Category: 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 3 Geometry 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 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

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.

Students can Download Maths Chapter 3 Geometry Ex 3.3 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 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.

Discontinue Series	Continuous Series
10-20	8-22
24-34	22-36
38-48	36-50
52-62	50-64

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

Students can Download Maths Chapter 3 Geometry Ex 3.1 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 3 Geometry Ex 3.1

Question 1.
Fill in the blanks:

1. The altitudes of a triangle intersect at………..
2. The medians of a triangle cross each other at………..
3. The meeting point of the angle bisectors of a triangle is………..
4. The perpendicular bisectors of the sides a triangle meet at………..
5. The centroid of a triangle divides each medians in the ratio………..

Solution:

1. Orthocentre
2. Centroid
3. Incentre
4. Circumcentre
5. 2 : 1

Question 2.
Say True or False:
(i) In any triangle the Centroid and the Incentre are located inside the triangle.
(ii) The centroid, orthocentre, and incentre of a triangle are collinear.
(iii) The incentre is equidistant from all the vertices of a triangle.
Solution:
(i) True
(ii) True
(iii) False

Question 3.
(a) Where does the circumcentre lie in the case of
(i) An acute angled triangle.
Solution:
Inside the triangle.

(ii) An obtuse-angled triangle.
Solution:
Exterior of the triangle.

(iii) A right angled triangle.
Solution:
On the hypotenuse.

(b) Where does the orthocentre lie in the case of
(i) An acute-angled triangle.
Solution:
Interior of the triangle.

(ii) An obtuse-angled triangle.
Solution:
Exterior of the triangle.

(iii) A right angled triangle.
Solution:
On the vertex containing 90°.

Question 4.
Fill in the blanks:
In the triangle ABC,

(i) The angle bisector is……….
(ii) The altitude is………..
(iii) The median is…………
Solution:
(i) BE
(ii) AD
(iii) CF

Question 5.
In right triangle ABC, what is the length of altitude drawn from the vertex A to BC?

Solution:
In this right angled triangle ΔABC, length of the altitude drawn from vertex A is the leg AB itself. By Pythagoras theorem.
AC² = AB² + BC²
13² = AB² + 12²
169 = AB² + 144
AB² = 169 – 144 = 25
AB² = 52
AB = 5cm

Question 6.
In triangle XYZ, YM is the angle bisector of ∠Y and ∠Y is 100°. Find ∠XYM and ∠ZYM.

Solution:
Given YM is the angle bisector of ∠Y.
also ∠Y = 100°
Angle -bisector divides the angle into two congruent angles.
∠XYM = ∠ZYM
∠Y = ∠XYM + ∠ZYM
100° = ∠XYM + ∠ZYM [∴ ∠XYM = ∠ZYM]
2 ∠XYM = 100°
∠XYM = \(\frac{1}{2}\) (100°)
∠XYM = 50°
∴ ∠ZYM = 50°

Question 7.
In triangle PQR, PS is a median and QS = 3.5 cm, then find QR?

Solution:
Given PS is the median and QS = 3.5 cm
Median is the line drawn from a vertex to the midpoint of the opposite side.
∴ QS = RS
so QS = RS = 3.5 cm
∴ QR = QS +SR = 3.5 + 3.5 = 7 cm
QR = 7 cm

Question 8.
In triangle ABC, line is a perpendicular bisector of BC. If BC = 12 cm, SM = 8 cm, find CS.

Solution:
Given l₁ is the perpendicular bisector of BC.
∴ ∠SMC = 90° and BM = MC
BC = 12 cm
⇒ BM + MC = 12 cm
MC + MC = 12 cm [∴ BM = MC]
2MC = 12
MC = \(\frac{12}{2}\)
MC = 6 cm
Given SM = 8 cm
By Pythagoras theorem SC² = SM² + MC²
SC² = 8² + 6²
SC² = 64 + 36
CS² = 100
CS² = 10²
CS = 10 cm

Students can Download Maths Chapter 2 Life Mathematics Ex 2.2 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 2 Life Mathematics Ex 2.2

Miscellaneous and Practice Problems

Question 1.
5 boys or 3 girls can do a science project in 40 days. How long will it take for 15 boys and 6 girls to do the same project?
Solution:
Let B and G denote Boys and Girls respectively.
Given 5B = 3G ⇒ 1B = \(\frac{3}{5}\)G
now 15B + 6G = 15 × \(\frac{3}{5}\) G + 6G = 9G + 6G = 15G
If 3 girls can do the project in 40 days then 15 girls can do it in
3G × 40 ÷ 15G = 3G × 40 × \(\frac{1}{15G}\) = \(\frac{40}{5}\)
= 8 days.
∴ 15 boys and 6 girls can complete the project in 8 days.

Question 2.
If 32 men working 12 hours a day can do a work in 15 days, how many men working 10 hours a day can do double that work in 24 days?
Solution:
Let the required number of men be x.

Let P₁= 32, H₁ = 12, D₁ = 12, W₁ = 1
P₂ = x, H₂ = 10, D₂ = 24, W₂ = 1

x = 24 persons
To complete the same work 24 men needed.
To complete double the work 24 × 2 = 48 men are required.

Question 3.
Amutha can weave a saree in 18 days. Anjali is twice as good a weaver as Amutha. If both of them weave together, in how many days can they complete weaving the saree?
Solution:
Amutha can weave a saree in 18 days. Anjali is twice as good as Amutha.
ie. If Amutha weave for 2 days, Anjali do the same work in 1 day.
If Anjali weave the saree she will take
\(\frac{18}{2}\) = 9 days
Hence time taken by them together = \(\frac{ab}{a + b}\) days = \(\frac{18 × 9}{18 + 9}\) = \(\frac{18 × 9}{27}\) = 6 days

In 6 days they complete weaving the saree.

Question 4.
A, B and C can complete a work in 5 days. If A and C can complete the same work in 7½ days and A alone in 15 days then, in how many days can B and G finish the work?
Solution:
A + B + C complete the work in 5 days.
∴ (A + B + C)’s 1 day work = \(\frac{1}{5}\)
(A + C) complete the work in 7 \(\frac{1}{2}\) days = \(\frac{15}{2}\) days
∴ (A + C)’s 1 day work = \(\frac{1}{\frac{15}{2}}\) = \(\frac{2}{15}\)
∴ B’s 1 day work = (A + B + C)’s 1 day work – (A + C)’s 1 day work.
\(\frac{1}{5}\) – \(\frac{2}{15}\) = \(\frac{3}{15}\) – \(\frac{2}{15}\)
C’s 1 day work = (A + C)’s 1 day work – A’s 1 day work
= \(\frac{2}{5}\) – \(\frac{1}{15}\) = \(\frac{1}{15}\)
Now (A + C)’s 1 day work = B’s 1 day work + C’s 1 day work
= \(\frac{1}{15}\) + \(\frac{1}{15}\) = \(\frac{2}{15}\)
∴ (B + C) can complete the work in \(\frac{1}{\frac{2}{15}}\) days. = \(\frac{15}{2}\) days = 7\(\frac{1}{2}\) days
∴B and C finish the work in 7\(\frac{1}{2}\) days.

Question 5.
P and Q can do a piece of work in 12 days and 15 days respectively. P started the work alone and then, after 3 days Q joined him till the work was completed. How long did the work last?
Solution:
p can do a piece of work in 12 days.
∴ p’s 1 day work = \(\frac{1}{12}\)
p’s 1 day work = 3 × \(\frac{1}{12}\) = \(\frac{3}{12}\)
Q can do a piece of work in 15 days.
∴ Q’s 1 day work = \(\frac{1}{15}\)
Remaining work after 3 days = 1 – \(\frac{3}{12}\) = \(\frac{9}{12}\)
(P + Q)’s 1 day work = \(\frac{1}{12}\) + \(\frac{1}{15}\) = \(\frac{5}{60}\) + \(\frac{4}{60}\) = \(\frac{9}{60}\)
Number of days required to finish the remaining work
= \(\frac{Remaining work}{(P + Q)’s 1 day work}\) = \(\frac{\frac{9}{12}}{\frac{9}{60}}\) = \(\frac{9}{12}\) × \(\frac{60}{9}\) = 5
Remaining work lasts for 5 days. Total work lasts for 3 + 5 = 8 days.

Challenging Problems

Question 6.
A camp had provisions for 490 soldiers for 65 days. After 15 days, more soldiers arrived and the remaining provisions lasted for 35 days. How many soldiers joined the camp?
Solution:

Now as the soldiers increases food last for less days.
∴ It is inverse proportion.
The proportion is (490 + x): 490 : : 50 : 35
Product of the extremes = Product of the means
(490 + x) × 35 = 490 × 50
(490 + x) = \(\frac{490 × 50}{35}\)

x = 700 – 490
x = 210
∴ 210 soldiers joined the camp.

Question 7.
A small – scale company undertakes an agreement to produce 540 motor pumps in 150 days and employs 40 men for the work. After 75 days, the company could make only 180 motor pumps. How many more men should the company employ so that the work is completed on time as per the agreement?
Solution:
Let the number of men to be appointed more be x.

To produce more pumps more men required
∴ It is direct variation.
∴ The multiplying factor is \(\frac{360}{180}\)
More days means less employees needed.
∴ It is Indirect proportion.
∴ The multiplying factor is \(\frac{75}{75}\)
Now 40 + x = 40 × \(\frac{360}{180}\) × \(\frac{75}{75}\)
40 + x = 80
x = 80 – 40
x = 40
40 more man should be employed to complete the work on time as per the agreement.

Question 8.
A can do a work in 45 days. He works at it for 15 days and then, B alone finishes the remaining work in 24 days. Find the time taken to complete 80% of the work if they work together.
Solution:
A can do a work in 45 days.
A’s 1 day work = \(\frac{1}{45}\)
∴ A’s 15 days work = 15 × \(\frac{1}{45}\) = \(\frac{1}{3}\)
Remaining work = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
B alone completes the remaining \(\frac{2}{3}\) work in 24 days
∴ B completes the whole work in \(\frac{24}{\frac{2}{3}}\) days = 24 × \(\frac{3}{2}\) = 36 days.
∴ B’s 1 day’s work = \(\frac{1}{36}\)
∴ (A + B)’s together complete the work in \(\frac{ab}{a + b}\) days = \(\frac{45 × 36}{45 + 36}\) = \(\frac{45 × 36}{81}\)

Whole wok will be completed by (A + B) in = 20 days.
∴ 80% of the work will be completed in \(\frac{80 × 20}{100}\) = 16 days.

Question 9.
P alone can do \(\frac{1}{2}\) of a work in 6 days and Q alone can do \(\frac{2}{3}\) of the same work in 4 days. In how many days working together, will they finish \(\frac{3}{4}\) of the work?
Solution:
\(\frac{1}{2}\) of the work is done by P in 6 days
∴ Full work is done by P in \(\frac{6}{\frac{1}{2}}\) = 6 × 2 = 12 days
\(\frac{2}{3}\) of work done by Q in 4 days.
∴ Full work done by Q in \(\frac{4}{\frac{2}{3}}\) = 4 × \(\frac{3}{2}\) = 6 days
(P + Q) will finish the whole work in \(\frac{ab}{a + b}\) days = \(\frac{12 × 6}{12 + 6}\) = \(\frac{12 × 6}{18}\) = 4 days
(P + Q) will finish \(\frac{3}{4}\) of the work in 4 × \(\frac{3}{4}\) = 3 days.

Question 10.
X alone can do a piece of work in 6 days and Y alone in 8 days. X and Y undertook the work for Rs 4800. With the help of Z, they completed the work in 3 days. How much is Z’s share?
Solution:
X can do the work in 6 days.
X’s 1 day work = \(\frac{1}{6}\)
X’s share for 1 day = \(\frac{1}{6}\) × 48000 = Rs 800
X’s share for 3 days = 3 × 800 = 2400
Y can complete the work in 8 days.
Y’s 1 day work = \(\frac{1}{8}\)
Y’s 1 day share = \(\frac{1}{8}\) × 4800 = 600
Y’s 3 days share = 600 × 3 = 1800
(X + Y)’s 3 days share = 2400 + 1800 = 4200
Remaining money is Z’s share
∴ Z’s share = 4800 – 4200 = 600

Students can Download Maths Chapter 1 Numbers Ex 1.5 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 1 Numbers Ex 1.5

Miscellaneous and Practice Problems

Question 1.
A square carpet covers an area of 1024 m² of a big hall. It is placed in the middle of the hall. What is the length of a side of the carpet?
Solution:
Area of the carpet = 1024 m²
side × side = 1024 m²
(side)² = 1024 m²

(side)² = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 2² × 2² × 2² × 2² × 2²
= (2 × 2 × 2 × 2 × 2)²
(side)² = 32²
side = 32
Length of a side of the carpet = 32 m

Question 2.
There is a large square portrait of a leader that covers an area of 4489 cm². If each side has a 2 cm liner, what would be its area?
Solution:
Area of the square = 4489 cm²
(side)² = 4489 cm²

(side)² = 67 x 67
side² = 67²
Length of a side = 67

Length of a side Length of a side with liner = 67 + 2 + 2 cm = 71 cm
Area of the larger square = 71 x 71 cm²
= 5041 cm²
Area of the liner = Area of big square-Area of small square
= (5041 – 4489) cm² = 552 cm²

Question 3.
2401 plants are planted in a garden such that each contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.
Solution:
Given number of plants in a row = Number of rows.
Number of rows × number of plants in a row = Total plants
Total plants = 2401

= 7 × 7 × 7 × 7 = 7² × 7²
= 49 × 49
∴ number of rows = 49
number of plants in a row = 49

Question 4.
If \(\sqrt[3]{1906624} \times \sqrt{x}\) = 3100, find x.
Solution:

\(\sqrt{x}\) = 25
Squaring on both sides \((\sqrt{x})^{2}\) = 25²
x = 625

Question 5.
If (625)^x = 15625, find x² and x³
Solution:
(625)^x = 15625

(5 x 5 x 5 x 5)^x = 5 x 5 x 5 x 5 x 5 x 5
(5⁴)^x = 5⁶
5^4x = 5⁶
5^4x = 5⁶
Comparing the powers of 5 both sides
4x = 6
x = \(\frac{6}{4}\)
x = \(\frac{3}{2}\)

Question 6.
If 2^m-1 + 2^m+1 = 640, then find ‘m’
Solution:
Given 2^m-1 + 2^m+1 = 640
2^m-1 + 2^m+1 = 128 + 512 [consecutive powers of 2]
2^m-1 + 2^m+1 = 2⁷+ 2⁹ [powers of 2: 2, 4, 8, 16, 32, 64, 128, 256, 512, …..]
m – 1 = 7
m = 7 + 1
m = 8

Question 7.
Simplify \(\frac{16 \times 10^{2} \times 64}{4^{2} \times 2^{4}}\)
Solution:

\(\frac{16 \times 10^{2} \times 64}{4^{2} \times 2^{4}}\) = \(\frac{2^{4} \times 10^{2} \times 2^{6}}{\left(2^{2}\right)^{2} \times 2^{4}}=\frac{2^{4+6} \times 10^{2}}{2^{4} \times 2^{4}}=\frac{2^{10} \times 100}{2^{8}}\)
= 2^10-8 × 100= 2² × 100 = 400

Question 8.
Give the answer in scientific notation:
A human heart beats at an average of 80 beats per minute. How many times does it beat in
(i) an hour?
(ii) a day?
(iii) a year?
(iv) 100 years?
Solution:
Heart beat per minute = 80 beats
(i) an hour One hour = 60 minutes
Heart beat in an hour = 60 x 80 = 4800 = 4.8 x 10³

(ii) In a day
One day = 24 hours = 24 x 60 minutes
∴ Heart beat in one day = 24 x 60 x 80 = 24 x 4800 = 115200 = 1.152 x 10⁵

(iii) a year
One year = 365 days = 365 x 24 hours = 365 x 24 x 60 minutes
∴ Heart beats in a year = 365 x 24 x 60 x 80 = 42048000 = 4.2048 x 10⁷

(iv) 100 years
Heart beats in one year = 4.2048 x 10⁷
Heart beats in 100 years = 4.2048 x 10⁷ x 100 = 4.2048 x 10⁷ x 10²
= 4.2048 x 10⁹

Challenging Problems

Question 9.
A greeting card has an area 90 cm². Between what two whole numbers is the length of its side?
Solution:
Area of the greeting card = 90 cm²
(side)² = 90 cm²

(side)² = 2 x 5 x 3 x 3 = 2 x 5 x 3²
\(\sqrt{({side})^{2}}=\sqrt{2 \times 5 \times 3^{2}}\)

side = 3 \(\sqrt{2 × 5}\)
side = \(\sqrt{10}\) cm
side = 3 × 3.2 cm
side = 9.6 cm
∴ Side lies between the whole numbers 9 and 10.

Question 10.
225 square shaped mosaic tiles, each of area 1 square decimetre exactly cover a square shaped verandah. How long is each side of the square shaped verandah?
Solution:
Area of one tile = 1 sq. decimeter
Area of 225 tiles = 225 sq.decimeter
225 square tiles exactly covers the square shaped verandah.
∴ Area of 225 tiles = Area of the verandah
Area of the verandah = 225 sq.decimeter
side x side = 15 x 15 sq.decimeter
side = 15 decimeters
Length of each side of verandah = 15 decimeters.

Question 11.
A group of 1536 cadets wanted to have a parade forming a square design. Is it possible? If it is not possible how many more cadets would be required?
Solution:
Number of cadets to form square design

1536 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3
The numbers 2 and 3 are unpaired
It is impossible to have the parade forming square design with 1536 cadets.

39 x 39 = 1521
Also 40 x 40 = 1600
∴ We have to add (1600 – 1536) = 64 to make 1536 a perfect square.
∴ 64 more cadets would be required to form the square design.

Question 12.
Find the decimal fraction which when multiplied by itself gives 176.252176.
Solution:
We will find the square root of 176.252176

176.252176 = 13.276 x 13.276
∴ The required number is 13.276

Question 13.
Evaluate \(\sqrt{286225}\) and use it to compute \(\sqrt{2862.25}\) + \(\sqrt{28.6225}\)
Solution:

Question 14.
The speed of light in glass is about 2 x 10⁸ m/sec. Use the formula, time = \(\frac{distence}{speed}\) to find the time (in hours) for a pulse of light to travel 7200 km in glass.
Solution:

Required time = 10^-5 hours

Question 15.
Simplify : (3.769 x 10⁵) + (4.21 x 10⁵)
Solution:

(3.769 x 10⁵) + (4.21 x 10⁵) = 3,76,900 + 4,21,00
= 7,97,900 = 7.979 x 10⁵

Question 16.
Order the following from the least to the greatest: 1625, 8100, 3500, 4400, 2600
Solution:
16²⁵ = (2⁴)²⁵ = 2¹⁰⁰
8¹⁰⁰ = (2³)¹⁰⁰ = 2³⁰⁰
4⁴⁰⁰ = (2²)⁴⁰⁰ = 2⁸⁰⁰
2⁶⁰⁰ = 2⁶⁰⁰
Comparing the powers we have, 2¹⁰⁰ < 2³⁰⁰ < 2⁶⁰⁰< 2⁸⁰⁰
∴ The required order : 16²⁵, 8¹⁰⁰, 2⁶⁰⁰, 3⁵⁰⁰, 4⁴⁰⁰