling. Not necessarily cutting at the mid point of either.
Solution:
Trapezium

Students can Download Maths Chapter 4 Statistics Ex 4.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 4 Statistics Ex 4.1

Question 1.
Fill in the blanks

1. Data has already been collected by some other person is _______ data.
2. The upper limit of the class interval (25-35) is _______
3. The range of the data 200, 15, 20, 103, 3, 197, is _______
4. If a class size is 10 and range is 80 then the number of classes are _______
5. Pie chart is a _______ graph.

Answers:

1. Secondary
2. 35
3. 197
4. 8
5. circular

Question 2.
Say True or False

1. Inclusive series is a continuous series.
2. Pie charts are easy to understand.
3. Same pie chart can be used for different samples.
4. Media and business people use pie charts.
5. A pie diagram is a circle broken down into component sectors.

Answers:

1. False
2. True
3. False
4. True
5. True

Question 3.
The continuous series of

Solution:
Difference in the gap = 24 – 20 = 4
Here half of the gap = \(\frac { 1 }{ 2 } \) (4) = 2
∴ 2 is the adjustment factor. So we subtract 2 from the lower limit and add 2 to the upper limit to make it as a continuous series.

Question 4.
Represent the following data in ungrouped frequency table which gives the number of children in 25 families.
1, 3, 0, 2, 5, 2, 3, 4, 1, 0, 5, 4, 3, 1, 3, 2, 5, 2, 1, 1, 2, 6, 2, 1, 4
Solution:
The data given is raw data. Ascending order : 0, 1, 2, 3, 4, 5, 6

∴ Tabulating in frequency distribution table we get

Question 5.
Form a continuous frequency distribution table for the marks obtained by 30 students in a X std public examination.
328, 470, 405, 375, 298, 326, 276, 362, 410, 255, 391, 370, 455, 229, 300, 183, 283, 366, 400, 495, 215, 157, 374, 306, 280, 409, 321, 269, 398, 200.
Solution:
Maximum mark obtained = 495
Minimum marks obtained = 157
Range = Maximum value – Minimum value
Range = 495 – 157 = 338
If we take the class size as 50 then the number of class intervals possible

Question 6.
Apaint company asked a group of students about their favourite colours and made a pie chart of their iindings. Use the information to answer the following questions.

(i) What percentage of the students like red colour?
(ii) How many students liked green colour?
(iii) What fraction of the students liked blue?
(iv) How many students did not like red colour?
(v) How many students liked pink or blue?
(vi) How many students were asked about their favourite colours?
Solution:
Total percentage of students = 100%
∴ 50 students = 100 % – (30% + 20% + 25% + 15%)
= 100 % – 90%
50 students = 10%
of total students = 50
∴ \(\frac { 10 }{ 100 } \) (Total students) = 50
Total students = \(\frac{50 \times 100}{10}\) = 500.
Total students = 500

(i) 20% of the students like red colour.

(ii) 15% of the students liked green colour.
\(\frac { 15 }{ 100 } \) × 500 = 75 students liked green colour.

(iii) 25% students liked blue ⇒ \(\frac { 25 }{ 100 } \) students liked blue.
⇒ \(\frac { 1 }{ 4 } \) students liked blue.

(iv) Percentage of students liked red colour = 20 %
Percentage of students did not like red colour
= 100% – 20% = 80%
∴ Number of students did not like red colour
= 80% of 500 = \(\frac { 80 }{ 100 } \) × 500 = 400
400 students did not like red colour.

(v) Students liked pink or blue = students liked pink + students liked blue.
= 30% of 500 + 25% of 500
= \(\frac { 30 }{ 100 } \) × 500 + × 500 = 150 + 125 = 275

(vi) Total number of students = 500.
500 students were asked about their favourite colour.

Question 7.
Write any five points from the given pie chart information regarding pollutants entering in the oceans.

Solution:

1. The pie chart gives the information about the amount of pollutants entering in oceans from various sources.
2. 30% of ocean is polluted by sewage.
3. 20% of ocean is polluted by air.
4. Farm run off equally contribute as much as air in polluting ocean.
5. The observation is about off shore oil, litter, Industrial waste water, Maritime transportation, Air pollution, Farm run off and sewage.

Question 8.
A survey gives the following information of food items preferred by people. Draw a Pie chart.

Solution:
Total number of people = 160 + 90 + 80 + 50 + 30 + 40 = 450
Converting the number of people prefer various food items into components part of 360°


Food items preferred by people.

Question 9.
Draw a pie chart for the following information.

Solution:


Water in Ocean

Question 10.
Income from various sources for Government of India from a rupee is given below. Draw a pie chart.

Solution:

Income from various sources for Government of India in a rupee.

Question 11.
Monthly expenditure of Kumaran’s family is given below. Draw a suitable Pie chart.

Also
1. Find the amount spent for education if Kumaran spends ₹ 6000 for Rent.
2. What is the total salary of Kumaran?
3. How much did he spend more for food than education?
Solution:

Monthly expenditure of Kumaran’s family

1. Given Kumaran spends ₹ 6000 for Rent.
∴ 15% oftotal expenditure
\(\frac { 15 }{ 100 } \) (Total Expenditure) = 6000
Total Expenditure = \(\frac{6000 \times 100}{15}\)
Total Expenditure = ₹ 40,000
Amount spend for education = 20% of total expenditure.
\(\frac { 20 }{ 100 } \) × 40,000 = ₹ 8000

2. Total salary of Kumaran = ₹ 40,000

3. Amount spend for food = 50% of (40,000)
Amount spend for the food than education = 20,000 – 8,000 = ₹ 12,000