Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions

Students can download Maths Chapter 8 Statistics Additional 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 Additional Questions

I. Choose the Best Answer

Question 1.
The Arithmetic mean of all the factors of 10 is ………
(a) 4.5
(b) 5.5
(c) 10
(d) 55
Solution:
(a) 4.5

Question 2.
The mean of five numbers is 27, if one number is excluded, then mean is 25. Then the excluded number is ……..
(a) 0
(b) 15
(c) 25
(d) 35
Solution:
(d) 35

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions

Question 3.
The mean of 8 numbers is 15. If each number is multiplied by 2, then the new mean will be ……..
(a) 7.5
(b) 30
(c) 10
(d) 25
Solution:
(b) 30

Question 4.
The median of 11, 8, 4, 9, 7, 5, 2, 4, 10 is ……..
(a) 1
(b) 8
(c) 4
(d) 11
Solution:
(a) 1

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions

Question 5.
Median is …….
(a) the most frequent value
(b) the least frequent value
(c) middle most value
(d) mean of first and last values
Solution:
(c) middle most value

Question 6.
The mode of the distribution is …….
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 1
(a) 3
(b) 4
(c) 6
(d) 14
Solution:
(b) 4

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions

Question 7.
Mode is …….
(a) the middle value
(b) extreme value
(c) minimum value
(d) the most repeated value
Solution:
(d) the most repeated value

Question 8.
The mode of the data 72, 33, 44, 72, 81, 72, 15 is ……..
(a) 12
(b) 33
(c) 81
(d) 15
Solution:
(a) 12

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions

Question 9.
The Arithmetic mean of 10 number is -7. If 5 is added to every number, then the new arithmetic mean is ……..
(a) 17
(b) 12
(c) -2
(d) -7
Solution:
(c) -2

Question 10.
The Arithmetic mean of integers from -5 to 5 is ……..
(a) 25
(b) 10
(c) 3
(d) 0
Solution:
(d) 0

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions

II. Answer the Following Questions

Question 1.
Find the Arithmetic mean for the following data.
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 2
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 3
Arithmetic mean (\(\bar { X }\)) = \(\frac{Σfx}{Σf}\)
= \(\frac{5540}{96}\)
= 57.7
∴ Arithmetic mean = 57.7

Question 2.
Calculate the Arithmetic mean of the following data using step deviation method.
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 4
Solution:
Assumed mean = 35
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 5
Arithmetic mean (\(\bar { X }\)) = A + \(\frac{Σfd}{Σf}\) × c
= 35 + \(\frac{(-20)}{100}\) × 10
= 35 – 2
= 33
∴ Arithmetic mean = 33

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions

Question 3.
Find the median for the following data.
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 6
Solution:
The given class intervals is inclusive type we convert it into exclusive type.
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 7
\(\frac{N}{2}\) = \(\frac{70}{2}\)
= 35
Median class is 25.5 – 30.5
Here l = 25.5, f = 26, m = 30, c = 5
Median = l + \(\frac{\frac{N}{2}-m}{f}\) × c
= 25.5 + \(\frac{35-30}{26}\) × 5
= 25.5 + \(\frac{5×5}{26}\)
= 25.5 + 0.96
= 26.46
∴ Median = 26.46

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions

Question 4.
Calculate the mode of the following data.
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 8
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 9
The highest frequency is 30. Corresponding class interval is the modal class.
Here l = 25, f = 30, f1 = 18, f2 = 20 and c = 5
Mode
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 10
∴ Arithmetic mean = 25 + 2.727
= 25 + 2.73
= 27.73
mode = 27.73

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions

Question 5.
Find the mean, median and mode of marks obtained by 20 students in an examination. The marks are given below.
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 11
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 12
Arithmetic mean:
Here Σfx = 560, Σf = 20
\(\bar { X }\) = \(\frac{Σfx}{Σf}\)
= \(\frac{560}{20}\)

Median:
Median class is 20 – 30
Here l = 20, \(\frac{N}{2}\) = 10, m = 5, c = 10, f = 5
Median
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 13
= 20 + 10
= 30

Mode:
The highest frequency is 8
Modal class is 30 – 40
Here, l = 30, f = 8, f1 = 5, f2 = 2, c = 10
Mode
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 14
= 33.33
Mean = 28, median = 30, mode = 33.33

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions

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 10th Maths Guide Chapter 4 Geometry Ex 4.3

Students can download Maths Chapter 4 Geometry Ex 4.3 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 4.3

Question 1.
A man goes 18 m due east and then 24 m due north. Find the distance of his current position from the starting point?
Solution:
Let the initial position of the man be “O” and his final
position be “B”.
By Pythagoras theorem
In the right ∆ OAB,
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 1
OB2 = OA2 + AB2
= 182 + 242
= 324 + 576 = 900
OB = \(\sqrt { 900 }\) = 30
The distance of his current position is 30 m

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3

Question 2.
There are two paths that one can choose to go from Sarah’s house to James house. One way is to take C street, and the other way requires to take A street and then B street. How much shorter is the direct path along C street? (Using figure).
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 2
Solution:
Distance between Sarah House and James House using “C street”.
AC2 = AB2 + BC2
= 22 + 1.52
= 4 + 2.25 = 6.25
AC = \(\sqrt { 6.25 }\)
AC = 2.5 miles
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 3
Distance covered by using “A Street” and “B Street”
= (2 + 1.5) miles = 3.5 miles
Difference in distance = 3.5 miles – 2.5 miles = 1 mile

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3

Question 3.
To get from point A to point B you must avoid walking through a pond. You must walk 34 m south and 41 m east. To the nearest meter, how many meters would be saved if it were possible to make a way through the pond?
Solution:
In the right ∆ABC,
By Pythagoras theorem
AC2= AB2 + BC2 = 342 + 412
= 1156 + 1681 = 2837
AC = \(\sqrt { 2837 }\)
= 53.26 m
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 4
Through A one must walk (34m + 41m) 75 m to reach C.
The difference in Distance = 75 – 53.26
= 21.74 m

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3

Question 4.
In the rectangle WXYZ, XY + YZ = 17 cm, and XZ + YW = 26 cm.
Calculate the length and breadth of the rectangle?
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 5
Solution:
Let the length of the rectangle be “a” and the breadth of the rectangle be “b”.
XY + YZ = 17 cm
b + a = 17 …….. (1)
In the right ∆ WXZ,
XZ2 = WX2 + WZ2
(XZ)2 = a2 + b2
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 6
XZ = \(\sqrt{a^{2}+b^{2}}\)
Similarly WY = \(\sqrt{a^{2}+b^{2}}\) ⇒ XZ + WY = 26
2 \(\sqrt{a^{2}+b^{2}}\) = 26 ⇒ \(\sqrt{a^{2}+b^{2}}\) = 13
Squaring on both sides
a2 + b2 = 169
(a + b)2 – 2ab = 169
172 – 2ab = 169 ⇒ 289 – 169 = 2 ab
120 = 2 ab ⇒ ∴ ab = 60
a = \(\frac { 60 }{ b } \) ….. (2)
Substituting the value of a = \(\frac { 60 }{ b } \) in (1)
\(\frac { 60 }{ b } \) + b = 17
b2 – 17b + 60 = 0
(b – 2) (b – 5) = 0
b = 12 or b = 5
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 7
If b = 12 ⇒ a = 5
If b = 6 ⇒ a = 12
Lenght = 12 m and breadth = 5 m

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3

Question 5.
The hypotenuse of a right triangle is 6 m more than twice of the shortest side. If the third side is 2 m less than the hypotenuse, find the sides of the triangle.
Solution:
Let the shortest side of the right ∆ be x.
∴ Hypotenuse = 6 + 2x
Third side = 2x + 6 – 2
= 2x + 4
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 8
In the right triangle ABC,
AC2 = AB2 + BC2
(2x + 6)2 = x2 + (2x + 4)2
4x2 + 36 + 24x = x2 + 4x2 + 16 + 16x
0 = x2 – 24x + 16x – 36 + 16
∴ x2 – 8x – 20 = 0
(x – 10) (x + 2) = 0
x – 10 = 0 or x + 2 = 0
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 9
x = 10 or x = -2 (Negative value will be omitted)
The side AB = 10 m
The side BC = 2 (10) + 4 = 24 m
Hypotenuse AC = 2(10) + 6 = 26 m

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3

Question 6.
5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4m high. If the foot of the ladder is moved 1.6 m towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall.
Solution:
“C” is the position of the foot of the ladder “A” is the position of the top of the ladder.
In the right ∆ABC,
BC2 = AC2 – AB2 = 52 – 42
= 25 – 16 = 9
BC = \(\sqrt { 9 }\) = 3m.
When the foot of the ladder moved 1.6 m toward the wall.
The distance between the foot of the ladder to the ground is
BE = 3 – 1.6 m
= 1.4 m
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 10
Let the distance moved upward on the wall be “h” m
The ladder touch the wall at (4 + h) M
In the right triangle BED,
ED2 = AB2 + BE2
52 = (4 + h)2 + (1.4)2
25 – 1.96= (4 + h)2
∴ 4 + h = \(\sqrt { 23.04 }\)
4 + h = 4. 8 m
h = 4.8 – 4
= 0.8 m
Distance moved upward on the wall = 0.8 m

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3

Question 7.
The perpendicular PS on the base QR of a ∆PQR intersects QR at S, such that QS = 3 SR. Prove that 2PQ2 = 2PR2 + QR2.
Solution:
Given QS = 3SR
QR = QS + SR
= 3SR + SR = 4SR
SR = \(\frac { 1 }{ 4 } \) QR …..(1)
QS = 3SR
SR = \(\frac { QS }{ 3 } \) ……..(2)
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 11
From (1) and (2) we get
\(\frac { 1 }{ 4 } \) QR = \(\frac { QS }{ 3 } \)
∴ QS = \(\frac { 3 }{ 4 } \) QR ………(3)
In the right ∆ PQS,
PQ2 = PS2 + QS2 ……….(4)
Similarly in ∆ PSR
PR2 = PS2 + SR2 ………..(5)
Subtract (4) and (5)
PQ2 – PR2 = PS2 + QS2 – PS2 – SR2
= QS2 – SR2
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 12
PQ2 – PR2 = \(\frac { 1 }{ 2 } \) QR2
2PQ2 – 2PR2 = QR2
2PQ2 = 2PR2 + QR2
Hence the proved.

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3

Question 8.
In the adjacent figure, ABC is a right angled triangle with right angle at B and points D, E trisect BC. Prove that 8AE2 = 3AC2 + 5AD2.
Solution:
Since the Points D, E trisect BC.
BD = DE = CE
Let BD = DE = CE = x
BE = 2x and BC = 3x
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 13
In the right ∆ABD,
AD2 = AB2 + BD2
AD2 = AB2 + x2 ……….(1)
In the right ∆ABE,
AE2 = AB2 + 2BE2
AE2 = AB2 + 4X2 ………..(2) (BE = 2x)
In the right ∆ABC
AC2 = AB2 + BC2
AC2 = AB2 + 9x2 …………… (3) (BC = 3x)
R.H.S = 3AC2 + 5AD2
= 3[AB2 + 9x2] + 5 [AB2 + x2] [From (1) and (3)]
= 3AB2 + 27x2 + 5AB2 + 5x2
= 8AB2 + 32x2
= 8 (AB2 + 4 x2)
= 8AE2 [From (2)]
= R.H.S.
∴ 8AE2 = 3AC2 + 5AD2

Samacheer Kalvi 9th Maths Guide Chapter 7 Mensuration Additional Questions

Students can download Maths Chapter 7 Mensuration Additional 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 7 Mensuration Additional Questions

I. Choose the Correct Answer

Question 1.
If the sides of a triangles are 5 cm, 6 cm and 7 cm then the area is ……..
(a) 18 cm²
(b) 6 √2 cm²
(c) 6 √6 cm²
(d) 6 √3 cm²
Solution:
(c) 6 √6 cm²

Question 2.
The perimeter of an equilateral triangle is 60 cm then the area is ………
(a) 60 √3 cm²
(b) 20 √3 cm²
(c) 50 √3 cm²
(d) 100 √3 cm²
Solution:
(d) 100 √3 cm²

Samacheer Kalvi 9th Maths Guide Chapter 7 Mensuration Additional Questions

Question 3.
The total surface area of the cuboid with dimension 20 cm × 30 cm × 15 cm is ………
(a) 2700 cm²
(b) 1500 cm²
(c) 2500 cm²
(d) 3000 cm²
Solution:
(a) 2700 cm²

Question 4.
The number of bricks each measuring 70 cm × 80 cm × 40 cm that will be required to build a wall whose dimensions are 7 m × 8 m × 4 m is ……..
(a) 4000
(b) 3000
(c) 2000
(d) 1000
Solution:
(d) 1000

Question 5.
The volume of a cube is 4913 m² then the length of its side is ……..
(a) 13 m
(b) 17 m
(c) 34 m
(d) 27 m
Solution:
(b) 17 m

Samacheer Kalvi 9th Maths Guide Chapter 7 Mensuration Additional Questions

II. Answer the Following Questions

Question 6.
A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non parallel sides are 14 m and 13 m. Find the area of the field.
Solution:
The non parallel sides are 13 m and 14 m. Draw BE || AD. Such that BE = 13 m
∴ ABED is a parallelogram
Samacheer Kalvi 9th Maths Guide Chapter 7 Mensuration Additional Questions 1
To find Area of a ΔBCE
a = 13 m, b = 15 m and c = 14 m
s = \(\frac{a+b+c}{2}\)
= \(\frac{13+15+14}{2}\)
= \(\frac{42}{2}\)
= 21 m
s – a = 21 – 13 = 8 m
s – b = 21 – 15 = 6 m
s – c = 21 – 14 = 7 m
Area of a ΔBCE
Samacheer Kalvi 9th Maths Guide Chapter 7 Mensuration Additional Questions 2
= 2² × 3 × 7
= 84 m²
Let the height of the triangle BF be x
Area of the ΔBEC = 84 m²
= \(\frac{1}{2}\) × b × h = 84
= \(\frac{1}{2}\) × 15 × h = 84
x = \(\frac{84×2}{15}\)
= \(\frac{56}{5}\) m
= 11.2 m
Area of parallelogram ABED = base × height sq. units
= 10 × 11.2 m²
= 112 m²
∴ Area of the field = Area of ΔBCE + Area of parallelogram ABED
= 84 m² + 112 m²
= 196 m²
(OR)
Area of the field = Area of the trapezium ABCD
= \(\frac{1}{2}\) h (a + b)
= \(\frac{1}{2}\) × 11.2 (25 + 10)
= \(\frac{1}{2}\) × 11.2 (35)
= 196 m²

Samacheer Kalvi 9th Maths Guide Chapter 7 Mensuration Additional Questions

Question 7.
Find the area of a quadrilateral ABCD in which AB = 8 cm, BC = 6 cm, CD = 8 cm, DA = 10 cm and AC = 10 cm and ⌊B = 90°.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 7 Mensuration Additional Questions 3
In ΔABC, ⌊B = 90°
∴ ABC is a right angle triangle
Area of the right ΔABC = \(\frac{1}{2}\) × AB × BC sq.units
= \(\frac{1}{2}\) × 8 × 6 cm²
= 24 cm²
In ΔACD a = 10 cm, b = 8 cm and c = 10 cm
s = \(\frac{a+b+c}{2}\)
= \(\frac{10+8+10}{2}\)
= \(\frac{28}{2}\)
= 14 cm
s – a = 14 – 10 = 4 cm
s – b = 14 – 8 = 6 cm
s – c = 14 – 10 = 4 cm
Area of ΔACD = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{14×4×6×4}\)
= \(\sqrt{2×7×4×2×3×4}\)
= 4 × 2 \(\sqrt{21}\) cm²
= 8\(\sqrt{21}\) cm²
= 8 × 4.58
= 36.64 cm²
Area of the quadrilateral ABCD
= Area of ΔABC + Area of ΔACD
= 24 cm² + 36.64 cm²
= 60.64 cm²
Area of the quadrilateral = 60.64 cm²

Samacheer Kalvi 9th Maths Guide Chapter 7 Mensuration Additional Questions

Question 8.
The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs 7.50 per m².
Solution:
Length of the room (l) = 5 m
Breadth of the room (b) = 4 m
Height of the room (h) = 3 m
Area for white washing = Lateral surface area of four walls + Area of the ceiling
= 2(l + b) × h + (l × b)
= 2(5 + 4) × 3 + (5 × 4) m²
= (2 × 9 × 3 + 20) m²
= (54 + 20) m²
= 74 m²
Cost of white washing for one m² = Rs 7.50
Cost of white washing for 74 m² = Rs 74 × 7.50
= Rs 555
The required cost = Rs 555

Samacheer Kalvi 9th Maths Guide Chapter 7 Mensuration Additional Questions

Question 9.
How many hollow blocks of size 30 cm × 15 cm × 20 cm are needed to construct a wall 60 m in length 0.3 m in breadth and 2 m in height.
Solution:
Length of a wall = 60 m = 6000 cm
Breadth of a wall = 0.3 m = 30 cm
Height of a wall = 2 m = 200 cm
Volume of the wall = l × b × h sq. unit
= 6000 × 30 × 200 cm³
For hollow block
l = 30 cm, b = 15 cm, h = 20 cm
Volume of one hollow block = l × b × h
= 30 × 15 × 20 cm²
Number of hollow blocks required
Samacheer Kalvi 9th Maths Guide Chapter 7 Mensuration Additional Questions 4
= 4000
∴ Number of bricks = 4000

Samacheer Kalvi 9th Maths Guide Chapter 7 Mensuration Additional Questions

Question 10.
Find the number of cubes of side 3 cm that can be cut from a cuboid of dimensions 10 cm × 9 cm × 6 cm.
Solution:
Side of a cube = 3 cm
Volume of a cube = a³ cm
= 3 × 3 × 3 cm³
Length of the cuboid (l) = 10 cm
Breadth of the cuboid (b) = 9 cm
Height of the cuboid (h) = 6 cm
Volume of the cuboid = l × b × h cu. unit
= 10 × 9 × 6 cm
Number of cubes
Samacheer Kalvi 9th Maths Guide Chapter 7 Mensuration Additional Questions 5
∴ Number of cubes = 20

Samacheer Kalvi 9th Maths Guide Chapter 7 Mensuration Ex 7.4

Students can download Maths Chapter 7 Mensuration Ex 7.4 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 7 Mensuration Ex 7.4

Question 1.
The semi-perimeter of a triangle having sides 15 cm, 20 cm and 25 cm is ……..
(a) 60 cm
(b) 45 cm
(c) 30 cm
(d) 15 cm
Solution:
(c) 30 cm
Hint:
l = 15 cm, b = 20 cm, h = 25 cm
Semi-perimeter = \(\frac{a+b+c}{2}\)
= \(\frac{15+20+25}{2}\)
= 30 cm

Samacheer Kalvi 9th Maths Guide Chapter 7 Mensuration Ex 7.4

Question 2.
If the sides of a triangle are 3 cm, 4 cm and 5 cm, then the area is ………
(a) 3 cm²
(b) 6 cm²
(c) 9 cm²
(d) 12 cm²
Solution:
(b) 6 cm²
Hint:
a- 3 cm, b = 4 cm, c = 5 cm
s = \(\frac{a+b+c}{2}\)
= \(\frac{3+4+5}{2}\)
= 6 cm
Area of the triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{6×3×2×1}\)
= \(\sqrt{36}\)
= 6 cm²

Samacheer Kalvi 9th Maths Guide Chapter 7 Mensuration Ex 7.4

Question 3.
The perimeter of an equilateral triangle is 30 cm. The area is ……..
(a) 10 √3 cm²
(b) 12 √3 cm²
(c) 15 √3 cm²
(d) 25 √3 cm²
Solution:
(d) 25 √3 cm²
Hint:
Perimeter of an equilateral triangle = 30 cm
3a = 30 cm
a = \(\frac{30}{3}\)
= 10 cm
Area of an equilateral triangle = \(\frac{√3}{4}\) a² sq.units
= \(\frac{√3}{4}\) × 10 × 10
= 25 √3 cm²

Samacheer Kalvi 9th Maths Guide Chapter 7 Mensuration Ex 7.4

Question 4.
The lateral surface area of a cube of side 12 cm is ……..
(a) 144 cm²
(b) 196 cm²
(c) 576 cm²
(d) 664 cm²
Solution:
(c) 576 cm²
Hint:
Side of a cube (a) = 12 cm
L.S.A. of a cube = 4a² sq.units
= 4 × 12 × 12 cm²
= 576 cm²

Samacheer Kalvi 9th Maths Guide Chapter 7 Mensuration Ex 7.4

Question 5.
If the lateral surface area of a cube is 600 cm², then the total surface area is ………
(a) 150 cm²
(b) 400 cm²
(c) 900 cm²
(d) 1350 cm²
Solution:
(c) 900 cm²
Hint:
L.S.A. of a cube = 600 cm²
4a² = 600
a² = \(\frac{600}{4}\)
= 150
Total surface area of a cube = 6a² sq.units
= 6 × 150 cm²
= 900 cm²

Samacheer Kalvi 9th Maths Guide Chapter 7 Mensuration Ex 7.4

Question 6.
The total surface area of a cuboid with dimension 10 cm × 6 cm × 5 cm is ………
(a) 280 cm²
(b) 300 cm²
(c) 360 cm²
(d) 600 cm²
Solution:
(a) 280 cm²
Hint:
T.S.A. of a cuboid = 2(lb + bh + lh) sq.units
= 2(10 × 6 + 6 × 5 + 10 × 5) cm²
= 2(60 + 30 + 50) cm²
= 2 × 140 cm²
= 280 cm²

Question 7.
If the ratio of the sides of two cubes are 2 : 3, then ratio of their surface areas will be ………
(a) 4 : 6
(b) 4 : 9
(c) 6 : 9
(d) 16 : 36
Solution:
(b) 4 : 9
Hint:
Ratio of the surface area of cubes = 4a12 : 4a22
= a12 : a22
= 4² : 9²
= 4 : 9

Samacheer Kalvi 9th Maths Guide Chapter 7 Mensuration Ex 7.4

Question 8.
The volume of a cuboid is 660 cm and the area of the base is 33 cm². Its height is ………
(a) 10 cm
(b) 12 cm
(c) 20 cm
(d) 22 cm
Solution:
(c) 20 cm
Hint:
Volume of a cuboid = 660 cm³
l × b × h = 660
33 × h = 660 (Area of the base = l × b)
h = \(\frac{660}{33}\)
= 20 cm

Question 9.
The capacity of a water tank of dimensions 10 m × 5 m × 1.5 m is ………
(a) 75 litres
(b) 750 litres
(c) 7500 litres
(d) 75000 litres
Solution:
(d) 75000 litres
Hint:
The capacity of a tank = l × b × h cu.units
= (10 × 5 × 1.5) m³
= 75 m³
= 75 × 1000 litres [1m³ = 1000 lit]
= 75000 litres

Samacheer Kalvi 9th Maths Guide Chapter 7 Mensuration Ex 7.4

Question 10.
The number of bricks each measuring 50 cm × 30 cm × 20 cm that will be required to build a wall whose dimensions are 5 m x 3 m x 2 m is ………
Solution:
(a) 1000
(b) 2000
(c) 3000
(d) 5000
Solution:
(a) 1000
Hint:
Volume of one brick = 50 × 30 × 20 cm³
Volume of the wall = l × b × h
[l = 5m = 500 cm]
[b = 3m = 300 cm]
[h = 2m = 200 cm]
= 500 × 300 × 200 cm³
No. of bricks
Samacheer Kalvi 9th Maths Guide Chapter 7 Mensuration Ex 7.4 1
= 10 × 10 × 10
= 1000 bricks

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2

Students can download Maths Chapter 4 Geometry Ex 4.2 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 4.2

Question 1.
In ∆ABC, D and E are points on the sides AB and AC respectively such that DE || BC
(i) If \(\frac { AD }{ DB } \) = \(\frac { 3 }{ 4 } \) and AC = 15 cm find AE.
(ii) If AD = 8x – 7 , DB = 5x – 3 , AE = 4x – 3 and EC = 3x – 1, find the value of x.
Solution:
(i) Let AE be x
∴ EC = 15 – x
In ∆ABC we have DE || BC
By Basic proportionality theorem, we have
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 1
\(\frac { AD }{ DB } \) = \(\frac { AE }{ EC } \)
\(\frac { 3 }{ 4 } \) = \(\frac { x }{ 15-x } \)
4x = 3 (15 – x)
4x = 45 – 3x
7x = 45 ⇒ x = \(\frac { 45 }{ 7 } \) = 6.43
The value of x = 6.43

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2

(ii) Given AD = 8x – 7; BD = 5x – 3; AE = 4x – 3; EC = 3x – 1
In ∆ABC we have DE || BC
By Basic proportionality theorem
\(\frac { AD }{ DB } \) = \(\frac { AE }{ EC } \)
\(\frac { 8x-7 }{ 5x-3 } \) = \(\frac { 4x-3 }{ 3x-1 } \)
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 2
(8x – 7) (3x – 1) = (4x – 3) (5x – 3)
24x2 – 8x – 21x + 7 = 20x2 – 12x – 15x + 9
24x2 – 20x2 – 29x + 27x + 7 – 9 = 0
4x2 – 2x – 2 = 0
2x2 – x – 1 = 0 (Divided by 2)
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 3
2x2 – 2x + x – 1 = 0
2x(x -1) + 1 (x – 1) = 0
(x – 1) (2x + 1) = 0
x – 1 = 0 or 2x + 1 = 0
x = 1 or 2x = -1 ⇒ x = – \(\frac { 1 }{ 2 } \) (Negative value will be omitted)
The value of x = 1

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2

Question 2.
ABCD is a trapezium in which AB || DC and P,Q are points on AD and BC respectively, such that PQ || DC if PD = 18 cm, BQ
Solution:
Join AC intersecting PQ at S.
Let AP be x
∴ AD = x + 18
In the ∆ABC, QS || AB
By basic proportionality theorem.
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 4
\(\frac { AS }{ SC } \) = \(\frac { BQ }{ QC } \)
\(\frac { AS }{ SC } \) = \(\frac { 35 }{ 15 } \) ………(1)
In the ∆ACD; PS || DC
By basic proportionality theorem.
\(\frac { AS }{ SC } \) = \(\frac { AP }{ PD } \)
\(\frac { AS }{ SC } \) = \(\frac { x }{ 18 } \) ………..(2)
From (1) and (2) we get
\(\frac { 35 }{ 15 } \) = \(\frac { x }{ 18 } \)
15x = 35 × 18 ⇒ x = \(\frac{35 \times 18}{15}\) = 42
AD = AP + PD
= 42 + 18 = 60
The value of AD = 60 cm

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2

Question 3.
In ∆ABC, D and E are points on the sides AB and AC respectively. For each of the following cases show that DE || BC.
(i) AB = 12 cm, AD = 8 cm, AE = 12 cm and AC = 18 cm.
(ii) AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm.
Solution:
(i) Here AB = 12 cm; BD =12 – 8 = 4 cm; AE =12 cm; EC = 18 – 12 = 6 cm
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 5
∴ \(\frac { AD }{ DB } \) = \(\frac { 8 }{ 4 } \) = 2
\(\frac { AE }{ EC } \) = \(\frac { 12 }{ 6 } \) = 2
\(\frac { AD }{ DB } \) = \(\frac { AE }{ EC } \)
By converse of basic proportionality theorem DE || BC

(ii) Here AB = 5.6 cm; AD = 1.4 cm;
BD = AB – AD
= 5.6 – 1.4 = 4.2
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 6
AC = 7.2 cm; AE = 1.8 cm
EC = AC – AE
= 7.2 – 1.8
EC = 5.4 cm
\(\frac { AD }{ DB } \) = \(\frac { 1.4 }{ 4.2 } \) = \(\frac { 1 }{ 3 } \)
\(\frac { AE }{ EC } \) = \(\frac { 1.8 }{ 5.4 } \) = \(\frac { 1 }{ 3 } \)
\(\frac { AE }{ EC } \) = \(\frac { AD }{ DB } \)
By converse of basic proportionality theorem DE || BC

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2

Question 4.
In fig. if PQ || BC and BC and PR || CD prove that
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 7
(i) \(\frac { AR }{ AD } \) = \(\frac { AQ }{ AB } \)
(ii) \(\frac { QB }{ AQ } \) = \(\frac { DR }{ AR } \)
Solution:
(i) In ∆ABC, We have PQ || BC
By basic proportionality theorem
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 8
\(\frac { AQ }{ AB } \) = \(\frac { AP }{ AC } \) ……(1)
In ∆ACD, We have PR || CD
basic proportionality theorem
\(\frac { AP }{ AC } \) = \(\frac { AR }{ AD } \) ………..(2)
From (1) and (2) we get
\(\frac { AQ }{ AB } \) = \(\frac { AR }{ AD } \) (or) \(\frac { AR }{ AD } \) = \(\frac { AQ }{ AB } \)

(ii) In ∆ABC, PQ || BC (Given)
By basic proportionality theorem
\(\frac { AP }{ PC } \) = \(\frac { AQ }{ QB } \) ………..(1)
In ∆ADC, PR || CD (Given)
By basic proportionality theorem
\(\frac { AP }{ PC } \) = \(\frac { AR }{ RD } \) ………(2)
From (1) and (2) we get
\(\frac { AQ }{ QB } \) = \(\frac { AP }{ RD } \) (or) \(\frac { QB }{ AQ } \) = \(\frac { RD }{ AR } \)

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2

Question 5.
Rhombus PQRB is inscribed in ∆ABC such that ∠B is one of its angle. P, Q and R lie on AB, AC and BC respectively. If AB = 12 cm and BC = 6 cm, find the sides PQ, RB of the rhombus.
Solution:
Let the side of the rhombus be “x”. Since PQRB is a Rhombus PQ || BC
By basic proportionality theorem
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 9
\(\frac { AP }{ AB } \) = \(\frac { PQ }{ BC } \) ⇒ \(\frac { 12-x }{ BC } \) = \(\frac { x }{ 6 } \)
12x = 6 (12 – x)
12x = 72 – 6x
12x + 6x = 72
18x = 72 ⇒ x = \(\frac { 72 }{ 18 } \) = 4
Side of a rhombus = 4 cm
PQ = RB = 4 cm

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2

Question 6.
In trapezium ABCD, AB || DC , E and F are points on non-parallel sides AD and BC respectively, such that EF || AB.
Show that = \(\frac { AE }{ ED } \) = \(\frac { BF }{ FC } \)
Solution:
Given: ABCD is a trapezium AB || DC
E and F are the points on the side AD and BC
EF || AB
To Prove: \(\frac { AE }{ ED } \) = \(\frac { BF }{ FC } \)
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 10
Construction: Join AC intersecting AC at P
Proof:
In ∆ABC, PF || AB (Given)
By basic proportionality theorem
\(\frac { AP }{ PC } \) = \(\frac { BF }{ FC } \) ………..(1)
In the ∆ACD, PE || CD (Given)
By basic Proportionality theorem
\(\frac { AP }{ PC } \) = \(\frac { AE }{ ED } \) …………..(2)
From (1) and (2) we get
\(\frac { AE }{ ED } \) = \(\frac { BF }{ FC } \)

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2

Question 7.
In figure DE || BC and CD || EE Prove that AD2 = AB × AF.
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 11
Solution:
Given: In ∆ABC, DE || BC and CD || EF
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 12
To Prove: AD2 = AB × AF
Proof: In ∆ABC, DE || BC (Given)
By basic proportionality theorem
\(\frac { AB }{ AD } \) = \(\frac { AC }{ AE } \) ……….. (1)
In ∆ADC; FE || DC (Given)
By basic Proportionality theorem
\(\frac { AD }{ AF } \) = \(\frac { AC }{ AE } \) ……..(2)
From (1) and (2) we get
\(\frac { AB }{ AD } \) = \(\frac { AD }{ AF } \)
AD2 = AB × AF
Hence it is proved

Question 8.
In ∆ABC, AD is the bisector of ∠A meeting side BC at D, if AB = 10 cm, AC = 14 cm and BC = 6 cm, find BD and DC.
Solution:
In ∆AABC AD is the internal bisector of ∠A
Given BC = 6 cm
Let BD = x ∴ DC = 6 – x cm
By Angle bisector theorem
\(\frac { BD }{ DC } \) = \(\frac { AB }{ AC } \)
\(\frac { x }{ 6-x } \) = \(\frac { 10 }{ 14 } \)
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 13
14x = 60 – 10x
24x = 60
x = \(\frac { 60 }{ 24 } \) = \(\frac { 10 }{ 4 } \) = 2.5
BD = 2.5 cm;
DC = 6 – x ⇒ 2.5 = 3.5 cm

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2

Question 9.
Check whether AD is bisector of ∠A of ∆ABC in each of the following,
(i) AB = 5 cm, AC = 10 cm, BD = 1.5 cm and CD = 3.5 cm.
(ii) AB = 4 cm, AC 6 cm, BD = 1.6 cm and CD = 2.4 cm.
Solution:
(i) In ∆ABC, AB = 5 cm, AC = 10 cm, BD = 1.5 cm, CD = 3.5 cm
\(\frac { BD }{ DC } \) = \(\frac { 1.5 }{ 3.5 } \) = \(\frac { 15 }{ 35 } \) = \(\frac { 3 }{ 7 } \)
\(\frac { AB }{ AC } \) = \(\frac { 5 }{ 10 } \) = \(\frac { 1 }{ 2 } \)
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 14
\(\frac { BD }{ DC } \) ≠ \(\frac { AB }{ AC } \)
∴ AD is not a bisector of ∠A.

(ii) In ∆ABC, AB = 4 cm, AC = 6 cm, BD = 1.6 cm, CD = 2.4 cm
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 15
\(\frac { BD }{ DC } \) = \(\frac { 1.6 }{ 2.4 } \) = \(\frac { 16 }{ 24 } \) = \(\frac { 2 }{ 3 } \)
\(\frac { AB }{ AC } \) = \(\frac { 4 }{ 6 } \) = \(\frac { 2 }{ 3 } \)
∴ \(\frac { BD }{ DC } \) = \(\frac { AB }{ AC } \)
By angle bisector theorem; AD is the internal bisector of ∠A

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2

Question 10.
In figure ∠QPR = 90°, PS is its bisector.
If ST ⊥ PR, prove that ST × (PQ + PR) = PQ × PR.
Solution:
Given: ∠QPR = 90°; PS is the bisector of ∠P. ST ⊥ ∠PR
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 16
To prove: ST × (PQ + PR) = PQ × PR
Proof: In ∆ PQR, PS is the bisector of ∠P.
∴ \(\frac { PQ }{ QR } \) = \(\frac { QS }{ SR } \)
Adding (1) on both side
1 + \(\frac { PQ }{ QR } \) = 1 + \(\frac { QS }{ SR } \)
\(\frac { PR+PQ }{ PR } \) = \(\frac { SR+QS }{ SR } \)
\(\frac { PQ+PR }{ PR } \) = \(\frac { QR }{ SR } \) ……….(1)
In ∆ RST And ∆ RQP
∠SRT = ∠QRP = ∠R (Common)
∴ ∠QRP = ∠STR = 90°
(By AA similarity) ∆ RST ~ RQP
\(\frac { SR }{ QR } \) = \(\frac { ST }{ PQ } \)
\(\frac { QR }{ SR } \) = \(\frac { PQ }{ ST } \) ……..(2)
From (1) and (2) we get
\(\frac { PQ+PR }{ PR } \) = \(\frac { PQ }{ ST } \)
ST (PQ + PR) = PQ × PR

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2

Question 11.
ABCD is a quadrilateral in which AB = AD, the bisector of ∠BAC and ∠CAD intersect the sides BC and CD at the points E and F respectively. Prove that EF || BD.
Solution:
ABCD is a quadrilateral. AB = AD.
AE and AF are the internal bisector of ∠BAC and ∠DAC.
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 17
To prove: EF || BD.
Construction: Join EF and BD
Proof: In ∆ ABC, AE is the internal bisector of ∠BAC.
By Angle bisector theorem, we have,
∴ \(\frac { AB }{ AC } \) = \(\frac { BE }{ EC } \) ………(1)
In ∆ ADC, AF is the internal bisector of ∠DAC
By Angle bisector theorem, we have,
\(\frac { AD }{ AC } \) = \(\frac { DF }{ FC } \)
∴ \(\frac { AB }{ AC } \) = \(\frac { DF }{ FC } \) (AB = AD given) ………(2)
From (1) and (2), we get,
\(\frac { BE }{ EC } \) = \(\frac { DF }{ FC } \)
Hence in ∆ BCD,
BD || EF (by converse of BPT)

Question 12.
Construct a ∆PQR which the base PQ = 4.5 cm, R = 35° and the median from R to RG is 6 cm.
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 18

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 19
Steps of construction

  1. Draw a line segment PQ = 4.5 cm
  2. At P, draw PE such that ∠QPE = 60°
  3. At P, draw PF such that ∠EPF = 90°
  4. Draw the perpendicular bisect to PQ, which intersects PF at O and PQ at G.
  5. With O as centre and OP as radius draw a circle.
  6. From G mark arcs of radius 5.8 cm on the circle. Mark them at R and S
  7. Join PR and RQ. PQR is the required triangle.

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2

Question 13.
Construct a ∆PQR in which QR = 5 cm, ∠P = 40° and the median PG from P to QR is 4.4 cm. Find the length of the altitude from P to QR.
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 20
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 21
Steps of construction

  1. Draw a line segment RQ = 5 cm.
  2. At R draw RE such that ∠QRE = 40°
  3. At R, draw RF such that ∠ERF = 90°
  4. Draw the perpendicular bisector to RQ, which intersects RF at O and RQ at G.
  5. With O as centre and OP as radius draw a circle.
  6. From G mark arcs of radius 4.4 cm on the circle. Mark them as P and S.
  7. Join PR and PQ. Then ∆PQR is the required triangle.
  8. From P draw a line PN which is perpendicular to RQ it meets at N.
  9. Measure the altitude PN.
    PN = 2.2 cm.

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2

Question 14.
Construct a ∆PQR such that QR = 6.5 cm, ∠P = 60° and the altitude from P to QR is of length 4.5 cm.
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 22
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 23
Steps of construction

  1. Draw a line segment QR = 6.5 cm.
  2. At Q draw QE such that ∠RQE = 60°.
  3. At Q, draw QF such that ∠EQF = 90°.
  4. Draw the perpendicular of QR which intersects QF at O and QR at G.
  5. With O as centre and OQ as radius draw a circle.
  6. X Y intersects QR at G. On X Y, from G mark an arc at M. Such that GM = 4.5 cm.
  7. Draw AB through M which is parallel to QR.
  8. AB Meets the circle at P and S.
  9. join QP and RP.
    PQR is the required triangle.

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2

Question 15.
Construct a ∆ABC such that AB = 5.5 cm, ∠C = 25° and the altitude from C to AB is
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 24
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 25
Steps of construction

  1. Draw a line segment AB = 5.5 cm.
  2. At A draw AE such that ∠BAE = 25°.
  3. At A draw AF such that ∠EAF = 90°.
  4. Draw the perpendicular bisector of AB which intersects AF at O and AB at G.
  5. With O as centre and OB as radius draw a circle.
  6. X Y intersects AB at G. On X Y, from G mark an arc at M. Such that GM = 4 cm.
  7. Through M draw a line parallel to AB intersect the circle at C and D.
  8. Join AC and BC.
    ABC is the required triangle.

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2

Question 16.
Draw a triangle ABC of base BC = 5.6 cm, ∠A = 40° and the bisector of ∠A meets BC at D such that CD = 4 cm.
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 26
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 27
Steps of construction

  1. Draw a line segment BC = 5.6 cm.
  2. At B draw BE such that ∠CBE = 40°.
  3. At B draw BF such that ∠EBF = 90°.
  4. Draw the perpendicular bisector to BC which intersects BF at O and BC at G.
  5. With O as centre and OB as radius draw a circle.
  6. From C mark an arc of 4 cm on CB at D.
  7. The perpendicular bisector intersects the circle at I. Joint ID.
  8. ID produced meets the circle at A. Now Join AB and AC.
    This ABC is the required triangle.

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2

Question 17.
Draw ∆PQR such that PQ = 6.8 cm, vertical angle is 50° and the bisector of the vertical angle meets the base at D where PD = 5.2 cm
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 28
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 29
Steps of construction

  1. Draw a line segment PQ = 6.8 cm.
  2. At P draw PE such that ∠QPE = 50°.
  3. At P draw PF such that ∠EPF = 90°.
  4. Draw the perpendicular bisector to PQ which intersects PF at O and PQ at G.
  5. With O as centre and OP as radius draw a circle.
  6. From P mark an arc of 5.2 cm on PQ at D.
  7. The perpendicular bisector intersects the circle at I. Join ID.
  8. ID produced meets the circle at A. Now Joint PR and QR. This PQR is the required triangle.

Samacheer Kalvi 9th Maths Guide Chapter 7 Mensuration Ex 7.3

Students can download Maths Chapter 7 Mensuration Ex 7.3 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 7 Mensuration Ex 7.3

Question 1.
Find the volume of a cuboid whose dimensions are
(i) length = 12 cm, breadth = 8 cm, height = 6 cm
(ii) length = 60 m, breadth = 25 m, height = 1.5 m
Solution:
(i) Here l = 12 cm, b = 8 cm, h = 6 cm
Volume of a cuboid = l × b × h
= (12 × 8 × 6) cm³
= 576 cm³

(ii) Here l = 60 m, b = 25 m. h = 1.5 m
Volume of a cuboid = l × b × h
= 60 × 25 × 1.5 m³
= 2250 m³

Samacheer Kalvi 9th Maths Guide Chapter 7 Mensuration Ex 7.3

Question 2.
The dimensions of a match box are 6 cm × 3.5 cm × 2.5 cm. Find the volume of a packet containing 12 such match boxes.
Solution:
Length of a match box (l) = 6 cm
Breadth of a match box (b) = 3.5 cm
Height of a match box (h) = 2.5 cm
Volume of one match box = l × b × h cu. units
= 6 × 3.5 × 2.5 cm³
= 52.5 cm³
Volume of 12 match box = 12 × 52.5 cm³
= 630 cm³

Samacheer Kalvi 9th Maths Guide Chapter 7 Mensuration Ex 7.3

Question 3.
The length, breadth and height of a chocolate box are in the ratio 5 : 4 : 3. If its volume is 7500 cm³, then find its dimensions.
Solution:
Let the length of a chocolate be 5x, the breadth of a chocolate be 4x, and the height of a chocolate be 3x.
Volume of a chocolate = 7500 cm³
l × b × h = 7500
5x × 4x × 3x = 7500
5 × 4 × 3 × x³ = 7500
x³ = \(\frac{7500}{5×4×3}\)
x³ = 125 ⇒ x³ = 5³
x = 5
∴ Length of a chocolate = 5 × 5 = 25 cm
Breath of a chocolate = 4 × 5 = 20 cm
Height of a chocolate = 3 × 5 = 15 cm

Samacheer Kalvi 9th Maths Guide Chapter 7 Mensuration Ex 7.3

Question 4.
The length, breadth and depth of a pond are 20.5 m, 16 m and 8 m respectively. Find the capacity of the pond in litres.
Solution:
Length of a pond (l) = 20.5 m
Breadth of a pond (b) = 16 m
Depth of a pond (h) = 8 m
Volume of the pond = l × b × h cu.units
= 20.5 × 16 × 8 m³
= 2624 m³ (1 cu. m = 1000 lit)
= (2624 × 1000) litres
= 2624000 lit

Samacheer Kalvi 9th Maths Guide Chapter 7 Mensuration Ex 7.3

Question 5.
The dimensions of a brick are 24 cm × 12 cm × 8 cm. How many such bricks will be required to build a wall of 20 m length, 48 cm breadth and 6 m height?
Solution:
Length of a brick (l) = 24 cm
Breadth of a brick (b) = 12 cm
Depth of a brick (h) = 8 cm
Volume of a brick = lbh cu.units
Volume of one brick = 24 × 12 × 8 cm³
Length of a wall (l) = 20 m = 2000 cm
Breadth of a wall (b) = 48 cm
Height of a wall (h) = 6 m = 600 cm
Volume of a wall = l × b × h cu. units
= 2000 × 48 × 600 cm³
Number of bricks
Samacheer Kalvi 9th Maths Guide Chapter 7 Mensuration Ex 7.3 1
= 500 × 50 ( ÷ by 4)
= 25000 bricks
∴ Number of bricks = 25000

Samacheer Kalvi 9th Maths Guide Chapter 7 Mensuration Ex 7.3

Question 6.
The volume of a container is 1440 m³. The length and breadth of the container are 15 m and 8 m respectively. Find its height.
Solution:
Let the height of the container be “h”
Length of the container (l) = 15 m
Breadth of the container (b) = 8 m
Volume of the container = 1440 m³
l × b × h = 1440
15 × 8 × h = 1440
h = \(\frac{1440}{15×8}\)
= 12 m
∴ Height of the container = 12 m

Samacheer Kalvi 9th Maths Guide Chapter 7 Mensuration Ex 7.3

Question 7.
Find the volume of a cube each of whose side is
(i) 5 cm
(ii) 3.5 m
(iii) 21 cm
Solution:
(i) Side of a cube (a) = 5 cm
Volume of a cube = a³ cu. units
= 5 × 5 × 5 cm³
= 125 cm³

(ii) Side of a cube (a) = 3.5 m a³ cu. units
Volume of a cube = 3.5 × 3.5 × 3.5 m³
= 42.875 m³

(iii) Side of a cube (a) = 21 cm
Volume of a cube = a³ cu. units
= 21 × 21 × 21 cm³
= 9261 cm³

Samacheer Kalvi 9th Maths Guide Chapter 7 Mensuration Ex 7.3

Question 8.
A cubical milk tank can hold 125000 litres of milk. Find the length of its side in metres.
Solution:
Volume of the cubical tank = 125000 liters
= \(\frac{125}{1000}\) m³ (1 cu.m = 1000 lit)
= 125 m³
a³ = 125 ⇒ a³ = 5³
a = 5
Side of a cube = 5 m

Samacheer Kalvi 9th Maths Guide Chapter 7 Mensuration Ex 7.3

Question 9.
A metallic cube with side 15 cm is melted and formed into a cuboid. If the length and height of the cuboid is 25 cm and 9 cm respectively then find the breadth of the cuboid.
Solution:
Side of a cube (a) = 15 cm
Length of a cuboid (l) = 25 cm
Height of a cuboid (h) = 9 cm
Volume of the cuboid = Volume of the cube
l × b × h = a³
25 × b × 9 = 15 × 15 × 15
b = \(\frac{15 × 15 × 15}{25 × 9}\)
= 15 cm
Breadth of the cuboid = 15 cm