Class 8

Students can Download Maths Chapter 4 Statistics 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 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.

Discontinuous series	Continuous series
15-25	13.5-26.5
28-38	26.5-39.5
41-51	39.5-52.5
54 – 64	52.5-65.5

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.

Students can Download Maths Chapter 2 Life Mathematics Additional 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 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 2 Life Mathematics Ex 2.1 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 2 Life Mathematics Ex 2.1

Question 1.
Fill in the blanks:
(i) A can finish a job in 3 days whereas B finishes it in 6 days. The time taken to complete the job together is………..days.
(ii) If 5 persons can do 5 jobs in 5 days, then 50 persons can do 50 jobs in………..days.
(iii) A can do a work in 24 days. A and B together can finish the work in 6 days. Then B alone can finish the work in…………days.
(iv) A alone can do a piece of work in 35 days. If B is 40% more efficient than A, then B will finish the work in………..days.
(v) A alone can do a work in 10 days and B alone in 15 days. They undertook the work for Rs 200000. The amount that A will get is………
Solution:
(i) 2 days
(ii) 5
(iii) 8
(iv) 25
(v) Rs 1,20,000

Question 2.
210 men working 12 hours a day can finish a job in 18 days. How many men are required to finish the job in 20 days working 14 hours a day?
Solution:
Let the required number of men be x.

More working hours ⇒ less men required.
∴ It is inverse proportion.
∴ Multiplying factor is \(\frac{12}{14}\)
Also more number of days ⇒ less men
∴ It is an inverse proportion.
∴ Multiplying factor is \(\frac{18}{20}\)
∴ x = 210 × \(\frac{12}{14}\) × \(\frac{18}{20}\)= 162 men

162 men are required.

Question 3.
A cement factory makes 7000 cement bags in 12 days with the help of 36 machines. How many bags can be made in 18 days using 24 machines?
Solution:
Let the required number of cement bags be x.

Number of days more ⇒ More cement bags.
∴ It is direct variation.
∴ The multiplying factor = \(\frac{18}{12}\)
Number of machines more ⇒ More cement bags.
∴ It is direct variation.
∴ The multiplying factor = \(\frac{24}{36}\)
∴ x = 7000 × \(\frac{18}{12}\) × \(\frac{24}{36}\)

x = 7000 cement bags
7000 cement bags can be made.

Question 4.
A soap factory produces 9600 soaps in 6 days working 15 hours a day. In how many days will it produce 14400 soaps working 3 hours more a day?
Solution:
Let the required number of days be x.

To produce more soaps more days required.
∴ It is direct proportion.
∴ Multiplying factor = \(\frac{14400}{9600}\)
If more hours spend, less days required.
∴ It is indirect proportion.
∴ Multiplying factor = \(\frac{15}{18}\)
∴ x = 6 × \(\frac{14400}{9600}\) × \(\frac{15}{18}\)

x = \(\frac{15}{2}\)
\(\frac{15}{2}\) days will be needed.

Question 5.
If 6 container lorries transport 135 tonnes of goods in 5 days, how many more lorries are required to transport 180 tonnes of goods in 4 days?
Solution:
Let the number of lorries required more = x.

As the goods are more ⇒ More lorries are needed to transport.
∴ It is direct proportion.
∴ Multiplying factor = \(\frac{180}{135}\)
Again if more days ⇒ less number of lorries enough.
∴ It is direct proportion.
∴ Multiplying factor = \(\frac{5}{4}\)
∴ 6 + x = 6 × \(\frac{180}{135}\) × \(\frac{5}{4}\)

6 + x = 10
x = 10 – 6
x = 4
∴ 4 more lorries are required.

Question 6.
A can do a piece of work in 12 hours, B and C can do it 3 hours whereas A and C can do it in 6 hours. How long will B alone take to do the same work?
Solution:
Time taken by A to complete the work =12 hrs.
∴ A’s 1 hr work = \(\frac{1}{12}\)…………(1)
(B + C) complete the work in 3 hrs.
∴ (B + C)’s 1 hour work = \(\frac{1}{3}\)…………(2)
(1) + (2) ⇒
∴ (A + B + C)’s 1 hour work = \(\frac{1}{12}\) + \(\frac{1}{3}\) = \(\frac{1+4}{12}\) = \(\frac{5}{12}\)
Now (A + C) complete the work in 6 hrs.
∴(A + C)’s 1 hour work = \(\frac{1}{6}\)
∴ B’s 1 hour work = (A + B + C)’s 1 hour work – (A + C)’s 1 hour work
= \(\frac{5}{12}\) – \(\frac{1}{6}\) = \(\frac{5-2}{12}\) = \(\frac{3}{12}\) = \(\frac{1}{4}\)
∴ B alone take 4 days to complete the work.

Question 7.
A and B can do a piece of work in 12 days, while B and C can do it in 15 days whereas A and C can do it in 20 days. How long would each take to do the same work?
Solution:
(A + B) complete the work in 12 days.
∴ (A + B)’s 1 day work = \(\frac{1}{12}\)……….(1)
(B + C) complete the work in 15 days
∴ (B + C)’s 1 day work = \(\frac{1}{15}\)……….(2)
(A + C) complete the work in 20 days
∴ (A + C)’s 1 day work = \(\frac{1}{20}\)……….(3)
Now (1) + (2) + (3) ⇒
[(A + B) + (B + C) + (A + C)]’s 1 day work = \(\frac{1}{12}\) + \(\frac{1}{15}\) + \(\frac{1}{20}\)

(2A + 2B + 2C)’s 1 day work = \(\frac{5}{60}\) + \(\frac{4}{60}\) + \(\frac{3}{60}\)
2(A + B + C)’s 1 day work = \(\frac{5+4+3}{60}\)
(A + B + C)’s 1 day work = \(\frac{12}{60×2}\)
(A + B + C)’s 1 day work = \(\frac{1}{10}\)
Now A’s 1 day’s work = (A + B + C)’s 1 day work – (B + C)’s 1 day work
\(\frac{1}{10}\) – \(\frac{1}{15}\) = \(\frac{3}{30}\) – \(\frac{2}{30}\) = \(\frac{1}{30}\)
∴ A takes 30 days to complete the work.
B’s 1 day work = (A + B + C)’s 1 day’s work – (A + C)’s 1 day’s work
= \(\frac{1}{10}\) – \(\frac{1}{20}\) = \(\frac{6}{60}\) – \(\frac{3}{60}\)
\(\frac{6-3}{60}\) = \(\frac{3}{60}\) = \(\frac{1}{20}\)
B takes 20 days to complete the work.
C’s 1 day work = (A + B + C)’s 1 day work – (A + B)’s 1 day work
\(\frac{1}{10}\) – \(\frac{1}{12}\) = \(\frac{6}{60}\) – \(\frac{5}{60}\) = \(\frac{6-5}{60}\) = \(\frac{1}{60}\)
∴ C takes 60 days to complete the work.

Question 8.
Carpenter A takes 15 minutes to fit the parts of a chair while Carpenter B takes 3 more minutes than A to do the same work. Working together, how long will it take for them to fit the parts for 22 chairs?
Solution:
Time taken by A to fit a chair = 15 minutes
Time taken by B = 3 minutes more than A
= 15 + 3 = 18 minutes
∴ A’s 1 minute work = \(\frac{1}{15}\)
B’s 1 minute work = \(\frac{1}{18}\)
(A + B)’s 1 minutes work = \(\frac{1}{15}\) + \(\frac{1}{18}\)

\(\frac{12}{180}\) + \(\frac{22}{180}\) = \(\frac{22}{180}\) = \(\frac{11}{90}\)
∴ Time taken by (A + B) to fit a chair
= \(\frac{1}{\frac{11}{90}}\) = \(\frac{90}{11}\) minutes
∴ Time taken by (A + B) to fit 22 chairs
= \(\frac{90}{11}\) × 22 = 180 minutes
= \(\frac{180}{60}\) = 3 hours

Question 9.
A man takes 10 days to finish a job where as a woman takes 6 days to finish the same job. Together they worked for 3 days and then the woman left. In how many days will the man complete the remaining job?
Solution:
Man can finish the work in 10 days and women can finish the same work in 6 days.
∴ Man’s 1 day work = \(\frac{1}{10}\)
Woman’s 1 day work = \(\frac{1}{6}\)
(Man + Woman)s 1 day work = \(\frac{1}{10}\) + \(\frac{1}{6}\) = \(\frac{6}{60}\) + \(\frac{10}{60}\) = \(\frac{16}{60}\)
(Man + Woman)s 3 days work

In 3 days \(\frac{4}{5}\) th of the whole work is completed.
Remaining work = 1 – \(\frac{4}{5}\) = \(\frac{5}{5}\) – \(\frac{4}{5}\) = \(\frac{1}{5}\)
Complete work is done by the man by 10 days
∴ \(\frac{1}{5}\) of the work is done by man in \(\frac{1}{5}\) × 10 = 2 days.

Question 10.
A is thrice as fast as B. If B can do a piece of work in 24 days then, find the number of days they will take to complete the work together.
Solution:
If B does the work in 3 days, A will do it in 1 day.
B complete the work in 24 days.
∴ A complete the same work in \(\frac{24}{3}\) = 8 days.
∴ (A + B) complete the work in \(\frac{ab}{a+b}\) days = \(\frac{24×8}{24+8}\) days = \(\frac{24×8}{32}\)days = 6 days

They together complete the work in 6 days.

Students can Download Maths Chapter 4 Statistics Additional 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 Additional Questions

Question 1.
The following information shows the number of students opting different subjects in a college.

Draw a pie-chart to represent the information.
Solution:

Students opting different subjects.

Question 2.
The pie chart represents the expenditures of a family on different items. Find the percentage expenditures on different items by reading the pie-chart.

Solution:
Percentage value of the component =
∴ Percentage expenditures on various items are given as :

Question 3.
The following table gives the marks of 100 students represent in the form of a histogram.

Solution:
We take marks on X axis and frequency Y axis.

Students can Download Maths Chapter 4 Statistics Ex 4.3 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 4.3

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 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

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

Exercise 2.1
Try These (Text book Page No. 33)

Classify the given examples as direct or inverse proportion:
(i) Weight of pulses to their cost.
Solution:
As weight increases cost also increases.
∴ Weight and cost are direct proportion.

(ii) Distance travelled by bus to the price of ticket.
Solution:
As the distance increases price to travel also increases.
∴ Distance and price are direct proportion.

(iii) Speed of the athelete to cover a certain distance.
Solution:
As the speed increases, the time to cover the distance become less.
So speed and time are in indirect proportion.

(iv) Number of workers employed to complete a construction in a specified time.
Solution:
As the number of workers increases, the amount of work become less, so they are in indirect proportion.

(v) Volume of water flown through a pipe to its pressure.
Solution:
As the pressure increases, volume also increases.
∴ They are direct proportions.

(vi) Area of a circle to its radius.
Solution:
If the radius of the circle increases its area also increases.
∴ Area and radius of circles are direct proportion.

Use the concept of direct and inverse proportions and try to answer the following questions:
Question 1.
A student can type 21 pages in 15 minutes. At the same rate, how long will it take the student to type 84 pages?
Solution:
Direct proportion
No. of minutes = x
k = \(\frac{21}{15}\)
\(\frac{21}{15}\) = \(\frac{84}{x}\)

Question 2.
The weight of an iron pipe varies directly with its length. If 8 feet of an iron pipe weighs 3.2 kg, find the proportionality constant k and determine the weight of a 36 feet iron pipe.
Solution:

Weight of 36 feet iron pipe = x
\(\frac{36}{x}\) = 2.5

Question 3.
A car covers a distance of 765 km in 51 litres of petrol. How much distance would it cover in 30 litres of petrol?
Solution:
Direct proportion
k = \(\frac{51}{765}\)
Distance cover = x km
\(\frac{30}{x}\) = \(\frac{51}{765}\)

Question 4.
If x and y vary inversely and x = 24 when y = 8, find x when y = 12.
Solution:
k = xy = 24 × 8 = 192
∴ 12 × x = 192

Question 5.
If 35 women can do a piece of work in 16 days, in how many days will 28 women do the same work?
Solution:
Inverse proportion
No. of days = x
k = 35 × 16
∴ 28 × x = 35 × 16

Question 6.
A farmer has food for 14 cows which can last for 39 days. How long would the food last, if 7 more cows join his cattle?
Solution:
Inverse variation
k = xy = 14 × 39
No. of cow = 14 + 7 = 21
No. of days = x
21 × x = 14 × 39

Question 7.
Identify the type of proportion and fill in the blank boxes:

Solution:
Direct proportion
\(\frac{x}{y}\) = k = \(\frac{1}{20}\)
(i) x = 2; y = ?
\(\frac{2}{y}\) = \(\frac{1}{20}\) ⇒ y = 2 × 20 = 40

(ii) x = ?; y = 60
\(\frac{x}{60}\) = \(\frac{1}{20}\) ⇒ x = \(\frac{60}{20}\) = 3

(iii) x = 4; y = ?
\(\frac{4}{y}\) = \(\frac{1}{20}\) ⇒ y = 80

(iv) x = 4; y = ?
\(\frac{8}{y}\) = \(\frac{1}{20}\) ⇒ y = 20 × 8 = 160

(v) x = ?; y = 180
\(\frac{x}{180}\) = \(\frac{1}{20}\)
x = \(\frac{180}{20}\) = 9

(vi) x = 12; y = ?
\(\frac{12}{y}\) = \(\frac{1}{20}\)
y = 12 × 20 = 240

(vii) x = ?; y = 360
\(\frac{x}{360}\) = \(\frac{1}{20}\) ⇒ x = \(\frac{360}{20}\) = 18

(viii) x = 24; y = ?
\(\frac{24}{y}\) = \(\frac{1}{20}\) ⇒ y = 24 × 20 = 480

Question 8.
Identify the type of proportion and fill in the blank boxes:

Solution:
Inverse proportion
k = xy = 1 × 144 = 144
(i) x = 2; y = ?
2y = 144
y = 72

(ii) X = ?; y = 48
48x = 144
x = \(\frac{144}{48}\) = 3

(iii) x = 4; y = ?
4y = 144
y = \(\frac{144}{4}\) = 36

(iv) x = 8; y = ?
8 y = 144
y = \(\frac{144}{8}\) = 18

(v) x = ?; y = 16
16x = 144
y = \(\frac{144}{16}\) = 9

(vi) x = 12; y = ?
12y = 144
y = \(\frac{144}{12}\) = 12

(vii) x = ?; y = 9
9x = 144
x = \(\frac{144}{9}\) = 16

(viii) x = 24; y = ?
24y = 144
y = \(\frac{144}{24}\) = 6

Try These (Text book Page No. 38)

Question 1.
When x = 5 and y = 5 find k, if x and y vary directly.
Solution:
If x and y vary directly then \(\frac{x}{y}\) = k
Here x = 5; y = 5
∴ k = \(\frac{5}{5}\)
k = 1

Question 2.
When x and y vary inversely, find the constant of variation when x = 64 and y = 0.75
Solution:
Given
x =64, y = 0.75
and also given x and y vary inversely.
∴ xy = k. the constant of variation.
∴ Constant = 64 × 0.75
Constant of variation = 48

Think (Text book Page No. 38)

(i) When x and y are in direct proportion and if y is doubled, then what happens to x?
Solution:
If x and y are in direct proportion \(\frac{x}{y}\) = k, constant.
if y is doubled, then \(\frac{x}{2}\) must be equal to k. So x also to be doubled.

(ii) if \(\frac{x}{y-x}\) = \(\frac{6}{7}\) What is \(\frac{x}{y}\)?
Solution:
if \(\frac{x}{y-x}\) = \(\frac{6}{7}\)
\(\frac{y-x}{x}\) = \(\frac{7}{6}\)
\(\frac{y}{x}\) – \(\frac{x}{x}\) = \(\frac{7}{6}\)
\(\frac{y}{x}\) = \(\frac{7}{6}\) + \(\frac{x}{x}\)
\(\frac{y}{x}\) = \(\frac{7}{6}\) + 1
\(\frac{y}{x}\) = \(\frac{7+6}{6}\)
\(\frac{y}{x}\) = \(\frac{13}{6}\)
\(\frac{x}{y}\) = \(\frac{6}{13}\)

Try These (Text book Page No. 40)

Identify the different variations present in the following questions:
Question 1.
24 men can make 48 articles in 12 days. Then, 6 men can make …………. articles in 6 days.
Solution:
Let the required no. of articles be x

(i) Mens and days are Indirect variables.
(ii) Men and Articles are direct variables
(iii) Days and articles are also direct variables using formula.
Let P₁ = 24, D₁ = 12, W₁ = 48
P₂ = 6, D₂ = 6, W₂ = x

Question 2.
15 workers can lay a road of length 4 km In 4 hours. Then, …………. workers can lay a road of length 8 km in 8 hours.
Solution:
Let the required number of workers be x

Length and workers are direct variable as more length need more workers.
The proportion is 4 : 8 : : 15 : x ……….(1)
Hours and workers are indirect variables as more working hours need less men.
∴ The proportion is 8 : 4 : : 15 : x ………..(2)
Combining (1) and (2)

Product of the extremes = product of the means
4 × 8 × x = 8 × 4 × 15
x = \(\frac{8×4×15}{4×8}\)
x = 15 workers

Question 3.
25 women working 12 hours a day can complete a work in 36 days. Then, 20 women must ……….. work hours to complete the same work in 30 days.
Solution:
Let the required hours be x

As women increases hours to work decreases
∴ It is an inverse proportion.
∴ Multiplying factor is \(\frac{25}{20}\)
As days increases hours needed become less
∴ It is also an indirect variation.
∴ Multiplying factor is \(\frac{36}{30}\)
∴ x = 12 × \(\frac{25}{20}\) × \(\frac{36}{30}\)
x = 18 hours

Question 4.
In a camp, there are 420 kg of rice sufficient for 98 persons for 45 days. The number of days that 60 kg of rice will last for 42 persons is…………
Solution:
Let the required number of days be x.

If amount of rice is more it will last for more days.
∴ It is Direct Proportion.
∴ Multiplying factor is \(\frac{60}{420}\)
If men increases number of days the rice lasts decreases
∴ It is an inverse proportion.
∴ Multiplying factor is \(\frac{98}{42}\)
x = 45 × \(\frac{60}{420}\) × \(\frac{98}{42}\)

x = 15 days

Try These (Text book Page No. 44)

Question 1.
Vikram can do one-third of work in p days. He can do \(\frac{3}{4}\)th of work in ………… days.
Solution:
\(\frac{1}{3}\) of the work will be done in p days
∴ Full work will be completed in 3p days
\(\frac{3}{4}\)th of the work will be done in = 3p × \(\frac{3}{4}\) = \(\frac{9}{4}\)p = 2\(\frac{1}{4}\)p days

Question 2.
If m persons can complete a work in n days, then 4m persons can complete the same work in ……….. days and \(\frac{m}{4}\) persons can complete the same work in…….. days
Solution:
Given m persons complete a work in n days
(i) Then work measured in terms of Man days = mn
4 m men do the work it will be completed in \(\frac{mn}{4m}\) days = \(\frac{n}{4}\) days.
(ii) \(\frac{m}{4}\) persons can complete the same work in \(\frac{mn}{\frac{m}{4}}\) days = \(\frac{4mn}{m}\) = 4n days

Students can Download Maths Chapter 4 Statistics Ex 4.2 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 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

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