Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.4

Students can download Maths Chapter 8 Statistics Ex 8.4 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 8 Statistics Ex 8.4

Question 1.
Let m be the mid point and b be the upper limit of a class in a continuous frequency distribution. The lower limit of the class is …….
(a) 2m – b
(b) 2m + b
(c) m – b
(d) m – 2b
Solution:
(a) 2m – b

Question 2.
The mean of a set of seven numbers is 81. If one of the numbers is discarded, the mean of the remaining numbers is 78. The value of discarded number is ……..
(a) 101
(b) 100
(c) 99
(d) 98
Solution:
(c) 99
Hint:
Total of 8 numbers = 81 × 7 = 567
Total of 7 numbers = 78 × 6 = 468
The number is = 567 – 468
= 99

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.4

Question 3.
A particular observation which occurs maximum number of times in a given data is called its ………
(a) frequency
(b) range
(c) mode
(d) median
Solution:
(c) mode

Question 4.
For which set of numbers do the mean, median and mode all have the same values?
(a) 2, 2, 2, 4
(b) 1, 3, 3, 3, 5
(c) 1, 1, 2, 5, 6
(d) 1, 1, 2, 1, 5
Solution:
(b) 1, 3, 3, 3, 5

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.4

Question 5.
The algebraic sum of the deviations of a set of n values from their mean is ……..
(a) 0
(b) n – 1
(c) n
(d) n + 1
Solution:
(a) 0

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.4

Question 6.
The mean of a, b, c, d and e is 28. If the mean of a, c and e is 24, then mean of b and d is ……..
(a) 24
(b) 36
(c) 26
(d) 34
Solution:
(d) 34
Hint:
Mean = 28
a + b + c + d + e = 28 × 5 = 140
= 140 …… (1)
But \(\frac{a+c+e}{3}\) = 24
a + c + e = 72
a + b + c + d + e = 140
b + d + 72 = 140
b + d = 140 – 72
= 68
Mean = \(\frac{68}{2}\)
= 34

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.4

Question 7.
If the mean of five observations x, x + 2, x + 4, x + 6, x + 8, is 11, then the mean of first three observations is …….
(a) 9
(b) 11
(c) 13
(d) 15
Solution:
(a) 9
Hint:
Mean = \(\frac{x+x+2+x+4+x+6+x+8}{5}\)
11 = \(\frac{5x+20}{5}\)
5x + 20 = 55
5x = 55 – 20
= 35
x = \(\frac{35}{5}\)
= 7
Mean of first 3 observation is = \(\frac{7+9+11}{3}\)
= 9

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.4

Question 8.
The mean of 5, 9, x, 17 and 21 is 13, then find the value of x.
(a) 9
(b) 13
(c) 17
(d) 21
Solution:
(b) 13
Hint:
Mean = \(\frac{5+9+x+17+21}{5}\)
13 = \(\frac{52+x}{5}\)
65 = 52 + x
x = 65 – 52
= 13

Question 9.
The mean of the square of first 11 natural numbers is
(a) 26
(b) 46
(c) 48
(d) 52
Solution:
(b) 46
Hint:
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.4 1
= 46

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.4

Question 10.
The mean of a set of numbers is \(\bar { X }\). If each number is multiplied by z, the mean is ……..
(a) \(\bar { X }\) + z
(b) \(\bar { X }\) – z
(c) z\(\bar { X }\)
(d) \(\bar { X }\)
Solution:
(c) z\(\bar { X }\)
Hint:
If each observation is multiplied by k, k ≠ 0 then the arithmetic mean is also multiplied by the same quantity.

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4

Students can download Maths Chapter 4 Geometry Ex 4.4 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Ex 4.4

Question 1.
The length of the tangent to a circle from a point P, which is 25 cm away from the centre is 24 cm. What is the radius of the circle?
Solution:
Let the radius AB be r. In the right ∆ ABO,
OB2 = OA2 + AB2
252 = 242 + r2
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 1
252 – 242 = r2
(25 + 24) (25 – 24) = r2
r = \(\sqrt { 49 }\) =7
Radius of the circle = 7 cm

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4

Question 2.
∆ LMN is a right angled triangle with ∠L = 90°. A circle is inscribed in it. The lengths of the sides containing the right angle are 6 cm and 8 cm. Find the radius of the circle.
Solution:
LN = 6; ML = 8. In the right ∆ LMN,
MN2 = LN2 + LM2
= 62 + 82 = 36 + 64 = 100
MN = \(\sqrt { 100 }\) = 10
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 2
OA= OB = OC = r
AN = CN (Tangent of the circle)
LN – AL= CN
LN – r = CN
8 – r = CN ……(1)
MC = MB (tangent of the circle)
MC = ML – LB
MC = 6 – r …….(2)
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 3
Add (1) and (2)
MC + CN = (6 – r) + (8 – r)
MN = 14 – 2r
10 = 14 – 2r
2r = 14 – 10 = 4
r = \(\frac { 4 }{ 2 } \) = 2 cm
radius of the circle = 2 cm

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4

Question 3.
A circle is inscribed in ∆ ABC having sides 8 cm, 10 cm and 12 cm as shown in figure, Find AD, BE and CF.
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 4
Solution:
AD = AF = x (tangent of the circle)
BD = BE = y (tangent of the circle)
CE = CF = z (tangent of the circle)
AB = AD + DB
x + y = 12 ……(1)
BC = BE + EC
y + z= 8 …….(2)
AC = AF + FC
x + z = 10 ……(3)
Add (1) (2) and (3)
2x + 2y + 2z = 12 + 8 + 10
x + y + z = \(\frac { 30 }{ 2 } \) = 15 …….(4)
By x + y = 12 in (4)
z = 3
y + z = 8 in (4)
x = 7
x + z = 10 in (4)
y = 5
AD = 7 cm; BE = 5 cm and CF = 3 cm

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4

Question 4.
PQ is a tangent drawn from a point P to a circle with centre O and QOR is a diameter of the circle such that ∠POR = 120° . Find ∠OPQ.
Solution:
∠POQ = 180° – 120° = 60°
In ∆OPQ, we know
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 5
∠POQ + ∠OQP + ∠OPQ = 180°
(Sum of the angles of a ∆ is 180°)
60° + 90° + ∠OPQ = 180°
∠OPQ = 180° – 150° = 30°
∠OPQ = 30°

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4

Question 5.
A tangent ST to a circle touches it at B. AB is a chord such that ∠ABT = 65°. Find ∠AOB, where “O” is the centre of the circle.
Solution:
Given ∠ABT = 65°
∠OBT = 90°(TB is the tangent of the circle)
∠ABO = 90° – 65° = 25°
∠ABO + ∠BOA + ∠OAB = 180°
25° + x + 25° = 180° (Sum of the angles of a ∆)
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 6
(OA and OB are the radius of the circle.
∴ ∠ABO = ∠BAO = 25°
x + 50 = 180°
x = 180° – 50° = 130°
∴ ∠BOA = 130°

Question 6.
In figure, O is the centre of the circle with radius 5 cm. T is a point such that OT = 13 cm and OT intersects the circle E, if AB is the tangent to the circle at E, find the length of AB.
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 7
Solution:
In the right ∆ OTP,
PT2 = OT2 – OP2
= 132 – 52
= 169 – 25
= 144
PT = \(\sqrt { 144 }\) = 12 cm
Since lengths of tangent drawn from a point to circle are equal.
∴ AP = AE = x
AT = PT – AP
= (12 – x) cm
Since AB is the tangent to the circle E.
∴ OE ⊥ AB
∠OEA = 90°
∠AET = 90°
In ∆AET, AT2 = AE2 + ET2
In the right triangle AET,
AT2 = AE2 + ET2
(12 – x)2 = x2 + (13 – 5)2
144 – 24x + x2 = x2 + 64
24x = 80 ⇒ x = \(\frac { 80 }{ 24 } \) = \(\frac { 20 }{ 6 } \) = \(\frac { 10 }{ 3 } \)
BE = \(\frac { 10 }{ 3 } \) cm
AB = AE + BE
= \(\frac { 10 }{ 3 } \) + \(\frac { 10 }{ 3 } \) = \(\frac { 20 }{ 3 } \)
Lenght of AB = \(\frac { 20 }{ 3 } \) cm

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4

Question 7.
In two concentric circles, a chord of length 16 cm of larger circle becomes a tangent to the smaller circle whose radius is 6 cm. Find the radius of the larger circle.
Solution:
Here AP = PB = 8 cm
In ∆OPA,
OA2 = OP2 + AP2
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 8
= 62 + 82
= 36 + 64
= 100
OA = \(\sqrt { 100 }\) = 10 cm
Radius of the larger circle = 10 cm

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4

Question 8.
Two circles with centres O and O’ of radii 3 cm and 4 cm, respectively intersect at two points P and Q, such that OP and O’ P are tangents to the two circles. Find the length of the common chord PQ.
Solution:
In ∆ OO’P
(O’O)2 = OP2 + O’P2
= 32 + 42
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 9
= 9 + 16
(OO’)2 = 25
∴ OO’ = 5cm
Since the line joining the centres of two intersecting circles is perpendicular bisector of their common chord.
OR ⊥ PQ and PR = RQ
Let OR be x, then O’R = 5 – x again Let PR = RQ = y cm
In ∆ ORP, OP2 = OR2 + PR2
9 = x2 + y2 …(1)
In ∆ O’RP, O’P2 = O’R2 + PR2
16 = (5 – x)2 + y2
16 = 25 + x2 – 10x + y2
16 = x2 + y2 + 25 – 10x
16 = 9 + 25 – 10x (from 1)
16 = 34 – 10x
10x = 34 – 16 = 18
x = \(\frac { 18 }{ 10 } \) = 1.8 cm
Substitute the value of x = 1.8 in (1)
9 = (1.8)2 + y2
y2 = 9 – 3.24
y2 = 5.76
y = \(\sqrt { 5.76 }\) = 2.4 cm
Hence PQ = 2 (2.4) = 4.8 cm
Length of the common chord PQ = 4.8 cm

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4

Question 9.
Show that the angle bisectors of a triangle are concurrent.
Solution:
Given: ABC is a triangle. AD, BE and CF are the angle bisector of ∠A, ∠B, and ∠C.
To Prove: Bisector AD, BE and CF intersect
Proof: The angle bisectors AD and BE meet at O. Assume CF does not pass through O. By angle bisector theorem.
AD is the angle bisector of ∠A
\(\frac { BD }{ DC } \) = \(\frac { AB }{ AC } \) …..(1)
BE is the angle bisector of ∠B
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 10
\(\frac { CE }{ EA } \) = \(\frac { BC }{ AB } \) …….(2)
CF is the angle bisector ∠C
\(\frac { AF }{ FB } \) = \(\frac { AC }{ BC } \) …….(3)
Multiply (1) (2) and (3)
\(\frac { BD }{ DC } \) × \(\frac { CE }{ EA } \) × \(\frac { AF }{ FB } \) = \(\frac { AB }{ AC } \) × \(\frac { BC }{ AB } \) × \(\frac { AC }{ BC } \)
So by Ceva’s theorem.
The bisector AD, BE and CF are concurrent.

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4

Question 10.
In ∆ABC , with ∠B = 90° , BC = 6 cm and AB = 8 cm, D is a point on AC such that AD = 2 cm and E is the midpoint of AB. Join D to E and extend it to meet at F. Find BF.
Solution:
Consider ∆ABC. Then D, E and F are respective points on the sides AC, AB and BC.
By construction D, E, F are collinear.
By Menelaus’ theorem \(\frac { AE }{ EB } \) × \(\frac { BF }{ FC } \) × \(\frac { CD }{ DA } \) = 1 ……(1)
AD = 2 cm; AE = EB = 4 cm; BC = 6 cm; FC = FB + BC = x + 6
In ∆ABC, By Pythagoras theorem.
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 11
AC2= AB2 + BC2
AC2 = 82 + 62 = 64 + 36 = 100
AC = \(\sqrt { 100 }\) = 10
CD = AC – AD
= 10 – 2 = 8 cm
Substituting the values in (1) we get
\(\frac { 4 }{ 4 } \) × \(\frac { x }{ x+6 } \) × \(\frac { 8 }{ 2 } \) = 1
\(\frac { x }{ x+6 } \) × 4 = 1
4x = x + 6
3x = 6 ⇒ x = \(\frac { 6 }{ 3 } \) = 2
The value of BF = 2 cm

Question 11.
An artist has created a triangular stained glass window and has one strip of small length left before completing the window. She needs to figure out the length of left out portion based on the lengths of the other sides as shown in the figure.
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 12
Solution:
Given that AE = 3 cm, EC = 4 cm, CD = 10 cm, DB = 3 cm, AF = 5 cm.
Let FB be x
Using Ceva’s theorem we have
\(\frac { AE }{ EC } \) × \(\frac { CD }{ DB } \) × \(\frac { BF }{ AF } \) = 1
\(\frac { 3 }{ 4 } \) × \(\frac { 10 }{ 3 } \) × \(\frac { x }{ 5 } \) = 1
\(\frac { 2x }{ 4 } \) = 1
2x = 4 ⇒ x = \(\frac { 4 }{ 2 } \) = 2
The value of BF = 2

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4

Question 12.
Draw a tangent at any point R on the circle of radius 3.4 cm and centre at P ?
Answer:
Given Radius = 3.4 cm
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 13
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 14
Steps of construction:

  1. Draw a circle with centre “O” of radius 3.4 cm.
  2. Take a point P on the circle Join OP.
  3. Draw a perpendicular line TT’ to OP which passes through P.
  4. TT’ is the required tangent.

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4

Question 13.
Draw a circle of radius 4.5 cm. Take a point on the circle. Draw the tangent at that point using the alternate segment theorem.
Answer:
Radius of the circle = 4.5 cm
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 15
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 16
Steps of construction:

  1. With O as centre, draw a circle of radius 4.5 cm.
  2. Take a point L on the circle. Through L draw any chord LM.
  3. Take a point M distinct from L and N on the circle, so that L, M, N are in anti-clockwise direction. Join LN and NM.
  4. Through “L” draw tangent TT’such that ∠TLM = ∠MNL
  5. TT’ is the required tangent.

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4

Question 14.
Draw the two tangents from a point which is 10 cm away from the centre of a circle of radius 5 cm. Also, measure the lengths of the tangents.
Answer:
Radius = 5 cm; Distance = 10 cm
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 17
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 18
Steps of construction:

  1. With O as centre, draw a circle of radius 5 cm.
  2. Draw a line OP =10 cm.
  3. Draw a perpendicular bisector of OP, which cuts OP at M.
  4. With M as centre and MO as radius draw a circle which cuts previous circle at A and B.
  5. Join AP and BP. AP and BP are the required tangents.

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4

Verification: In the right ∆ OAP
PA2 = OP2 – OA2
= 102 – 52 = \(\sqrt { 100-25 }\) = \(\sqrt { 75 }\) = 8.7 cm
Lenght of the tangent is = 8.7 cm

Question 15.
Take a point which is 11 cm away from the centre of a circle of radius 4 cm and draw the two tangents to the circle from that point.
Answer:
Radius = 4 cm; Distance = 11 cm
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 19
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 20
Steps of construction:

  1. With O as centre, draw a circle of radius 4 cm.
  2. Draw a line OP = 11 cm.
  3. Draw a perpendicular bisector of OP, which cuts OP at M.
  4. With M as centre and MO as radius, draw a circle which cuts previous circle A and B.
  5. Join AP and BP. AP and BP are the required tangents.

This the length of the tangents PA = PB = 10.2 cm
Verification: In the right angle triangle OAP
PA2 = OP2 – OA2
= 112 – 42 = 121 – 16 = 105
PA = \(\sqrt { 105 }\) = 10.2 cm
Length of the tangents = 10.2 cm

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4

Question 16.
Draw the two tangents from a point which is 5 cm away from the centre of a circle of diameter 6 cm. Also, measure the lengths of the tangents.
Answer:
Radius = 3cm; Distance = 5cm.
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 21
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 22
Steps of construction:

  1. With O as centre, draw a circle of radius 3 cm.
  2. Draw a line OP = 5 cm.
  3. Draw a perpendicular bisector of OP, which cuts OP at M.
  4. With M as centre and MO as radius draw a circle which cuts previous circles at A and B.
  5. Join AP and BP, AP and BP are the required tangents.

The length of the tangent PA = PB = 4 cm
Verification: In the right angle triangle OAP
PA2 = OP2 – OA2
= 52 – 32
= 25 – 9
= 16 PA
= \(\sqrt { 16 }\) = 4 cm
Length of the tangent = 4 cm

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4

Question 17.
Draw a tangent to the circle from the point P having radius 3.6 cm, and centre at O. Point P is at a distance 7.2 cm from the centre.
Answer:
Radius = 3.6; Distance = 7.2 cm.
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 23
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 24
Steps of construction:

  1. With O as centre, draw a circle of radius 3.6 cm.
  2. Draw a line OP = 7.2 cm.
  3. Draw a perpendicular bisector of OP which cuts OP at M.
  4. With M as centre and MO as radius draw a circle which cuts the previous circle at A and B.
  5. Join AP and BP, AP and BP are the required tangents.

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4

Length of the tangents PA = PB = 6.26 cm
Verification: In the right triangle ∆OAP
PA2 = OP2 – OA2
= 7.22 – 3.62 =(7.2 + 3.6) (7.2 – 3.6)
PA2 = 10.8 × 3.6 = \(\sqrt { 38.88 }\)
PA = 6.2 cm
Length of the tangent = 6.2 cm

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3

Students can download Maths Chapter 8 Statistics Ex 8.3 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 8 Statistics Ex 8.3

Question 1.
The monthly salary of 10 employees in a factory are given below:
Rs 5000, Rs 7000, Rs 5000, Rs 7000, Rs 8000, Rs 7000, Rs 7000, Rs 8000, Rs 7000, Rs 5000
Find the mean, median and mode.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3 1
Mean = 6600
Median:
Arrange in ascending order we get.
5000, 5000, 5000, 7000, 7000, 7000, 7000, 7000, 8000, 8000
The number of values = 10
Median = Average of (\(\frac{10}{2})^{th}\) and (\(\frac{10}{2}+1)^{th}\)
= Average of 5th value and 6th value
= \(\frac{7000+7000}{2}\)
∴ Median = 7000
Mode: 7000 repeated 5 times
∴ Mode = 7000

Question 2.
Find the mode of the given data: 3.1, 3.2, 3.3, 2.1, 1.3, 3.3, 3.1
Solution:
3.1 occuring two times
3.3 occuring two times
∴ 3.1 and 3.3 are the mode (bimodal)

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3

Question 3.
For the data 11, 15, 17, x + 1, 19, x – 2, 3 if the mean is 14, find the value of x. Also find the mode of the data.
Solution:
Arithmetic mean
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3 2
∴ 2x + 64 = 14 × 7
2x = 98 – 64
2x = 34
x = \(\frac{34}{2}\)
= 17
The given numbers are 11, 15, 17, 18, 19, 15 and 3
15 occuring two times
∴ Mode = 15
The value of x = 17 and mode = 15

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3

Question 4.
The demand of track suit of different sizes as obtained by a survey is given below:
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3 3
Which size is demanded more?
Solution:
The highest frequency is 37
The corresponding value is the mode
∴ Mode = 40
Size 40 is demanded more.

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3

Question 5.
Find the mode of the following data:
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3 4
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3 5
The highest frequency is 46
20 – 30 is the modal class
Here l = 20, f = 46, f1 = 38, f2 = 34 and c = 10
Mode
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3 6
= 20 + 4
= 24
∴ Mode = 24

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3

Question 6.
Find the mode of the following distribution
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3 7
Solution:
In the given table the class intervals are in inclusive form; convert them into exclusive form.
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3 8
The highest frequency is 14
Modal class is 54.5 – 64.5
Here l = 54.5, f = 14, f1 = 10, f2 = 8 and c = 10
mode
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3 9
= 58.5
∴ Mode = 58.5

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.2

Students can download Maths Chapter 8 Statistics Ex 8.2 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 8 Statistics Ex 8.2

Question 1.
Find the median of the given values: 47, 53, 62, 71, 83, 21, 43, 47, 41
Solution:
Arrange the values in ascending order we get
21, 41, 43, 47, 47, 53, 62, 71, 83
The number of values = 9 which is odd
Median = (\(\frac{9+1}{2})^{th}\) variable
= 5th variable
∴ Median = 47

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.2

Question 2.
Find the median of the given data: 36, 44, 86, 31, 37, 44, 86, 35, 60, 51
Solution:
Arrange the values in ascending order we get
31, 35, 36, 37, 44, 44, 51, 60, 86, 86
The number of values = 10 which is even
Median = Average of (\(\frac{10}{2})^{th}\) and (\(\frac{10}{2}+1)^{th}\) value
= Average of 5th and 6th value
= \(\frac{44+44}{2}\)
= \(\frac{88}{2}\)
= 44
∴ Median = 44

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.2

Question 3.
The median of observation 11, 12, 14, 18, x + 2, x + 4, 30, 32, 35, 41 arranged in ascending order is 24. Find the values of x.
Solution:
The given observation is 11, 12, 14, 18, x + 2, x + 4, 30, 32, 35, 41 (is ascending order)
The number of values =10
Median = Average of (\(\frac{10}{2})^{th}\) and (\(\frac{10}{2}+1)^{th}\) value
= Average of 5th and 6th value
24 = \(\frac{x+2.+x+4}{2}\)
24 = \(\frac{2x+6}{2}\)
2x + 6 = 48
2x = 48 – 6
2x = 42
x = \(\frac{42}{2}\)
= 21
The value of x = 21

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.2

Question 4.
A researcher studying the behavior of mice has recorded the time (in seconds) taken by each mouse to locate its food by considering 13 different mice as 31, 33, 63, 33, 28, 29, 33, 27, 27, 34, 35, 28, 32. Find the median time that mice spent in searching its food.
Solution:
Arrange the value in ascending order we get
27, 27, 28, 28, 29, 31, 32, 33, 33, 33, 34, 35, 63
The number of values = 13 which is odd
Median = (\(\frac{13+1}{2})^{th}\) value
= (\(\frac{14}{2})^{th}\)
= 7th value
7th value is = 32
∴ Median = 32

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.2

Question 5.
The following are the marks scored by the students in the Summative Assessment exam.
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.2 1
Calculate the median.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.2 2
\(\frac{N}{2}\) = \(\frac{50}{2}\)
= 25
Here l = 30, f = 10; m = 24 and c = 10
Median = l + \(\frac{(\frac{N}{2}-m)×c}{f}\)
= 30 + \(\frac{(25-24)10}{10}\)
= 30 + 1
= 31
∴ Median = 31

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.2

Question 6.
The mean of five positive integers is twice their median. If four of the integers are 3, 4, 6, 9 and median is 6, then find the fifth integer.
Solution:
Let the 5th positive integer be x
\(\bar { x }\) = \(\frac{3+4+6+9+x}{5}\)
= \(\frac{22+x}{5}\)
Median = 6
Mean = 2 × median
\(\frac{22+x}{5}\) = 2 × 6
22 + x = 60
x = 60 – 22
= 38
The fifth integer is 38.

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.2

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.2

Students can download Maths Chapter 4 Geometry Ex 4.2 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 4 Geometry Ex 4.2

Question 1.
The angles of a quadrilateral are in the ratio 2 : 4 : 5 : 7. Find all the angles.
Solution:
Let the angles of a quadrilateral be 2x, 4x, 5x, and 7x.
Total angle of a quadrilateral = 360°
2x + 4x + 5x + 7x = 360°
18° = 360°
x = \(\frac{360°}{18}\)
= 20°
2x = 2 × 20° = 40°; 4x = 4 × 20° = 80°;
5x = 5 × 20° = 100°; 7x = 7 × 20° = 140°
The angles of a quadrilateral are 40°, 80°, 100° and 140°.

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.2

Question 2.
In a quadrilateral ABCD, ∠A = 72° and ∠C is the supplementary of ∠A. The other two angles are 2x – 10 and x + 4. Find the value of x and the measure of all the angles.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.2 1
∠A = 72°
∠C = 180° – 12° (∠A and ∠C are supplementary)
= 108°
∠A + ∠B + ∠C + ∠D = 360° (Total angles of quadrilateral)
72° + 2x – 10 + 108° + x + 4 = 360°
3x + 174° = 360°
x = \(\frac{186°}{3}\)
= 62°
The value of x is 62°
∠B = 2x – 10
= 2(62°) – 10
= 124° – 10°
= 114°
∠D = x + 4
= 62° + 4
= 66°
The other angles are 114°, 62° and 66°.

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.2

Question 3.
ABCD is a rectangle whose diagonals AC and BD intersect at O. If ∠OAB = 46°, find ∠OBC.
Solution:
Since the diagonals of a rectangle AC and BD are equal and bisect each other
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.2 2
∴ OA = OB
∠OAB = ∠OBA = 46°
Each angle of a rectangle measures 90°
∠ABC = 90°
∠ABO + ∠OBC = 90°
46° + ∠OBC = 90°
∠OBC = 90°-46°
∴ ∠OBC = 44°

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.2

Question 4.
The lengths of the diagonals of a Rhombus are 12 cm and 16 cm. Find the side of the rhombus.
Solution:
Since the diagonals of a rhombus bisect each other at right angles
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.2 3
AO = \(\frac{1}{2}\)AC = \(\frac{1}{2}\) × 12 = 6 cm
BO = \(\frac{1}{2}\)BD = \(\frac{1}{2}\) × 16 = 8 cm
In the right triangle AOD
AD² = AO² + DO²
= 6² + 8²
= 36 + 64
= 100
∴ AD = \(\sqrt{100}\)
= 10
∴ AB = BC = CD = AD = 10 cm.

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.2

Question 5.
Show that the bisectors of angles of a parallelogram form a rectangle.
Solution:
Given: A parallelogram in which bisector of angle A, B, C, D intersect at P, Q, R, S to form a quadrilateral PQRS.
To prove: Quadrilateral PQRS is a rectangle.
Proof: Since ABCD is a parallelogram. Therefore, AB || DC.
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.2 4
Now, AB || DC, and transversal AD cuts them, so we have
∠A + ∠D = 180°
\(\frac{1}{2}\)∠A + \(\frac{1}{2}\)∠D = \(\frac{180°}{2}\)
∠DAS + ∠ADS = 90°
But in ΔASD, we have
∠ADS + ∠DAS + ∠ASD = 180°
90° + ∠ASD = 180°
∠ASD = 90°
∠RSP = ∠ASD (vertically opposite angle)
∠RSP = 90°
Similarly, we can prove that
∠SRQ = 90°, ∠RQP = 90° and ∠QPS = 90°
Thus, PQRS is a quadrilateral each of whose angle is 90°.
Hence, PQRS is a rectangle.

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.2

Question 6.
If a triangle and a parallelogram lie on the same base and between the same parallels then prove that the area of the triangle is equal to half of the area of parallelogram.
Solution:
Let ΔAPB and parallelogram ABCD lie on base AB and between parallels AB and PC.
To show area ΔAPB = \(\frac{1}{2}\) Area (ABCD)
Now, draw BQ || AP. Then ABQP is a parallelogram.
Now area ABQP = Area ABCD
(They are on same base AB and between same parallels AB and PC)
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.2 5
⇒ ΔPAB ≅ ΔBQP
Area PAB = Area BQP
= \(\frac{1}{2}\) Area ABQP
= \(\frac{1}{2}\) Area ABCD

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.2

Question 7.
Iron rods a, b, c, d, e, and f are making a design in a bridge as shown in the figure.
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.2 6
If a || b, c || d, e || f, find the marked angles between
(i) b and c
(ii) d and e
(iii) d and f
(iv) c and f
Solution:
(i) Angle between b and c = 30°
(vertically opposite angles)

(ii) Angle between d and e = 180° – 75° = 105°
(sum of the adjacent angles of a parallelogram is 180°)

(iii) Angle between d and f = 75°
(opposite angles of a parallelogram)

(iv) Angle between c and f = 180° – 75° = 105°
(Adjacent angles of a parallelogram)

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.2

Question 8.
In the given figure, ∠A = 64°, ∠ABC = 58°. If BO and CO are the bisectors of ∠ABC and ∠ACB respectively of ΔABC, find x° and y°.
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.2 7
Solution:
In the given ΔABC
∠A = 64° and ∠B = 58°
∠C = 180°- (64° + 58°)
= 180° – 122°
= 58°
Since OC is the bisector of ∠C
y = \(\frac{58°}{2}\)
= 29°
Given ΔOBC
∠OCB = \(\frac{58°}{2}\) = 29°
∠OCB = 29°
∴ ∠BOC = 180°- (29° + 29°)
x = 180° – 58°
x = 122°
∠x = 122° and ∠y = 29°.

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.2

Question 9.
In the given figure, if AB = 2, BC = 6, AE = 6, BF = 8, CE = 7 and CF = 7, compute the ratio of the area of quadrilateral ABDE to the area of ΔCDF.
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.2 8
Solution:
Given: AB = 2 cm, BC = 6 cm, AE = 6 cm, BF = 8 cm, CE = 7 cm and CF = 7 cm
Consider ΔAEC and ΔBCF.
In ΔAEC, AE = 6 cm, EC = 7 cm and AC = 8 cm (2 + 6 = 8)
In ΔBCF, BC = 6 cm, CF = 7 cm and BF = 8 cm
∴ ΔAEC s ΔBCF
∴ Area of ΔAEC = Area of ΔBCF (Two triangles are similar areas are equal)
Subtract area of ΔBDC on both sides we get,
Area of ΔAEC – Area of ΔBDC = Area of ΔBCF – Area of ΔBDC
Area of quadrilateral ABDE = Area of ΔCDF

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.2

Question 10.
In the given figure, ABCD is a rectangle and EFGH is a parallelogram. Using the measurements given in the figure, what is the length ”d” of the segment that is perpendicular to \(\overline { HE }\) and \(\overline { FG }\)?
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.2 9
Solution:
In the given figure ABCD is a rectangle and EFGH is a parallelogram.
In the right triangle AEH
HE = \(\sqrt{AH^{2} + AE^{2}}\)
= \(\sqrt{3^{2} + 4^{2}}\)
= \(\sqrt{9 + 16}\)
= \(\sqrt{25}\)
HE = 5
∴ GF = 5 (HE and Gf are opposite sides of a parallelogram)
In the right triangle
GC = \(\sqrt{GF^{2} – FC^{2}}\)
= \(\sqrt{5^{2} – 3^{2}}\)
= \(\sqrt{25 – 9}\)
= \(\sqrt{16}\)
∴ DG = 10 – 6 = 4
Area of ΔAEH + Area of ΔBEF + Area of ΔFCG + Area of ΔHDG
= \(\frac{1}{2}\) × 3 × 4 + \(\frac{1}{2}\) × 6 × 5 + \(\frac{1}{2}\) × 3 × 4 + \(\frac{1}{2}\) × 5 × 6
= (6 + 15 + 6 + 15)
= 42
∴ Area of 4 triangles = 42
Area of the parallelogram = Area of the rectangle ABCD – Area of 4 triangles.
= 10 × 8 – 42
= 80 – 42
= 38
b × h = 38
5 × d = 38
d = \(\frac{38}{5}\)
= 7\(\frac{3}{5}\)
Length of d = 7\(\frac{3}{5}\) or 7.6

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.2

Question 11.
In parallelogram ABCD of the accompanying diagram, line DP is drawn bisecting BC at N and meeting AB (extended) at P. From vertex C, line CQ is drawn bisecting side AD at M and meeting AB (extended) at Q. Lines DP and CQ meet at O. Show that the area of triangle QPO is \(\frac{9}{8}\) of the area of the parallelogram ABCD.
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.2 10
Solution:
Draw OX perpendicular to QP.
In ΔADP, MN = \(\frac{1}{2}\) AP,
In ΔBCQ, MN = \(\frac{1}{2}\) QB
So, AP = BQ (or) AB + BP = AB + QA
∴ PB = QA
∴ QA = AB = BP (or) QP = QA + AB + BP = 3 AB
Area of ΔOQP = \(\frac{1}{2}\) × QP × OX
= \(\frac{1}{2}\) × 3 AB × OX
= \(\frac{3}{2}\) × AB × OX
= \(\frac{3}{2}\) AB (OY + YX)
= \(\frac{3}{2}\) × AB × OY + \(\frac{3}{2}\) × AB × YX (AB = MN)
= \(\frac{3}{2}\) × MN × OY + \(\frac{3}{2}\) × AB × YX
= 3 Area ΔOMN + \(\frac{3}{2}\) + Area ΔBNM
= 3[\(\frac{1}{4}\) area of MNCD] + \(\frac{3}{2}\)[\(\frac{1}{2}\) area of ABCD]
= \(\frac{3}{4}\)[\(\frac{1}{2}\) area of ABCD] + \(\frac{3}{4}\)[area of ABCD]
= \(\frac{3}{8}\) area of ABCD + \(\frac{3}{4}\) area of ABCD
= area of ABCD [\(\frac{3}{8}\) + \(\frac{3}{4}\)]
= area of ABCD (\(\frac{3+6}{8}\))
= \(\frac{9}{8}\) area of ABCD.
Hence it is proved.

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.2

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.13

Students can download Maths Chapter 3 Algebra Ex 3.13 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.13

Question 1.
Determine the nature of the roots for the following quadratic equations
(i) 15x2 + 11x + 2 = 0
Answer:
Here a = 15, b = 11, c = 2
∆ = b2 – 4ac
∆ = 112 – 4(15) × 2
= 121 – 120
∆ = 1 > 0
So the equation will have real and unequal roots

(ii) x2 – x – 1 = 0
Answer:
Here a = 1, b = -1, c = -1
∆ = b2 – 4ac
= (-1)2 – 4(1)(-1)
= 1 + 4 = 5
∆ = 1 > 0
So the equation will have real and unequal roots.

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.13

(iii) \(\sqrt { 2 }\) t2 – 3t + 3\(\sqrt { 2 }\) = 0
Answer:
Here a = \(\sqrt { 2 }\) , b = -3, c = 3\(\sqrt { 2 }\)
∆ = b2 – 4ac
= (-3)2 – 4(\(\sqrt { 2 }\)) (3\(\sqrt { 2 }\))
= 9 – 24 = -15
∆ = -15 < 0
So the equation will have no real roots.

(iv) 9y2 – 6\(\sqrt { 2y }\) + 2 = 0
Answer:
Here a = 9, b = -6\(\sqrt { 2 }\), c = 2
∆ = b2 – 4ac
= (-6\(\sqrt { 2 }\))2 – 4(9) (2)
= 72 – 72
= 0
So the equation will have real and equal roots.

(v) 9a2b2x2 – 24abcdx + 16c2d2 = 0, a ≠ 0, b ≠ 0
Answer:
Here a = 9a2b2; b = -24 abed, c = 16c2d2
∆ = b2 – 4ac
= (-24abcd)2 – 4(9a2b2) (16c2d2)
= 576a2b2c2d2 – 576a2b2c2d2
∆ = 0
So the equation will have real and equal roots.

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.13

Question 2.
Find the value(s) of ‘k’ for which the roots of the following equations are real and equal.
(i) (5k – 6)2 + 2kx + 1 = 0
(ii) kx2 + (6k + 2)x + 16 = 0
Solution:
(5k – 6)x2 + 2kx + 1 :
a = (5k – 6), b = 2, c = 1
Δ = b2 – 4ac
⇒ (2k)2 – 4 (5k – 6)(1)
⇒ 4k2 – 20k + 24 = 0 [∵ Since the roots are real and equal)
⇒ k2 – 5k + 6 = 0
⇒ (k – 3)(k – 2) = 0
k = 3, 2

(ii) kx2 + (6k + 2)x + 16 = 0
a = k, b = (6k + 2), c = 16
Δ = b2 – 4ac [∵ the roots are real and equal)
⇒ (6k + 2)2 – 4 × k × 16 = 0
⇒ 36k2 + 24k + 4 – 64k = 0
⇒ 36k2 – 40k + 4 =0
⇒ 36k2 – 36k – 4k + 4 =0
⇒ 36k (k – 1) – 4 (k – 1) = 0
⇒ 4 (k – 1) (9k – 1) =0
⇒ k = 1 or k = \(\frac{1}{9}\)

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.13

Question 3.
If the roots of (a – b)x2 + (b – c)x + (c – a) = 0 are real and equal, then prove that b, a, c are in arithmetic progression.
Answer:
(a – b) x2 + (b – c) x + (c – a) = 0
Here a = (a – b);b = b – c ; c = c – a
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.13 1
Since the equation has real and equal roots ∆ = 0
∴ b2 – 4ac = 0
(b – c)2 – 4(a – b)(c – a) = 0
b2 + c2 – 2bc -4 (ac – a2 – bc + ab) = 0
b2 + c2 – 2bc – 4ac + 4a2 + 4bc – 4ab = 0
b2 + c2 + 2bc -4a (b + c) + 4a2 = 0
(b + c)2 – 4a (b + c) + 4a2 = 0
[(b+c) – 2a]2 = 0 [using a2 – 2ab + b2 = (a – b)2]
b + c – 2a = 0
b + c = 2a
b + c = a + a
c – a = a – b (t2 – t1 = t3 – t2)
b,a,c are in A.P.

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.13

Question 4.
If a, b are real then show that the roots of the equation (a – b)2 – 6(a + b)x – 9(a – b) = 0 are real and unequal.
Solution:
(a – b)x2 – 6(a + b)x – 9(a – b) = 0
Δ = b2 – 4ac
= (-6(a + b)2 – 4(a – b)(-9(a – b))
= 36(a + b)2 + 36(a – b)2
= 36 (a2 + 2ab + b2) + 36(a2 – 2ab + b2)
= 72a2 + 12b2
= 72(a2 + b2) > 0
∴ The roots are real and unequal.

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.13

Question 5.
If the roots of the equation (c2 – ab)x2 – 2(a2 – bc)x + b2 – ac = 0 are real and equal prove that either a = 0 (or) a3 + b3 + c3 = 3abc
Answer:
(c2 – ab)x2 – 2(a2 – bc)x + b2 – ac = 0
Here a = c2 – ab ; b = – 2 (a2 – bc); c = b2 – ac
Since the roots are real and equal
∆ = b2 – 4ac
[-2 (a2 – bc)]2 – 4(c2 – ab) (b2 – ac) = 0
4(a2 – bc)2 – 4[c2 b2 – ac3 – ab3 + a2bc] = 0
Divided by 4 we get
(a2 – bc)2 – [c2 b2 – ac3 – ab3 + a2bc] = 0
a4 + b2 c2 – 2a2 bc – c2b2 + ac3 + ab3 – a2bc = 0
a4 + ab3 + ac3 – 3a2bc = 0
= a(a3 + b3 + c3) = 3a2bc
a3 + b3 + c3 = \(\frac{3 a^{2} b c}{a}\)
a3 + b3 + c3 = 3 abc
Hence it is proved

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7

Students can download Maths Chapter 4 Geometry Ex 4.7 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 4 Geometry Ex 4.7

Multiple Choice Questions

Question 1.
The exterior angle of a triangle is equal to the sum of two ……….
(a) Exterior angles
(b) Interior opposite angles
(c) Alternate angles
(d) Interior angles
Solution:
(b) Interior opposite angles

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7

Question 2.
In the quadrilateral ABCD, AB = BC and AD = DC Measure of ∠BCD is …………..
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7 1
(a) 150°
(b) 30°
(c) 105°
(d) 72°
Solution:
(c) 105°
Hint:
Join BD
∠DBC = 54°; ∠BDC = 21°
∴ ∠BCD = 180° – (54° + 21°)
= 180° – 75°
= 105°

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7

Question 3.
ABCD is a square, diagonals AC and BD meet at O. The number of pairs of congruent triangles with vertex O are ……….
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7 2
(a) 6
(b) 8
(c) 4
(d) 12
Solution:
(a) 6

Question 4.
In the given figure CE || DB then the value of x° is ………
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7 3
(a) 45°
(b) 30°
(c) 75°
(d) 85°
∠B = 360° – (∠A + ∠D + ∠BCD)
= 360° – (110° + 75° + 60°)
= 360° – 245°
= 115°
∠DBC = 115° – 30°
= 85°
∴ ∠x = 85° (BD || CE) Alternate angles are equal

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7

Question 5.
The correct statement out of the following is ……..
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7 4
(a) ΔABC ≅ ΔDEF
(b) ΔABC ≅ ΔEDF
(c) ΔABC ≅ ΔFDE
(d) ΔABC ≅ ΔFED
Solution:
(d) ΔABC ≅ ΔFED

Question 6.
If the diagonal of a rhombus are equal, then the rhombus is a ……..
(a) Parallelogram but not a rectangle
(b) Rectangle but not a square
(c) Square
(d) Parallelogram but not a square
Solution:
(c) Square

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7

Question 7.
If bisectors of ∠A and ∠B of a quadrilateral ABCD meet at O, then ∠AOB is ……..
(a) ∠C + ∠D
(b) \(\frac{1}{2}\) (∠C + ∠D)
(c) \(\frac{1}{2}\) ∠C + \(\frac{1}{3}\) ∠D
(d) \(\frac{1}{3}\) ∠C + \(\frac{1}{2}\) ∠D
Solution:
(b) \(\frac{1}{2}\) (∠C + ∠D)

Question 8.
The interior angle made by the side in a parallelogram is 90° then the parallelogram is a …….
(a) rhombus
(b) rectangle
(c) trapezium
(d) kite
Solution:
(b) rectangle
Hint:
Opposite sides are equal and each angle is 90° then it is a rectangle.

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7

Question 9.
Which of the following statement is correct?
(a) Opposite angles of a parallelogram are not equal
(b) Adjacent angles of a parallelogram are complementary.
(c) Diagonals of a parallelogram are always equal.
(d) Both pairs of opposite sides of a parallelogram are always equal.
Solution:
(d) Both pairs of opposite sides of a parallelogram are always equal.

Question 10.
The angles of the triangle are 3x – 40, x + 20 and 2x – 10 then the value of x is ………
(a) 40°
(b) 35°
(c) 50°
(d) 45°
Solution:
(b) 35°
Hint:
3x – 40 + x + 20 + 2x – 10 = 180° (Sum of the angles of a triangle is 180°)
6x – 30 = 180°
6x = 180°+ 30°
x = \(\frac{210°}{6°}\)
= 35°

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7

Question 11.
PQ and RS are two equal chords of a circle with centre O such that ∠POQ = 70°, then ∠ORS = ……….
(a) 60°
(b) 70°
(c) 55°
(d) 80°
Solution:
(c) 55°
Hint:
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7 5
∠POQ = 70° (Vertically opposite Angle)
∠ORS and ∠OSR are equal. (OR = OS radius of the circle)
∠ORS + ∠OSR + ∠ROS = 180°
x° + x° + 70° = 180°
2x + 70° = 180°
2x = 180° – 70° = 110°
x = \(\frac{110°}{2}\) = 55°
∴ ∠ORS = 55°

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7

Question 12.
A chord is at a distance of 15cm from the centre of the circle of radius 25cm. The length of the chord is ……….
(a) 25cm
(b) 20cm
(c) 40cm
(d) 18cm
Solution:
(c) 40cm
Hint:
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7 6
In the right triangle OAC,
AC² = OA² – OC²
= 25² – 15²
= (25 + 15) (25 – 15)
= 40 × 10
AC² = 400
AC = \(\sqrt{400}\)
= 20
Length of the chord AB = 20 + 20 = 40 cm.

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7

Question 13.
In the figure, O is the centre of the circle and ∠ACB = 40° then ∠AOB = ………
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7 7
(a) 80°
(b) 85°
(c) 70°
(d) 65°
Solution:
(a) 80°
Hint:
∠AOB = 2∠ACB (The angle subtended by an arc of the circle at the centre is double the angle subtended at the remaining part of the circle.)
= 2 × 40° = 80°

Question 14.
In a cyclic quadrilaterals ABCD, ∠A = 4x, ∠C = 2x the value of x is ……..
(a) 30°
(b) 20°
(c) 15°
(d) 25
Solution:
(a) 30°
Hint:
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7 8
∠A + ∠C = 180° (Sum of the opposite angle of cyclic quadrilateral is 180°)
4x + 2x = 180°
x = \(\frac{180°}{6}\)
= 30°

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7

Question 15.
In the figure, O is the centre of a circle and diameter AB bisects the chord CD at a point E such that CE = ED = 8 cm and EB = 4cm. The radius of the circle is ………
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7 9
(a) 8cm
(b) 4cm
(c) 6cm
(d) 10cm
Solution:
(i) 10cm
Hint:
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7 10
Let the radius OD be x.
OE = OB – BE
= x – 4 (OB radius of the circle)
In the ΔOED,
OD² = OE² + ED²
x² = (x – 4)² + 8²
x² = x² + 16 – 8x + 64
8x = 80
x = \(\frac{80}{8}\)
= 10 cm
Radius of the circle = 10 cm.

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7

Question 16.
In the figure, PQRS and PTVS are two cyclic quadrilaterals, If ∠QRS = 80°, then ∠TVS = ………
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7 11
(a) 80°
(b) 100°
(c) 70°
(d) 90°
Solution:
(a) 80°
Hint:
∠SPQ = 180° – 80° (Opposite angles of a cyclic quadrilateral PQRS)
= 100°
In the cyclic quadrilateral PVTS,
∠V = 180° – 100° = 80° (Opposite angles of a cyclic quadrilateral)

Question 17.
If one angle of a cyclic quadrilateral is 75°, then the opposite angle is ………
(a) 100°
(b) 105°
(c) 85°
(d) 90°
Solution:
(b) 105°
Hint:
Opposite angles = 180° – 75°
= 105° (Sum of the opposite angle is 180°)

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7

Question 18.
In the figure, ABCD is a cyclic quadrilateral in which DC produced to E and CF is drawn parallel to AB such that ∠ADC = 80° and ∠ECF = 20°, then BAD = ?
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7 12
(a) 100°
(b) 20°
(c) 120°
(d) 110°
Solution:
(c) 120°
Hint:
∠BAD = ∠ABC + ∠ECF (By exterior angle property)
= 100° + 20°
= 120°

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7

Question 19.
AD is a diameter of a circle and AB is a chord If AD = 30cm and AB = 24cm then the distance of AB from the centre of the circle is ………
(a) 10cm
(b) 9cm
(c) 8cm
(d) 6cm
Solution:
(b) 9cm
Hint:
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7 13
In ΔAOC,
AO = 15 cm
AC = \(\frac{1}{2}\) AB
= \(\frac{1}{2}\) × 24
= 12 cm
In ΔAOC,
OC² = AO² – AC²
= 15² – 12²
= 225 – 144
= 81
OC = \(\sqrt{81}\)
= 9 cm

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7

Question 20.
In the given figure, If OP = 17cm, PQ = 30cm and OS is perpendicular to PQ, then RS is ……….
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7 14
(a) 10cm
(b) 6cm
(c) 7cm
(d) 9cm
Solution:
(d) 9cm
Hint:
In ΔOPR, OP = 17cm, PR = 30/2 = 15 cm.
OR² = OP² – PR²
= 17² – 15²
= (17 + 15) (17 – 15)
= 32 × 2
OR² = \(\sqrt{64}\)
= 8 cm
RS = OS – OR
= 17 – 8
= 9 cm

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.12

Students can download Maths Chapter 3 Algebra Ex 3.12 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.12

Question 1.
If the difference between a number and its reciprocal is \(\frac { 24 }{ 5 } \), find the number.
Answer:
Let the number be “x” and its reciprocal is \(\frac { 1 }{ x } \)
By the given condition
x – \(\frac { 1 }{ x } \) = \(\frac { 24 }{ 5 } \)
\(\frac{x^{2}-1}{x}\) = \(\frac { 24 }{ 5 } \)
5x2 – 5 = 24x
5x2 – 24x – 5 = 0
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.12 1
5x2 – 25x + x – 5 = 0 ⇒ 5x (x – 5) + 1(x – 5) = 0
(x – 5) (5x – 1) = 0 ⇒ x – 5 = 0 or 5x + 1 = 0
x = 5 or 5x = -1 ⇒ x = \(\frac { -1 }{ 5 } \)
The number is 5 or \(\frac { -1 }{ 5 } \)

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.12

Question 2.
A garden measuring 12m by 16m is to have a pedestrian pathway that is meters wide installed all the way around so that it increases the total area to 285 m2. What is the width of the pathway?
Answer:
Let the width of the rectangle be “ω”
Length of the outer rectangle = 16 + (ω + ω)
16 + 2ω
Breadth of the outer rectangle = 12 + 2ω
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.12 2
By the given condition
(16 + 2ω) (12 + 2ω) = 285
192 + 32 ω + 24 ω + 4 ω2 = 285
4 ω2 + 56ω = 285 – 192
4 ω2 + 56 ω = 93
4 ω2+ 56 ω – 93 = 0
Here a = 4, b = 56, c = -93
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.12 3
= 1.5 or -15.5 (Width is not negative)
∴ Width of the path way = 1.5 m

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.12

Question 3.
A bus covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hour more it would have taken 30 minutes less for the journey. Find the original speed of the journey.
Answer:
Let the original speed of the bus be “x” km/hr
Time taken to cover 90 km = \(\frac { 90 }{ x } \)
After increasing the speed by 15 km/hr
Time taken to cover 90 km = \(\frac { 90 }{ x+15 } \)
By the given condition
\(\frac { 90 }{ x } \) – \(\frac { 90 }{ x+15 } \) = \(\frac { 1 }{ 2 } \)
\(\frac{90(x+15)-90 x}{x(x+15)}\) = \(\frac { 1 }{ 2 } \)
90x + 1350 – 90x = \(\frac{x^{2}+15 x}{2}\)
1350 = \(\frac{x^{2}+15 x}{2}\)
2700 = x2 + 15x
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.12 4
x2 + 15x – 2700 = 0
(x + 60) (x – 45) = 0
x + 60 = 0 or x – 45 = 0
x = – 60 or x = 45
The speed will not be negative
∴ Original speed of the bus = 45 km/hr

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.12

Question 4.
A girl is twice as old as her sister. Five years hence, the product of their ages – (in years) will be 375. Find their present ages.
Answer:
Let the age of the sister be “x”
The age of the girl = 2x
Five years hence
Age of the sister = x + 5
Age of the girl = 2x + 5
By the given condition
(x + 5) (2x + 5) = 375
2x2 + 5x + 10x + 25 = 375
2x2 + 15x – 350 = 0
a = 2, b = 15, c = -350
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.12 5
Age will not be negative
Age of the girl = 10 years
Age of the sister = 20 years (2 × 10)

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.12

Question 5.
A pole has to be erected at a point on the boundary of a circular ground of diameter 20 m in such a way that the difference of its distances from two diametricallyopposite fixed gates P and Q on the boundary is 4 m. Is it possible to do so? If answer is yes at what distance from the two gates should the pole be erected?
Answer:
Let “R” be the required location of the pole
Let the distance from the gate P is “x” m : PR = “x” m
The distance from the gate Q is (x + 4)m
∴ QR = (x + 4)m
In the right ∆ PQR,
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.12 6
PR2 + QR2 = PQ2 (By Pythagoras theorem)
x2 + (x + 4)2 = 202
x2 + x2 + 16 + 8x = 400
2x2 + 8x – 384 = 0
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.12 7
x2 + 4x – 192 = 0(divided by 2)
(x + 16) (x – 12) = 0
x + 16 = 0 or x – 12 = 0 [negative value is not considered]
x = -16 or x = 12
Yes it is possible to erect
The distance from the two gates are 12 m and 16 m

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.12

Question 6.
From a group of 2x2 black bees , square root of half of the group went to a tree. Again eight-ninth of the bees went to the same tree. The remaining two got caught up in a fragrant lotus. How many bees were there in total?
Answer:
Total numbers of black bees = 2x2
Half of the group = \(\frac { 1 }{ 2 } \) × 2x2 = x2
Square root of half of the group = \(\sqrt{x^{2}}\) = x
Eight – ninth of the bees = \(\frac { 8 }{ 9 } \) × 2x2 = \(\frac{16 x^{2}}{9}\)
Number ofbees in the lotus = 2
By the given condition
x + \(\frac{16 x^{2}}{9}\) + 2 = 2x2
(Multiply by 9) 9x + 16×2 + 18 = 18x2
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.12 8
18x2 – 16x2 – 9x – 18 = 0 ⇒ 2x2 – 9x – 18 = 0
2x2 – 12x + 3x – 18 = 0
2x(x – 6) + 3 (x – 6) = 0
(x – 6) (2x + 3) = 0
x – 6 = 0 or 2x + 3 = 0
x = 6 or 2x = -3 ⇒ x = \(\frac { -3 }{ 2 } \) (number of bees will not be negative)
Total number of black bees = 2x2 = 2(6)2
= 72

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.12

Question 7.
Music is been played in two opposite galleries with certain group of people. In the first gallery a group of 4 singers were singing and in the second gallery 9 singers were singing. The two galleries are separated by the distance of 70 m. Where should a person stand for hearing the same intensity of the singers voice?
(Hint: The ratio of the sound intensity is equal to the square of the ratio of their corresponding distances).
Answer:
Number of singers in the first group = 4
Number of singers in the second group = 9
Distance between the two galleries = 70 m
Let the distance of the person from the first group be x
and the distance of the person from the second group be 70 – x
By the given condition
4 : 9 = x2 : (70 – x)2 (by the given hint)
\(\frac { 4 }{ 9 } \) = \(\frac{x^{2}}{(70-x)^{2}}\)
\(\frac { 2 }{ 3 } \) = \(\frac { x }{ 70-x } \) [taking square root on both sides]
3x = 140 – 2x
5x = 140
x = \(\frac { 140 }{ 5 } \) = 28
The required distance to hear same intensity of the singers voice from the first galleries is 28m
The required distance to hear same intensity of the singers voice from the second galleries is (70 – 28) = 42 m

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.12

Question 8.
There is a square field whose side is 10 m. A square flower bed is prepared in its centre leaving a gravel path all round the flower bed. The total cost of laying the flower bed and gravelling the path at ₹3 and ₹4 per square metre respectively is ₹364. Find the width of the gravel path.
Answer:
Let the width of the gravel path be ‘x’
Side of the flower bed = 10 – (x + x)
= 10 – 2x
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.12 9
Area of the path way = Area of the field – Area of the flower bed
= 10 × 10 – (10 – 2x) (10 – 2x) sq.m
= 100 – (100 + 4x2 – 40x)
= 100 – 100 – 4x2 + 40x
= 40x – 4x2 sq.m
Area of the flower bed = (10 – 2x) (10 – 2x) sq.m.
= 100 + 4x2 – 40x
By the given condition
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.12 10
3(100 + 4x2 – 40x) + 4(40x – 4x2) = 364
300 + 12x2 – 120x + 160x – 16x2 = 364
-4x2 + 40x + 300 – 364 = 0
-4x2 + 40x – 64 = 0
(÷ by 4) ⇒ x2 – 10x + 16 = 0
[The width must not be equal to 8 m since the side of the field is 10m]
(x – 8) (x – 2) = 0
x – 8 = 0 or x – 2 = 0 x = 8 or x = 2
Width of the gravel path = 2 m

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.12

Question 9.
Two women together took 100 eggs to a market, one had more than the other. Both sold them for the same sum of money. The first then said to the second: “If I had your eggs, I would have earned ₹ 15”, to which the second replied: “If I had your eggs, I would have earned ₹ 6 \(\frac{2}{3}\) How many eggs did each had in the beginning?
Solution:
Let the no. of eggs with woman 1 be x and woman 2 be y.
∴ x + y = 100
Let w1 sell the eggs at ₹ ‘a’ per egg.
Let w2 sell the eggs at ₹ ‘b’ per egg.
Case 1:
They sold them for same money.
∴ ax = by
Case 2:
ay = 15 and bx = \(\frac{20}{3}\)
∴ One woman had 40 eggs and the other had 60 eggs.

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.12

Question 10.
The hypotenuse of a right angled triangle is 25 cm and its perimeter 56 cm. Find the length of the smallest side.
Answer:
Perimeter of a right angle triangle = 56 cm
Sum of the two sides + hypotenuse = 56
Sum of the two sides = 56 – 25
= 31 cm
Let one side of the triangle be “x”
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.12 11
The other side of the triangle = (31 – x) cm
By Pythagoras theorem
AB2 + BC2 = AC2
x2 + (31 – x)2 = 252
x2 + 961 + x2 – 62x = 625
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.12 12
2x2 – 62x + 961 – 625 = 0
2x2 – 62x + 336 = 0 ⇒ x2 – 31x + 168 = 0
(x – 24) (x – 7) = 0
x – 24 = 0 (or) x – 7 = 0
x = 24 (or) x = 7
Length of the smallest side is 7 cm

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.6

Students can download Maths Chapter 4 Geometry Ex 4.6 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 4 Geometry Ex 4.6

Question 1.
Draw a triangle ABC, where AB = 8 cm, BC = 6 cm and ∠B = 70° and locate its circumcentre and draw the circumcircle.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.6 1
Steps for construction:
Step 1: Draw the ΔABC with the given measures.
Step 2: Construct the perpendicular bisector of (AB and BC) any two sides and let them, meet at S which is the circumcenter.
Step 3: With S as centre and SA = SB = SC as radius draw the circumcircle to passes through A, B and C.
Circum radius = 4.3 cm.

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.6

Question 2.
Construct the right triangle PQR whose perpendicular sides are 4.5 cm and 6 cm. Also locate its circumcentre and draw circumcircle.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.6 2
Steps for construction:
Step 1: Draw the ΔPQR with the given measures.
Step 2: Construct the perpendicular bisector of (PQ and PR) any two sides and let them meet at S which is the circumcenter.
Step 3: With S as centre and SP = SQ = SR as radius draw the circumcircle to passes through P, Q and R.
Circum radius = 3.8 cm.

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.6

Question 3.
Construct ΔABC with AB = 5 cm ∠B = 100° and BC = 6 cm. Also locate its circumcentre draw circumcircle.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.6 3
Steps for construction:
Step 1: Draw the ΔABC with the given measures.
Step 2: Construct the perpendicular bisector of any two sides (AB and BC) and let them meet at S which is the circumcenter.
Step 3: With S as centre and SA = SB = SC as radius draw the circumcircle to passes through A, B and C.
Circum radius = 4.3 cm.

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.6

Question 4.
Construct an isosceles triangle PQR where PQ = PR and ∠Q = 50°, QR = 7cm. Also draw its circumcircle.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.6 4
Given PQ = PR
∴ ∠R = 50° (opposite angles are equal)
Steps for construction:
Step 1: Draw the ΔABC with the given measures.
Step 2: Construct the perpendicular bisector of any two sides (QR and PR) and let them meet at S. S is the circumcenter of ΔPQR.
Step 3: With S as centre SP = SQ = SR as radius. Draw the circumcircle.
Circum radius = 3.5 cm.

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.6

Question 5.
Draw an equilateral triangle of side 6.5 cm and locate its incentre. Also draw the incircle.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.6 5
Steps for construction:
Step 1: Draw the ΔABC with the each side measure 6.5 cm.
Step 2: Construct the angles bisectors of any two angles (A and B) and let them meet at I. Then I is the incentre of ΔABC.
Step 3: Draw perpendicular from I to any one of the side (AB) to meet AB at D.
Step 4: With I as centre and ID as radius draw the circle. This circle touches all the sides of the triangle internally.
Step 5: Measure of In-radius = 1.5 cm.

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.6

Question 6.
Draw a right triangle whose hypotenuse is 10 cm and one of the legs is 8 cm. Locate its incentre and also draw the incircle
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.6 6
Steps for construction:
Step 1: Draw the ΔABC with AB = 8cm, BC = 10 cm and ∠A = 90°.
Step 2: Construct the angle bisectors of any two angles (∠B and ∠C) and let them meet at I. Then I is the incentre of ΔABC.
Step 3: Draw perpendicular from I to any one of the side to meet AB at D.
Step 4: With I as centre and ID as radius draw the circle. This circle touches all the sides of the triangle internally.
Step 5: Measure of In-radius = 1.8 cm.

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.6

Question 7.
Draw ΔABC given AB = 9 cm, ∠CAB = 115° and ∠ABC = 40°. Locate its incentre and also draw the incircle. (Note: You can check from the above examples that the incentre of any triangle is always in its interior).
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.6 7
Step 1: Draw the ΔABC with AB = 9cm, ∠B = 40° and ∠A = 115°.
Step 2: Construct the angle bisectors of any two angles (A and B) and let them meet at I. Then I is the incentre of ΔABC.
Step 3: Draw perpendicular from I to any one of the side (AB) to meet AB at D.
Step 4: With I as centre and ID as radius draw the circle. This circle touches all the sides of the triangle internally.
Step 5: Measure of In-radius = 2.7 cm.

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.6

Question 8.
Construct ΔABC in which AB = BC = 6cm and ∠B = 80°. Locate its incentre and draw the incircle.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.6 8
Steps for construction:
Step 1: Draw the ΔABC with AB = 6 cm, ∠B = 80° and BC = 6 cm.
Step 2: Construct the angle bisectors of any two angles (A and B) and let them meet at I. Then I is the incentre of ΔABC.
Step 3: Draw perpendicular from I to any one of the side (AB) to meet AB at D.
Step 4: With I as centre and ID as radius draw the circle. This circle touches all the sides of the triangle internally.
Step 5: Measure of In-radius = 1.9 cm.

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.6

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.5

Students can download Maths Chapter 4 Geometry Ex 4.5 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 4 Geometry Ex 4.5

Question 1.
Construct the ΔLMN such that LM = 7.5 cm, MN = 5cm and LN = 8cm. Locate its centroid.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.5 1
Steps for construction:
Step 1: Draw the ΔLMN using the given measurement LM = 7.5 cm, MN = 5cm and LN = 8cm.
Step 2: Construct the perpendicular bisectors of any two sides LM and MN intersect LM at P and MN at Q respectively.
Step 3: Draw the median LQ and PN meet at G.
The point G is the centroid of the given ΔLMN.

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.5

Question 2.
Draw and locate the centroid of the triangle ABC where right angle at A, AB = 4cm and AC = 3 cm.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.5 2
Steps for construction:
Step 1: Draw the ΔABC using the given measurement AB = 4cm and AC = 3 cm and ZA = 90°.
Step 2: Construct the perpendicular bisectors of any two sides AB and AC to find the mid-points P and Q of AB and AC.
Step 3: Draw the medians PC and BQ intersect at G.
The point G is the centroid of the given ΔABC.

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.5

Question 3.
Draw the ΔABC, where AB = 6cm, ∠B = 110° and AC = 9cm and construct the centroid.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.5 3
Steps for construction:
Step 1: Draw the ΔABC using the given measurement AB = 6cm, AC = 9cm and ∠B =110°.
Step 2: Construct the perpendicular bisectors of any two sides AB and BC to find the mid-points P and Q of AB and BC.
Step 3: Draw the medians PC and AQ intersect at G.
The point G is the centroid of the given ΔABC.

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.5

Question 4.
Construct the ΔPQR such that PQ = 5cm, PR = 6cm and ∠QPR = 60° and locate its centroid.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.5 4
Steps for construction:
Step 1 : Draw ΔPQR using the given measurements PQ = 5cm, PR = 6cm and ∠P = 60°.
Step 2 : Construct the perpendicular bisectors of any two sides PQ and QR to find the mid-points of M and N respectively.
Step 3 : Draw the median PN and MR and let them meet at G.
The point G is the centroid of the given ΔPQR.

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.5

Question 5.
Draw ΔPQR with sides PQ = 7 cm, QR = 8 cm and PR = 5 cm and construct its Orthocentre.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.5 5
Steps for construction:
Step 1: Draw the ΔPQR with the given measurements.
Step 2: Construct altitudes from any two vertices P and Q to their opposite sides QR and PR respectively.
Step 3: The point of intersection of the altitude H is the orthocentre of the given ΔPQR.

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.5

Question 6.
Draw an equilateral triangle of sides 6.5 cm and locate its Orthocentre.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.5 6
Steps for construction:
Step 1: Draw the ΔABC with the given measurements.
Step 2: Construct altitudes from any two vertices A and C to their opposite sides BC and AB respectively.
Step 3: The point of intersection of the altitude H is the orthocentre of the given ΔABC.

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.5

Question 7.
Draw ΔABC, where AB = 6 cm, ∠B = 110° and BC = 5 cm and construct its Orthocentre.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.5 7
Steps for construction:
Step 1: Draw the ΔABC with the given measurements.
Step 2: Construct altitudes from any two vertices B and C to their opposite sides AC and BC respectively.
Step 3: The point of intersection of the altitude H is the orthocentre of the given ΔABC.

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.5

Question 8.
Draw and locate the Orthocentre of a right triangle PQR where PQ = 4.5 cm, QR = 6 cm and PR = 7.5 cm.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.5 8
Steps for construction:
Step 1: Draw the ΔPQR with the given measures.
Step 2: Construct altitude from any two vertices Q and R to their opposite side PR and PQ respectively.
Step 3: The point of intersection of the altitude H is the orthocentre of the given ΔPQR.

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.5