Category: Class 10

Students can download Maths Chapter 4 Geometry Ex 4.1 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 4.1

Question 1.
Check whether the which triangles are similar and find the value of x.


Solution:
(i) In ∆ABC and ∆AED
\(\frac { AB }{ AD } \) = \(\frac { AC }{ AE } \)
\(\frac { 8 }{ 3 } \) = \(\frac{11}{\frac{2}{2}}\)
\(\frac { 8 }{ 3 } \) = \(\frac { 11 }{ 4 } \) ⇒ 32 ≠ 33
∴ The two triangles are not similar.

(ii) In ∆ABC and ∆PQC
∠PQC = 70°
∠ABC = ∠PQC = 70°
∠ACB = ∠PCQ (common)
∆ABC ~ ∆PQC
\(\frac { 5 }{ X } \) = \(\frac { 6 }{ 3 } \)
6x = 15
x = \(\frac { 15 }{ 6 } \) = \(\frac { 5 }{ 2 } \)
∴ x = 2.5
∆ ABC and ∆PQC are similar. The value of x = 2.5

Question 2.
A girl looks the reflection of the top of the lamp post on the mirror which is 66 m away from the foot of the lamppost. The girl whose height is 12.5 m is standing 2.5 m away from the mirror. Assuming the mirror is placed on the ground facing the sky and the girl, mirror and the lamppost are in a same line, find the height of the lamp post.
Solution:
Let the height of the tower ED be “x” m. In ∆ABC and ∆EDC.
∠ABC = ∠CED = 90° (vertical Pole)
∠ACB = ∠ECD (Laws of reflection)
∆ ABC ~ ∆DEC
\(\frac { AB }{ DE } \) = \(\frac { BC }{ EC } \)
\(\frac { 1.5 }{ x } \) = \(\frac { 0.4 }{ 87.6 } \)

x = \(\frac{1.5 \times 87.6}{0.4}\) = \(\frac{1.5 \times 876}{4}\)
= 1.5 × 219 = 328.5
The height of the Lamp Post = 328.5 m

Question 3.
A vertical stick of length 6 m casts a shadow 400 cm long on the ground and at the same time a tower casts a shadow 28 m long. Using similarity, find the height of the tower.
Solution:
In ∆ABC and ∆PQR,
∠ABC = ∠PQR = 90° (Vertical Stick)
∠ACB = ∠PRQ (Same time casts shadow)
∆BCA ~ ∆QRP

\(\frac { AB }{ PQ } \) = \(\frac { BC }{ QR } \)
\(\frac { 6 }{ x } \) = \(\frac { 4 }{ 28 } \)
4x = 6 × 28 ⇒ x = \(\frac{6 \times 28}{4}\) = 42
Length of the lamp post = 42m

Question 4.
Two triangles QPR and QSR, right angled at P and S respectively are drawn on the same base QR and on the same side of QR. If PR and SQ intersect at T, prove that PT × TR = ST × TQ.
Solution:
In ∆PQT and ∆STR we have
∠P = ∠S = 90° (Given)
∠PTQ = ∠STR (Vertically opposite angle)

By AA similarity
∆PTQ ~ ∆STR we get
\(\frac { PT }{ ST } \) = \(\frac { TQ }{ TR } \)
PT × TR = ST × TQ
Hence it is proved.

Question 5.
In the adjacent figure, ∆ABC is right angled at C and DE ⊥ AB. Prove that ∆ABC ~ ∆ADE and hence find the lengths of AE and DE.

Solution:
In ∆ABC and ∆ADE
∠ACB = ∠AED = 90°
∠A = ∠A (common)
∴ ∆ABC ~ ∆ADE (By AA similarity)
\(\frac { BC }{ DE } \) = \(\frac { AB }{ AD } \) = \(\frac { AC }{ AE } \)
\(\frac { 12 }{ DE } \) = \(\frac { 13 }{ 3 } \) = \(\frac { 5 }{ AE } \)
In ∆ABC, AB² = BC² + AC²
= 122 + 52 = 144 + 25 = 169
AB = \(\sqrt { 169 }\) = 13
Consider, \(\frac { 13 }{ 3 } \) = \(\frac { 5 }{ AE } \)
∴ AE = \(\frac{5 \times 3}{13}\) = \(\frac { 15 }{ 13 } \)
AE = \(\frac { 15 }{ 13 } \) and DE = \(\frac { 36 }{ 13 } \)
Consider, \(\frac { 12 }{ DE } \) = \(\frac { 13 }{ 3 } \)
DE = \(\frac{12 \times 3}{13}\) = \(\frac { 36 }{ 13 } \)

Question 6.
In the adjacent figure, ∆ACB ~ ∆APQ. If BC = 8 cm, PQ = 4 cm, BA = 6.5 cm and AP = 2.8 cm, find CA and AQ.

Solution:
Given ∆ACB ~ ∆APQ
\(\frac { AC }{ AP } \) = \(\frac { BC }{ PQ } \) = \(\frac { AB }{ AQ } \)
\(\frac { AC }{ 2.8 } \) = \(\frac { 8 }{ 4 } \) = \(\frac { 6.5 }{ AQ } \)
Consider \(\frac { AC }{ 2.8 } \) = \(\frac { 8 }{ 4 } \)
4 AC = 8 × 2.8
AC = \(\frac{8 \times 2.8}{4}\) = 5.6 cm
Consider \(\frac { 8 }{ 4 } \) = \(\frac { 6.5 }{ AQ } \)
8 AQ = 4 × 6.5
AQ = \(\frac{4 \times 6.5}{8}\) = 3.25 cm
Length of AC = 5.6 cm; Length of AQ = 3.25 cm

Question 7.
If figure OPRQ is a square and ∠MLN = 90°. Prove that

(i) ∆LOP ~ ∆QMO
(ii) ∆LOP ~ ∆RPN
(iii) ∆QMO ~ ∆RPN
(iv) QR2 = MQ × RN.
Solution:
(i) In ∆LOP and ∆QMO
∠OLP = ∠OQM = 90°
∠LOP = ∠OMQ (Since OQRP is a square OP || MN)
∴ ∆LOP~ ∆QMO (By AA similarity)

(ii) In ∆LOP and ∆RPN
∠OLP = ∠PRN = 90°
∠LPO = ∠PNR (OP || MN) .
∴ ∆LOP ~ ∆RPN (By AA similarity)

(iii) In ∆QMO and ∆RPN
∠MQO = ∠NRP = 90°
∠RPN = ∠QOM (OP || MN)
∴ ∆QMO ~ ∆RPN (By AA similarity)

(iv) We have ∆QMO ~ ∆RPN
\(\frac { MQ }{ PR } \) = \(\frac { QO }{ RN } \)
\(\frac { MQ }{ QR } \) = \(\frac { QR }{ RN } \)
QR² = MQ × RN
Hence it is proved.

Question 8.
If ∆ABC ~ ∆DEF such that area of ∆ABC is 9cm² and the area of ∆DEF is 16 cm² and BC = 2.1 cm. Find the length of EF.
Solution:
Given ∆ABC ~ ∆DEF

\(\frac { 9 }{ 16 } \) = \(\frac{(2.1)^{2}}{\mathrm{E} \mathrm{F}^{2}}\)
(\(\frac { 3 }{ 4 } \))² = (\(\frac { 2.1 }{ EF } \))²
\(\frac { 3 }{ 4 } \) = \(\frac { 2.1 }{ EF } \)
EF = \(\frac{4 \times 2.1}{3}\) = 2.8 cm
Legth of EF = 2.8 cm

Question 9.
Two vertical poles of heights 6 m and 3 m are erected above a horizontal ground AC. Find the value of y.

Solution:
In the ∆PAC and ∆BQC
∠PAC = ∠QBC = 90°
∠C is common
∆PAC ~ QBC
\(\frac { AP }{ BQ } \) = \(\frac { AC }{ BC } \)
\(\frac { 6 }{ y } \) = \(\frac { AC }{ BC } \)
∴ \(\frac { BC }{ AC } \) = \(\frac { y }{ 6 } \) …..(1)
In the ∆ACR and ∆QBC
∠ACR = ∠QBC = 90°
∠A is common
∆ACR ~ ABQ
\(\frac { RC }{ QB } \) = \(\frac { AC }{ AB } \)
\(\frac { 3 }{ y } \) = \(\frac { AC }{ AB } \)
\(\frac { AB }{ AC } \) = \(\frac { y }{ 3 } \) ……..(2)
By adding (1) and (2)
\(\frac { BC }{ AC } \) + \(\frac { AB }{ AC } \) = \(\frac { y }{ 6 } \) + \(\frac { y }{ 3 } \)
1 = \(\frac{3 y+6 y}{18}\)
9y = 18 ⇒ y = \(\frac { 18 }{ 9 } \) = 2
The Value of y = 2m

Question 10.
Construct a triangle similar to a given triangle PQR with its sides equal to \(\frac { 2 }{ 3 } \) of the corresponding sides of the triangle PQR (scale factor \(\frac { 2 }{ 3 } \) ).
Solution:
Given ∆PQR, we are required to construct another triangle whose sides are \(\frac { 2 }{ 3 } \) of the corresponding sides of the ∆PQR

Steps of construction:
(i) Construct a ∆PQR with any measurement.
(ii) Draw a ray QX making an acute angle with QR on the side opposite to the vertex P.
(iii) Locate 3 points Q₁, Q₂ and Q₃ on QX.
So that QQ₁ = Q₁Q₂ = Q₂Q₃
(iv) Join Q₃ R and draw a line through Q₂ parallel to Q₃ R to intersect QR at R’.
(v) Draw a line through R’ parallel to the line RP to intersect QP at P’. Then ∆ P’QR’ is the required triangle.

Question 11.
Construct a triangle similar to a given triangle LMN with its sides equal to \(\frac { 4 }{ 5 } \) of the corresponding sides of the triangle LMN (scale factor \(\frac { 4 }{ 5 } \) ).
Solution:
Given a triangle LMN, we are required to construct another ∆ whose sides are \(\frac { 4 }{ 5 } \) of the corresponding sides of the ∆LMN.

Steps of Construction:

1. Construct a ∆LMN with any measurement.
2. Draw a ray MX making an acute angle with MN on the side opposite to the vertex L.
3. Locate 5 Points Q₁, Q₂, Q₃, Q₄, Q₅ on MX.
   So that MQ₁ = Q₁Q₂ = Q₂Q₃ = Q₃Q₄ = Q₄Q₅
4. Join Q₅ N and draw a line through Q₄. Parallel to Q₅N to intersect MN at N’.
5. Draw a line through N’ parallel to the line LN to intersect ML at L’.
   ∴ ∆L’ MN’ is the required triangle.

Question 12.
Construct a triangle similar to a given triangle ABC with its sides equal to \(\frac { 6 }{ 5 } \) of the corresponding sides of the triangle ABC (scale factor \(\frac { 6 }{ 4 } \)).
Solution:
Given triangle ∆ABC, we are required to construct another triangle whose sides are \(\frac { 6 }{ 5 } \) of the corresponding sides of the ∆ABC.
Steps of construction
(i) Construct an ∆ABC with any measurement.
(ii) Draw a ray BX making an acute angle with BC.
(iii) Locate 6 points Q₁, Q₂, Q₃, Q₄, Q₅, Q₆ on BX such that
BQ₁ = Q₁Q₂ = Q₂Q₃ = Q₃Q₄ = Q₅Q₆

(iv) Join Q₅ to C and draw a line through Q₆ parallel to Q₅ C intersecting the extended line BC at C’.
(v) Draw a line through C’ parallel to AC intersecting the extended line segment AB at A’.
∆A’BC’ is the required triangle.

Question 13.
Construct a triangle similar to a given triangle PQR with its sides equal to \(\frac { 7 }{ 3 } \) of the corresponding sides of the triangle PQR (scale factor \(\frac { 7 }{ 3 } \)).
Solution:
Given triangle ABC, we are required to construct another triangle whose sides are \(\frac { 7 }{ 3 } \) of the corresponding sides of the ∆ABC.
Steps of construction
(i) Construct a ∆PQR with any measurement.
(ii) Draw a ray QX making an acute angle with QR on the side opposite to the vertex P.
(iii) Locate 7 points Q₁, Q₂, Q₃, Q₄, Q₅, Q₆, Q₇ on QX.
So that
QQ₁ = Q₁Q₂ = Q₂Q₃ = Q₃Q₄ = Q₅Q₆ = Q₆Q₇

(iv) Join Q₃ to R and draw a line through Q₇ parallel to Q₃R intersecting the extended line segment QR at R’.
(v) Draw a line through parallel to RP.
Intersecting the extended line segment QP at P’.
∴ ∆P’QR’ is the required triangle.

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,

OB² = OA² + AB²
= 18² + 24²
= 324 + 576 = 900
OB = \(\sqrt { 900 }\) = 30
The distance of his current position is 30 m

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).

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

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

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

Through A one must walk (34m + 41m) 75 m to reach C.
The difference in Distance = 75 – 53.26
= 21.74 m

Question 4.
In the rectangle WXYZ, XY + YZ = 17 cm, and XZ + YW = 26 cm.
Calculate the length and breadth of the rectangle?

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

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

If b = 12 ⇒ a = 5
If b = 6 ⇒ a = 12
Lenght = 12 m and breadth = 5 m

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

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

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

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

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

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)

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

PQ² – PR² = \(\frac { 1 }{ 2 } \) QR²
2PQ² – 2PR² = QR²
2PQ² = 2PR² + QR²
Hence the proved.

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

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

Students can download Maths Chapter 3 Algebra Additional 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 Additional Questions

I. Choose the correct answer.

Question 1.
The HCF of x² – y²; x³ – y³, …………. xn – yn where n ∈ N is
(1) x – y
(2) x + y
(3) xn – yn
(4) do not intersect
Answer:
(1) x – y

Question 2.
Which of the following is correct.
(i) Every polynomial has finite number of multiples
(ii) LCM of two polynomials of degree “2” may be a constant
(iii) HCF of 2 polynomials may be a constant
(iv) Degree of HCF of two polynomials is always less than degree of L.C.M.
(1) (i) and (iii)
(2) (iii) and (iv)
(3) (iii) only
(4) (iv) only
Answer:
(3) (iii) only

Question 3.
The HCF of x² – 2xy + y² and x⁴ – y⁴ is …………….
(1) 1
(2) x + y
(3) x – y
(4) x² – y²
Answer:
(3) x – y

Question 4.
The L.C.M. of a^k a^k+3, a^k+5 where K ∈ N is …………
(1) a^k+5
(2) a^k
(3) a^k+6
(4) a^k+9
Answer:
(1) a^k+5

Question 5.
The LCM of (x + 1)² (x – 3) and
(x² – 9) (x + 1) is
(1) (x + 1)³ (x² – 9)
(2) (x + 1)² x² – 9)
(3) (x + 1)² (x – 3)
(4) (x – 9) (x + 1)
Answer:
(2) (x + 1)²(x² – 9)

Question 6.
If \(\frac{a^{3}}{a-b}\) is added with \(\frac{b^{3}}{b-a}\) then the new expressions is …………
(1) a² – ab + b²
(2) a² + ab + b²
(3) a³ + b³
(4) a³ – b³
Answer:
(2) a² + ab + b²

Question 7.
The solution set of x + \(\frac { 1 }{ x } \) = \(\frac { 5 }{ 2 } \) is ………….
(1) 2,\(\frac { 1 }{ 2 } \)
(2) 2,-\(\frac { 1 }{ 2 } \)
(3) -2, – \(\frac { 1 }{ 2 } \)
(4) -2, \(\frac { 7 }{ 2 } \)
Answer:
(1) 2,\(\frac { 1 }{ 2 } \)

Question 8.
On dividing \(\frac{x^{2}-25}{x+3}\) by \(\frac{x+5}{x^{2}-9}\) is equal to ……………….
(1) (x – 5)(x + 3)
(2) (x + 5) (x – 3)
(3) (x – 5)(x – 3)
(4) (x + 5)(x + 3)
Answer:
(3) (x – 5)(x – 3)

Question 9.
The square root of (x + 11)² – 44x is ………….
(1)|(x – 11)²
(2) |x + 11|
(3) |11 – x²|
(4) |x – 11|
Answer:
(4) |x – 11|

Question 10.
If α, β are the zeros of the polynomial p(x) = 4x² + 3x + 7 then \(\frac{1}{\alpha}\) + \(\frac{1}{\beta}\) is equal to …………
(1) \(\frac { 7 }{ 3 } \)
(2) – \(\frac { 7 }{ 3 } \)
(3) \(\frac { 3 }{ 7 } \)
(4) – \(\frac { 3 }{ 7 } \)
Answer:
(4) – \(\frac { 3 }{ 7 } \)

Question 11.
The value of is  ……….
(1) -5
(2) 5
(3) 4
(4) -3
Answer:
(2) 5

Question 12.
If α and β are the roots of the equation ax² + bx + c = 0 then (α + β)² is ……………..
(1) \(\frac{-b^{2}}{a^{2}}\)
(2) \(\frac{-c^{2}}{a^{2}}\)
(3) \(\frac{-b^{2}}{a^{2}}\)
(4) \(\frac { bc }{ a } \)
Answer:
(3) \(\frac{-b^{2}}{a^{2}}\)

Question 13.
The roots of the equation x² – 8x + 12 = 0 are
(1) real and equal
(2) real and rational
(3) real and irrational
(4) unreal
Answer:
(2) real and rational

Question 14.
If one root of the equation is the reciprocal of the other root in ax² + bx + c = 0 then …………
(1) a = c
(2) a = b
(3) b = c
(4) c = 0
Answer:
(1) a = c

Question 15.
If α and β are the roots of the equation x² + 2x + 8 = 0 then the value of \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\) is ………………
(1) \(\frac { 1 }{ 2 } \)
(2) 6
(3) \(\frac { 3 }{ 2 } \)
(4) –\(\frac { 3 }{ 2 } \)
Answer:
(4) –\(\frac { 3 }{ 2 } \)

Question 16.

are………
(1) 4, 6, 6
(2) 6, 6, 4
(3) 6, 4, 6
(4) 4, 4, 6
Answer:
(3) 6, 4, 6

Question 17.
If [-1 -2 4] then the value of “a” is ………….
(1) 2
(2) -4
(3) 4
(4) -2
Answer:
(4) -2

Question 18.
The matrix A given by (a_ij)_2×2 if a_ij = i – j is …………

Answer:

Question 19.
If A is of order 4 × 3 and B is of order 3 × 4 then the order of BA is ………………….
(1) 3 × 4
(2) 4 × 4
(3) 3 × 3
(4) 4 × 1
Answer:
(3) 3 × 3

Question 20.
If then “x” is ……………..
(1) 1
(2) 2
(3) 3
(4) 4
Answer:
(4) 4

II. Answer the following.

Question 1.
Solve x + y = 7; y + z = 4; z + x = 1
Answer:
x + y = 7 ……(1)
y + z = 4 ………(2)
z + x = 1 …………(3)
Adding (1); (2) and (3)
2x + 2y + 2z = 12
x + y + z = 6 ….(4)
From (1) ⇒ x + y = 7
7 + z = 6
z = 6 – 7 = -1
From (2) ⇒ x + 4 = 6
x = 6 – 4 = 2
From (3) ⇒ y + 1 = 6
y = 6 – 1 = 5
The value of x = 2, y = 5 and z = -1

Question 2.
Find the HCF of 25x⁴y⁷; 35x³y⁸; 45x³y³
Answer:
25x⁴y⁷ = 5 × 5 × x⁴ × y⁷
35x³y⁸ = 5 × 7 × x³ × y⁸
45 x³y³ = 3 × 3 × 5 × x³ × y³
H.C.F. = 5x³y³

Question 3.
Find the values of k for which the following equation has equal roots.
(k – 12)x² + 2(k – 12)x + 2 = 0
Solution:
\(\frac{(k-12)}{a} x^{2}+\frac{2(k-12)}{b} x+\frac{2}{c}=0\)
Δ = b² – 4ac = (2(k – 12))² – 4(6 – 12)(2)
= 4(k – 12)[(k – 12) – 2]
= 4(k – 12)(k – 14)
The given equation will have equal roots, if A = 0
⇒ 4(k – 12)(k – 14) = 0
k – 12 = 0 or k – 14 = 0
k = 12, 14

Question 4.
Find the LCM of x³ + y³; x³ – y³; x⁴ + x²y² + y⁴
Answer:
x³ + y³ = (x + y) (x² – xy + y²)
x³ – y³ = (x – y)(x² + xy + y²)
x⁴ + x²y² + y⁴ = (x² + y²)² – (xy)²
= (x² + y² + xy)
L.C.M. = (x + y)(x – y) (x² + xy + y²)
(x² – xy + y²)
= [(x + y) (x² – xy +y²)]
[(x – y) (x² + xy + y²)]
= (x³ + y³) (x³ – y³)
L.C.M. = x⁶ – y⁶

Question 5.
The sum of two numbers is 15. If the sum of their reciprocals is \(\frac{3}{10}\), find the numbers.
Solution:
Let the numbers be α, β
Sum of the roots = α + β = 15 ………….. (1)
sum of their reciprocals = \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{3}{10}\) ……….. (2)
\(\frac{\beta+\alpha}{\alpha \beta}=\frac{3}{10}\)
10(α + β) = 3αβ …………. (3)
3αβ = 10 × 15 = 150
Products of the roots = αβ = 50 ………….. (4)
∴ From (1) & (4), we have
x² – 15x + 50 = 0
(x – 10)(x – 5) = 0 ⇒ x = 10, 5
∴ he numbers are 10, 5.

Question 6.
For What value of k, the G.C.D. of [x² + x – (2k + 2)] and 2x² + kx – 12 is (x + 4)?
Answer:
p(x) = x² + x – (2k + 2)
g(x) = 2x² + kx – 12
G.C.D. = x + 4
when x + 4 is the G.C.D.
p(-4) = 0 or g(-4) = 0
[Hint: Take any one of the polynomial]
g(x) = 2x² + kx – 12 = 0
2(-4)² + k (-4) – 12 = 0
2(16) – 4x – 12 = 0
32 – 4k – 12 = 0
20 = 4k
k = \(\frac { 20 }{ 4 } \) = 5
The value of k = 5

Question 7.
Simplify \(\frac{x^{2}+x-6}{x^{2}+4 x+3}\)
Answer:
x² + x – 6 = (x + 3) (x – 2)
x² + 4x + 3 = (x + 3) (x + 1)

Question 8.
Multiply

Answer:

Question 9.
if P = \(\frac{x^{3}-36}{x^{2}-49}\) and Q = \(\frac { x+6 }{ x+7 } \) find the value of \(\frac { P }{ Q } \).
Answer:

Question 10.
Simplify
\(\frac { x }{ x+y } \) – \(\frac { y }{ x-y } \)
Answer:

Question 11.
Find the square root of (x + 11)² – 44x
Answer:

Question 12.
Find the square root of x⁴ + \(\frac{1}{x^{4}}\) + 2
Answer:

Question 13.
Solve the equation 2x – 1 – \(\frac { 2 }{ x-2 } \) = 3
Answer:
2x – 1 – \(\frac { 2 }{ x-2 } \) = 3

(x – 3) (x – 1) = 0
x – 3 = 0 or x – 1 = 0
x = 3 or x = 1
The solution set is (1,3)

Question 14.
Find the roots of \(\sqrt { 2 }\) x² + 7x + 5\(\sqrt { 2 }\) = 0
Answer:
\(\sqrt { 2 }\) x² + 7x + 5 \(\sqrt { 2 }\) = 0
\(\sqrt { 2 }\) x² + 2x + 5x + 5 \(\sqrt { 2 }\) = 0

\(\sqrt { 2 }\) x (x + \(\sqrt { 2 }\)) + 5 (x + \(\sqrt { 2 }\)) = 0
(x + \(\sqrt { 2 }\)) (\(\sqrt { 2 }\) x + 5) = 0
(x + \(\sqrt { 2 }\) ) = 0 or \(\sqrt { 2 }\) x + 5 = 0
x = – \(\sqrt { 2 }\) or \(\sqrt { 2 }\) x + 5 = 0
x = – \(\sqrt { 2 }\) or \(\sqrt { 2 }\) x = -5
x = \(\frac{-5}{\sqrt{2}}\)
The roots are and – \(\sqrt { 2 }\) and \(\frac{-5}{\sqrt{2}}\)

Question 15.
Solve \(\sqrt { x+5 }\) = 2x + 3 using formula method.
Answer:
\(\sqrt { x+5 }\) = 2x + 3
(\(\sqrt { x+5 }\))² = (2x + 3)²
x + 5 = 4x² + 9 + 12x
0 = 4x² + 12x – x + 9 – 5
0 = 4x² + 11x + 4
Here a = 4, b = 11, c = 4

Question 16.
The sum of a number and its reciprocal is \(\frac { 37 }{ 6 } \). Find the number.
Answer:
Let the require number be “x”
Its reciprocal is \(\frac { 1 }{ x } \)
By the given data

The required number is \(\frac { 1 }{ 6 } \) or 6

Question 17.
Determine the nature of the roots of the equation 2x² + x – 1 = 0
Answer:
2x² + x – 1 = 0
Here a = 2,b = 1,c = -1
∆ = b² – 4 ac
= 1² – 4(2) (-1)
= 1 + 8
= 9
Since b² – 4ac > 0 the roots are real and unequal

Question 18.
Find the value of k for which the given equation 9x² + 3kx + 4 = 0 has real and equal roots.
Answer:
9x² + 3 kx + 4 = 0
a = 9, b = 5k, c = 4
since the equation has real and equal roots
b² – 4ac = 0
(3k)² – 4(9) (4) = 0
9k² – 144 = 0
9k² = 144
k² = \(\frac { 144 }{ 9 } \) = 16
k = \(\sqrt { 16 }\)
k = ± 4

Question 19.
If one root of the equation
3x² – 10x + 3 = 0 is \(\frac { 1 }{ 3 } \) find the other root
Answer:
α and β are the roots of the equation 3x² – 10x + 3 = 0
Sum of the roots (α + β) = \(\frac { 10 }{ 3 } \)
Product of the roots (αβ) = \(\frac { 3 }{ 3 } \) = 1
one of the roots is \(\frac { 1 }{ 3 } \) (say α = \(\frac { 1 }{ 3 } \))
αβ = 1
\(\frac { 1 }{ 3 } \) × β = 1
β = 3
The other roots is 3

Question 20.
Form the quadratic equation whose roots are 3 + \(\sqrt { 7 }\); 3 – \(\sqrt { 7 }\)
Answer:
Sum of the roots = 3 + \(\sqrt { 7 }\) + 3 – \(\sqrt { 7 }\)
= 6
Product of the roots = (3 + \(\sqrt { 7 }\)) (3 – \(\sqrt { 7 }\) )
= 32 – (\(\sqrt { 7 }\))2
= 9 – 7
= 2
The required equation is
x² – (sum of the roots) x + product of the roots = 0
x² – (6)x + 2 = 0
x – 6x + 2 = 0

Question 21.
If α and β are the roots of the equation 3x² – 5x + 2 = 0, then find the value of α – β.
Answer:
α and β are the roots of the equation
3x² – 5x + 2 = 0
α + β = \(\frac { 5 }{ 3 } \), αβ = \(\frac { 2 }{ 3 } \)

Question 22.
Determine the matrix A = (a_ij)_3×2 if a_ij = 3i – 2j
Answer:

a_ij = 3_i – 2_j
a₁₁ = 3(1) – 2(1) = 3 – 2 = 1
a₁₂ = 3(1) – 2(1) = 3 – 4 = 1
a₂₁ = 3(2) – 2(1) = 6 – 2 = 4
a₂₂ = 3(2) – 2(2) = 6 – 4 = 2
a₃₁ = 3(3) – 2(1) = 9 – 2 = 7
a₃₂ = 3(3) – 2(2) = 9 – 4 = 5

Question 23.

Answer:

Question 24.
Find if

Answer:

Question 25.

find BA
Answer:

Question 26.
Find the unknowns a, b, c, d, x, y in the given matrix equation.

Answer:

Equating the corresponding elements of the two matrices we get
d + 1 = 2
d = 2 – 1 = 1
10 + a = 2a + 1
10 – 1 = 2a – a
9 = a
36 – 2 = b – 5
3b – b = -5 + 2
2b = -3 ⇒ b = \(\frac { -3 }{ 2 } \)
a – 4 = 4c ⇒ a – 4c = 4
9 – 4c = 4 ⇒ 4c = 4 – 9
-4c = -5 ⇒ c = \(\frac { 5 }{ 4 } \)
The value of a = 9, b = \(\frac { -3 }{ 2 } \), c = \(\frac { 5 }{ 4 } \) and d = 1

Question 27.
Prove that

multiplication is inverse to each other.
Answer:

AB = BA = I
Multiplication of matrices are iverse to each other.

III. Answer the following questions.

Question 1.
Solve x – \(\frac { y }{ 5 } \) = 6; y – \(\frac { z }{ 7 } \) = 8; z – \(\frac { x }{ 2 } \) = 10
Answer:
x – \(\frac { y }{ 5 } \) = 6
multiply by 5
5x – y = 30 …….(1)
y – \(\frac { z }{ 7 } \) = 8

Substitute the value of x = 8 in (1)
5(8) – y = 30
– y = 30 – 40 = -10
∴ y = 10
Substitute the value of x = 8 in (3)
2z – 8 = 20
2z = 20 + 8
z = \(\frac { 28 }{ 2 } \) = 14
The value of x = 8, y = 10 and z = 14

Question 2.
Solve for x,y and z using the given 3 equations
\(\frac { 2 }{ y } \) – \(\frac { 4 }{ z } \) + \(\frac { 3 }{ x } \) = 3; \(\frac { 5 }{ x } \) – \(\frac { 4 }{ y } \) – \(\frac { 8 }{ z } \) = 8 ; \(\frac { 6 }{ y } \) + \(\frac { 6 }{ z } \) +\(\frac { 1 }{ x } \) = 2
Answer:
Let \(\frac { 1 }{ x } \) = a, \(\frac { 1 }{ y } \) = b, \(\frac { 1 }{ z } \) = c
3a + 2b – 4c = 3 ………(1)
5a – 4b – 8c = 8 ………(2)
a + 6b + 6c = 2 ………(3)
(1) × 2 ⇒ 6a + 4b – 8c = 6 …..(1)

Question 3.
100 pencils are to be kept inside three types of boxes A, B and C. If 5 boxes of type A, 3 boxes of type B, 2 boxes of type C are used 6 pencils are left out. If 3 boxes of type A, 5 boxes of type B, 2 boxes of type C are used 2 pencils are left out. If 2 boxes of type A, 4 boxes of type B and 4 boxes of type C are used, there is a space for 4 pencils. Find the number of pencils that each box can hold.
Answer:
Let the number of pencil in the box A be “x”
Let the number of pencil in the box B be “y”
Let the number of pencil in the box C be “z”
By the given first condition
5x + 3y + 2z = 94 ….(1)
By the given second condition
3x + 5y + 2z = 98 ….(2)
By the given third condition
2x + 4y + 4z = 104 ….(3)
subtract (1) and (3)

substitute x = 8 and y = 10 in (1)
5(8) + 3(10) + 2z = 94
40 + 30 + 2z = 94
2z = 94 – 70
2z = 24
z = \(\frac { 24 }{ 2 } \) = 12
Number of pencil in box A = 8
Number of pencil in box B = 10
Number of pencil in box C = 12

Question 4.
What 2 masons earn in a day is earned by 3 male workers in a day. The daily wages of 15 female workers is ₹30 more than the total daily wages of 5 masons and 3 male workers. If one mason, one male worker and 2 female workers are engaged for a day, the builder has to pay ?160 as wages. Find the daily wages of a mason, a male worker and a female worker.
Answer:
Let the daily wage of a mason be ₹ x
Let the daily wage of a male worker be ₹ y
Let the daily wage of a female worker be ₹ z
By the given first condition
2x = 3y
2x – 3y = 0 …..(1)
By the given second condition
15z = 5x + 3y + 30
-5x – 3y + 15z = 30
5x + 3y – 15z = -30 ………(2)
By the given third condition

substitute the value of x = 60 in ….(1)
2(60) – 3y = 0
120 = 3y
y = \(\frac { 120 }{ 3 } \) = 40
substitute the value of x = 60 and y = 40 in (3)
60 + 40 + 2z = 160
2z = 160 – 100
2z = 60
z = \(\frac { 60 }{ 2 } \) = 30
Daily wages of a manson = ₹60
Daily wages of a male worker = ₹40
Daily wages of a female worker = ₹30

Question 5.
Find the G.C.D. of x³ – 10x² + 31x – 30 and 2x³ – 8x² + 2x + 12
Answer:
p(x) = x³ – 10x² + 31x – 30
g(x) = 2x³ – 8x² + 2x + 12
= 2(x³ – 4x² + x + 6)

G.C.D. = x² – 5x + 6

Question 6.
The G.C.D of x⁴ + 3x³ + 5x² + 26x + 56 and x⁴ + 2x³ – 4x² – x + 28 is x² + 5x + 7. Find their L.C.M.
Answer:
p(x) = x⁴ + 3x³ + 5x² + 26x + 56
g(x) = x⁴ + 2x³ – 4x² – x + 28
G.C.D. = x² + 5x + 7

Question 7.
Find the values of “a” and “b” given that p(x) = (x² + 3x + 2) (x² – 4x + a); g(x) = (x² – 6x + 9) × (x² + 4x + b) and their G.C.D is (x + 2) (x – 3)
Answer:
p(x) = (x² + 3x + 2) (x² – 4x + a)
= (x + 1) (x + 2) (x² – 4x + a)
G.C.D is given as (x + 2) (x – 3)
x – 3 is a factor of x² – 4x + a

p(3) = 0
9 – 4(3) + a = 0
9 – 12 + a = 0
– 3 + a =0
a = 3

g(x) = (x² – 6x + 9) (x² + 4x + 6)
= (x – 3) (x – 3) (x² + 4x + b)
But G.C.D. is (x + 2) (x – 3)
∴ x + 2 is a factor of x² + 4x + 6
g(-2) = 0
4 + 4(-2) + b = 0
4 – 8 + 6 = 0
-4 + b = 0
b = 4
The value of a = 3 and b = 4

Question 8.
Find the other polynomial g(x), given that LCM, HCF and p(x) as (x – 1) (x – 2) (x² – 3x + 3); x – 1 and x³ – 4x² + 6x – 3 respectively.
Answer:
LC.M. = (x – 1) (x – 2) (x² – 3x + 3)
HCF = (x – 1)
p(x) = x³ – 4x² + 6x – 3

p(x) = (x – 1) (x² – 3x + 3)
p(x) × g(x) = LCM × HCF
(x – 1) (x² – 3x + 3) × g(x)
= (x – 1) (x – 2) (x² – 3x + 3) (x – 1)

The other polynominal g(x) x² – 3x + 2

Question 9.
Divide

Answer:
2x² + x – 3 = 2x² + 3x – 2x – 3
= x(2x + 3) – 1 (2x + 3)
= (2x + 3) (x – 1)

2x² + 5x + 3 = 2x² + 3x + 2x + 3
= x(2x + 3) + 1 (2x + 3)
= (2x + 3) (x + 1)
x² -1 = (x + 1) (x – 1)

Question 10.
Simplify

Answer:
(x² – x – 6) = (x – 3) (x + 2)
2x² + 5x – 3 = 2×2 + 6x – x – 3

= 2x (x + 3) -1 (x + 3)
= (x + 3) (2x – 1)
x² + 5x + 6 = (x + 2) (x + 3)

Question 11.
Find the square root of (6x² + 5x – 6) (6x² – x – 2) (4x² + 8x + 3)
Answer:
6x² + 5x – 6 = 6x² + 9x – 4x – 6
= 3x(2x + 3) -2 (2x + 3)
= (2x + 3) (3x – 2)

6x² – x – 2 = 6x² – 4x + 3x – 2
= 2x (3x – 2) + 1 (3x – 2)
= (3x – 2) (2x + 1)

4x² + 8x + 3 = 4x² + 6x + 2x + 3
= 2x(2x + 3) + 1 (2x + 3)
= (2x + 3) (2x + 1)

Question 12.
Find the square root of the polynomial
\(\frac{4 x^{2}}{y^{2}}\) + \(\frac { 8x }{ y } \) + 16 + 12 \(\frac { y }{ x } \) + \(\frac{9 y^{2}}{x^{2}}\)
Answer:

Question 13.
If m – nx + 28x² + 12x³ + 9x⁴ is a perfect square, then find the values of m and n.
Answer:
Arrange the polynomial in descending power of x.
9x⁴ + 12x³ + 28x² – nx + m
Now,

Since the given polynomial is a perfect square,
-nx – 16x = 0
-x (n + 16) = 0
n + 16 = 0 ⇒ n = -16
m – 16 = 0 ⇒ m = 16
The value m = 16 and n = -16

Question 14.
If b + \(\frac { a }{ x } \) + \(\frac{13}{x^{2}}\) – \(\frac{6}{x^{3}}\) + \(\frac{1}{x^{4}}\) is a perfect square, find the values of “a” and “b”
Answer:
Arrange the values of “a” and “b”

Since it is a perfect square
\(\frac { a }{ x } \) + \(\frac { 12 }{ x } \) = 0
\(\frac { 1 }{ x } \) (a + 12) = 0
a + 12 = 0 ⇒ a = -12
b – 4 = 0 ⇒ b = 4
The value of a = -12 and b = 4

Question 15.
Solve
\(\frac { 1 }{ x + 1 } \) + \(\frac { 4 }{ 3x+6 } \) = \(\frac { 2 }{ 3 } \)
Answer:

6x² – 12x + 9x – 18 = 0
6x(x – 2) + 9(x – 2) = 0

(x – 2) (6x + 9) = 0
x – 2 = 0 or 6x + 9 = 0
x = 2 or 6x + 9 = 0
x = 2 or 6x = -9
x = – \(\frac { 9 }{ 6 } \) = \(\frac { -3 }{ 2 } \)
The solution is \(\frac { -3 }{ 2 } \) or 2

Question 16.
A two-digit number is such that the product of the digits is 14. When 45 is added to the number, the digits interchange their places. Find the number (solve by completing square method)
Answer:
Let the ten’s digit be “x”
∴ The unit digit = \(\frac { 14 }{ x } \)
∴ The number is 10x + \(\frac { 14 }{ x } \)
If the digits are interchanged the number is \(\frac { 140 }{ x } \) + x
By the given condition
10x + \(\frac { 14 }{ x } \) + 45 = \(\frac { 140 }{ x } \) + x
multiply by x
10x² + 14 + 45x = 140 + x²
9x² + 45x + 14 – 140 = 0
9x² + 45x – 126 = 0
Divided by 9
x² + 5x – 14 = 0
x² + 5x = 14

Since the digit of the number can not be negative
∴ x = 2
The number = 10x + \(\frac { 14 }{ x } \)
= 20 + \(\frac { 14 }{ 2 } \)
= 20 + 7
= 27
The number is 27

Question 17.
A rectangular garden 10 m by 16 m is to be surrounded by a concreate walk of uniform width. Given that the area of walk is 120 sqm assuming the width of walk be ‘V form the equation then solve it by formula method.
Answer:
Area of the garden = 16 × 10
= 160 sq.m

Area of the garden with walking area
= (1.6 + 2x) (10 + 2x)
= 160 + 32x + 20x + 4x²
= 4x² + 52x + 160
Area of the concrete walk = Area of the garden with walk – Area of garden
= 4x² + 52x + 160 – 160
120 = 4x² + 52x
4x² + 52x – 120 = 0
(÷ by 4) ⇒ x² + 13x – 30 = 0
Here a = 1, b = 13, c = -30
(comparing with ax² + bx + c = 0)

Since the width cannot be negative. Width of the garden = 2 m

Question 18.
If α and β are the roots of the equation 3x² – 5x + 2 = 0 find the value of
(i) \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\)
(ii) α – β
(iii) \(\frac{\alpha^{2}}{\beta}+\frac{\beta^{2}}{\alpha}\)
Answer:
Comparing with ax² + bx + c = 0
a = 3, b = -5, c = 2
α and β are the roots of the equation
3x² – 5x + 2 = 0



(iii) \(\frac{\alpha^{2}}{\beta}+\frac{\beta^{2}}{\alpha}=\frac{\alpha^{3}+\beta^{3}}{\alpha \beta}\)

Question 19.
If α and β are the roots of the equation 3x² – 6x + 1 = 0 from the equation whose roots are
(i) α² β;β²α
(ii) 2α + β; 2β + a
Answer:
α and β are the roots of 3x² – 6x + 1 = 0
α + β = \(\frac { 6 }{ 3 } \) = 2
αβ = \(\frac { 1 }{ 3 } \)
(i) Given the roots are α²β and β²α
Sum of the roots = α²β + β²α
= αβ (α + β)
= \(\frac { 1 }{ 3 } \)(2)
= \(\frac { 2 }{ 3 } \)
Product of the roots = (α²β) x (β²α)
= α²β²
= (αβ)³
= (\(\frac { 1 }{ 3 } \))³
= \(\frac { 1 }{ 27 } \)
The quadratic equation is
x² – (sum of the roots) x + product of the roots = 0
x² – (\(\frac { 2 }{ 3 } \)) x + \(\frac { 1 }{ 27 } \) = 0
multiply by 27
27x² – 18x + 1 = 0

(ii) Given the roots are 2α + β; 2 β + α
Sum of the roots = 2α + β + 2 β + α
= 2(α + β) + (α + β)
= 2(2) + 2
= 6

The quadratic polynomial is
x² – (sum of the roots) x + product of the roots = 0
x² – 6x + \(\frac { 25 }{ 3 } \) = 0
3x² – 18x + 25 = 0

Question 20.
Find X and Y if

Answer:

Question 21.
Solve for x,y

Answer:

Question 22.

Show that A² – 7A + 101₃ = 0
Answer:

Question 23.
Verify that (AB)^T = B^T A^T if

Answer:

From (1) and (2) we get
(AB)^T = B^TA^T

Question 24.
Draw the graph of y = x² and hence solve x² – 4x – 5 = 0.
Answer:
Given equations are y = x² and x² – 4x – 5 = 0
(i) Assume the values of x from – 4 to 5.

(ii) Plot the points (- 4,16), (- 3, 9), (- 2,4), (-1, 1), (0,0), (1, 1), (2,4), (3, 9), (4,16), (5,25).
(iii) Join the points by a smooth curve.
(iv) Solve the given equations


(v) The points of intersection of the line and the parabola are (-1, 1) and (5, 25).
The x-coordinates of the points are -1 and 5.
Thus solution set is {- 1, 5}.

Question 25.
Draw the graph of y = 2x² + x – 6 and hence solve 2x² + x – 10 = 0.
Answer:
Given equations are y = x² and x² – 4x – 5 = 0
(i) Assume the values of x from – 4 to 5.

(ii) Plot the points (- 4, 22), (- 3, 9), (- 2, 0), (-1, -5), (0, -6), (1, -3), (2, 4), (3, 15), (4, 30).
(iii) Join the points by a smooth curve.
(iv) Solve the given equations: Subtract 2x² + x – 10 = 0

y = 4 is a straight line parallel to X-axis
(v) The straight line and parabola intersect at point (-2.5, 4) and (2, 4).
The x-coordinates of the points are -2.5 and 2.
The solution set is {- 2.5, 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

\(\frac { AD }{ DB } \) = \(\frac { AE }{ EC } \)
\(\frac { 3 }{ 4 } \) = \(\frac { x }{ 15-x } \)
4x = 3 (15 – x)
4x = 45 – 3x
7x = 45 ⇒ x = \(\frac { 45 }{ 7 } \) = 6.43
The value of x = 6.43

(ii) Given AD = 8x – 7; BD = 5x – 3; AE = 4x – 3; EC = 3x – 1
In ∆ABC we have DE || BC
By Basic proportionality theorem
\(\frac { AD }{ DB } \) = \(\frac { AE }{ EC } \)
\(\frac { 8x-7 }{ 5x-3 } \) = \(\frac { 4x-3 }{ 3x-1 } \)

(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)

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

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.

\(\frac { AS }{ SC } \) = \(\frac { BQ }{ QC } \)
\(\frac { AS }{ SC } \) = \(\frac { 35 }{ 15 } \) ………(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

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

∴ \(\frac { AD }{ DB } \) = \(\frac { 8 }{ 4 } \) = 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

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

Question 4.
In fig. if PQ || BC and BC and PR || CD prove that

(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

\(\frac { AQ }{ AB } \) = \(\frac { AP }{ AC } \) ……(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) \(\frac { QB }{ AQ } \) = \(\frac { RD }{ AR } \)

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

\(\frac { AP }{ AB } \) = \(\frac { PQ }{ BC } \) ⇒ \(\frac { 12-x }{ BC } \) = \(\frac { x }{ 6 } \)
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

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: \(\frac { AE }{ ED } \) = \(\frac { BF }{ FC } \)

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
\(\frac { AE }{ ED } \) = \(\frac { BF }{ FC } \)

Question 7.
In figure DE || BC and CD || EE Prove that AD² = AB × AF.

Solution:
Given: In ∆ABC, DE || BC and CD || EF

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
\(\frac { BD }{ DC } \) = \(\frac { AB }{ AC } \)
\(\frac { x }{ 6-x } \) = \(\frac { 10 }{ 14 } \)

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

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
\(\frac { BD }{ DC } \) = \(\frac { 1.5 }{ 3.5 } \) = \(\frac { 15 }{ 35 } \) = \(\frac { 3 }{ 7 } \)
\(\frac { AB }{ AC } \) = \(\frac { 5 }{ 10 } \) = \(\frac { 1 }{ 2 } \)

\(\frac { BD }{ DC } \) ≠ \(\frac { AB }{ AC } \)
∴ AD is not a bisector of ∠A.

(ii) In ∆ABC, AB = 4 cm, AC = 6 cm, BD = 1.6 cm, CD = 2.4 cm

\(\frac { BD }{ DC } \) = \(\frac { 1.6 }{ 2.4 } \) = \(\frac { 16 }{ 24 } \) = \(\frac { 2 }{ 3 } \)
\(\frac { AB }{ AC } \) = \(\frac { 4 }{ 6 } \) = \(\frac { 2 }{ 3 } \)
∴ \(\frac { BD }{ DC } \) = \(\frac { AB }{ AC } \)
By angle bisector theorem; AD is the internal bisector of ∠A

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

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

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.

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:

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.

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:


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.

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:


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.

Question 15.
Construct a ∆ABC such that AB = 5.5 cm, ∠C = 25° and the altitude from C to AB is
Answer:


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.

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:


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.

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:


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.

Students can download Maths Chapter 4 Geometry Ex 4.5 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 4.5

Question 1.
If in triangles ABC and EDF, \(\frac { AB }{ DE } \) = \(\frac { BC }{ FD } \) then they will be similar, when ……….
(1) ∠B = ∠E
(2) ∠A = ∠D
(3) ∠B = ∠D
(4) ∠A = ∠F
Answer:
(3) ∠B = ∠D
Hint:

Question 2.
In ∆LMN, ∠L = 60°, ∠M = 50°. If ∆LMN ~ ∆PQR then the value of ∠R is ……………
(1) 40°
(2) 70°
(3) 30°
(4) 110°
Answer:
(2) 70°
Hint:
Since ∆LMN ~ ∆PQR
∠N = ∠R
∠N = 180 – (60 + 50)
= 180 – 110°
∠N = 70° ∴ ∠R = 70°

Question 3.
If ∆ABC is an isosceles triangle with ∠C = 90° and AC = 5 cm, then AB is ………
(1) 2.5 cm
(2) 5 cm
(3) 10 cm
(4) 5 \(\sqrt { 2 }\) cm
Answer:
(4) 5 \(\sqrt { 2 }\) cm
Hint:

AB² = AC² + BC²
AB² = 5² + 5²
(It is an isosceles triangle)
AB = \(\sqrt { 50 }\) = \(\sqrt{25 \times 2}\)
AB = 5 \(\sqrt { 2 }\)

Question 4.
In a given figure ST || QR, PS = 2 cm and SQ = 3 cm. Then the ratio of the area of ∆PQR to the area of ∆PST is …………….

(1) 25 : 4
(2) 25 : 7
(3) 25 : 11
(4) 25 : 13
Answer:
(1) 25 : 4
Hint. Area of ∆PQR : Area of ∆PST
\(\frac{\text { Area of } \Delta \mathrm{PQR}}{\text { Area of } \Delta \mathrm{PST}}=\frac{\mathrm{PQ}^{2}}{\mathrm{PS}^{2}}=\frac{5^{2}}{2^{2}}=\frac{25}{4}\)
Area of ∆PQR : Area of ∆PST = 25 : 4

Question 5.
The perimeters of two similar triangles ∆ABC and ∆PQR are 36 cm and 24 cm respectively. If PQ =10 cm, then the length of AB is ………….
(1) 6 \(\frac { 2 }{ 3 } \) cm
(2) \(\frac{10 \sqrt{6}}{3}\)
(3) 66 \(\frac { 2 }{ 3 } \) cm
(4) 15 cm
Answer:
(4) 15 cm
Hint:

Question 6.
If in ∆ABC, DE || BC. AB = 3.6 cm, AC = 2.4 cm and AD = 2.1 cm then the length of AE is ………..
(1) 1.4 cm
(2) 1.8 cm
(3) 1.2 cm
(4) 1.05 cm
Answer:
(1) 1.4 cm
Hint:

Question 7.
In a ∆ABC, AD is the bisector of ∠BAC. If AB = 8 cm, BD = 6 cm and DC = 3 cm.
The length of the side AC is ………….
(1) 6 cm
(2) 4 cm
(3) 3 cm
(4) 8 cm
Answer:
(2) 4 cm
Hint:

Question 8.
In the adjacent figure ∠BAC = 90° and AD ⊥ BC then ………..

(1) BD.CD = BC²
(2) AB.AC = BC²
(3) BD.CD = AD²
(4) AB.AC = AD²
Answer:
(3) BD CD = AD²
Hint:

Question 9.
Two poles of heights 6 m and 11m stand vertically on a plane ground. If the distance between their feet is 12 m, what is the distance between their tops?
(1) 13 m
(2) 14 m
(3) 15 m
(4) 12.8 m
Answer:
(1) 13 m
Hint:

AC² (Distance between the two tops)
= AE² + EC²
= 5² + 122
= 25 + 144= 169
AC = \(\sqrt { 169 }\) = 13 cm

Question 10.
In the given figure, PR = 26 cm, QR = 24 cm, ∠PAQ = 90°, PA = 6 cm and QA = 8 cm. Find ∠PQR.

(1) 80°
(2) 85°
(3) 75°
(4) 90°
Answer:
(4) 90°
Hint.
PR = 26 cm, QR = 24 cm, ∠PAQ = 90°
In the ∆PQR,

In the right ∆ APQ
PQ² = PA² + AQ²
= 6² + 8²
= 36 + 64 = 100
PQ = \(\sqrt { 100 }\) = 10
∆ PQR is a right angled triangle at Q. Since
PR² = PQ² + QR²
∠PQR = 90°

Question 11.
A tangent is perpendicular to the radius at the
(1) centre
(2) point of contact
(3) infinity
(4) chord
Solution:
(2) point of contact

Question 12.
How many tangents can be drawn to the circle from an exterior point?
(1) one
(2) two
(3) infinite
(4) zero
Answer:
(2) two

Question 13.
The two tangents from an external points P to a circle with centre at O are PA and PB.
If ∠APB = 70° then the value of ∠AOB is ……….
(1) 100°
(2) 110°
(3) 120°
(4) 130°
Answer:
(2) 110°
Hint.

∠OAP = 90°
∠APO = 35°
∠AOP = 180 – (90 + 35)
= 180 – 125 = 55
∠AOB = 2 × 55 = 110°

Question 14.
In figure CP and CQ are tangents to a circle with centre at 0. ARB is another tangent touching the circle at R. If CP = 11 cm and BC = 7 cm, then the length of BR is …….

(1) 6 cm
(2) 5 cm
(3) 8 cm
(4) 4 cm
Answer:
(4) 4 cm
Hint.
BQ = BR = 4 cm (tangent of the circle)
PC = QC = 11 cm (tangent of the circle)
QC = 11 cm
QB + BC = 11
QB + 7 = 11
QB = 11 – 7 = 4 cm
BR = BQ = 4 cm

Question 15.
In figure if PR is tangent to the circle at P and O is the centre of the circle, then ∠POQ is ………….

(1) 120°
(2) 100°
(3) 110°
(4) 90°
Answer:
(1) 120°
Hint.
Since PR is tangent of the circle.
∠QPR = 90°
∠OPQ = 90° – 60° = 30°
∠OQB = 30°
In ∆OPQ
∠P + ∠Q + ∠O = 180°
30 + 30° + ∠O = 180°
(OP and OQ are equal radius)
∠O = 180° – 60° = 120°

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

25² – 24² = r²
(25 + 24) (25 – 24) = r²
r = \(\sqrt { 49 }\) =7
Radius of the circle = 7 cm

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

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)

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

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.

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

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

∠POQ + ∠OQP + ∠OPQ = 180°
(Sum of the angles of a ∆ is 180°)
60° + 90° + ∠OPQ = 180°
∠OPQ = 180° – 150° = 30°
∠OPQ = 30°

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 ∆)

(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.

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

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

= 6² + 8²
= 36 + 64
= 100
OA = \(\sqrt { 100 }\) = 10 cm
Radius of the larger circle = 10 cm

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

= 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

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

\(\frac { CE }{ EA } \) = \(\frac { BC }{ AB } \) …….(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.

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.

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.

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

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


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.

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


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.

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


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.

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


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

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.


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

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.


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.

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

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.

(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.

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 = \(\frac{1}{9}\)

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

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.

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.

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

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

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 \(\frac { -1 }{ 5 } \)

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

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

= 1.5 or -15.5 (Width is not negative)
∴ Width of the path way = 1.5 m

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

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

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

Age will not be negative
Age of the girl = 10 years
Age of the sister = 20 years (2 × 10)

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,

PR² + QR² = PQ² (By Pythagoras theorem)
x² + (x + 4)² = 20²
x² + x² + 16 + 8x = 400
2x² + 8x – 384 = 0

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

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

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

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

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

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

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

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.

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”

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

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

Students can download Maths Chapter 3 Algebra Ex 3.11 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 3.11

Question 1.
Solve the following quadratic equations by completing the square method
(i) 9x² – 12x + 4 = 0
Answer:
9x² – 12x + 4 = 0
x² – \(\frac { 12x }{ 9 } \) + \(\frac { 4 }{ 9 } \) = 0 (Divided by 9)
x² – \(\frac { 4x }{ 3 } \) = \(\frac { -4 }{ 9 } \)
Add [\(\frac { 1 }{ 2 } \) (\(\frac { 4 }{ 3 } \))]² on both sides

The solution is \(\frac { 2 }{ 3 } \)

(ii) \(\frac { 5x+7 }{ x-1 } \) = 3x + 2
Answer:
(3x + 2) (x – 1) = 5x + 7
3x² – 3x + 2x – 2 = 5x + 7 ⇒ 3x² – x – 5x – 2 – 7 = 0
3x² – 6x – 9 = 0 ⇒ x² – 2x – 3 = 0 (divided by 3)
x² – 2x = 3
Adding (\(\frac { 1 }{ 2 } \) × 2)² on both sides
x² – 2x + 1 = 3 + 1
(x – 1)² = 4 ⇒ x – 1 = \(\sqrt { 4 }\)
x – 1 = ±2
x – 1 = 2 or x – 1 = -2
x = 3 or x = -1
The solution set is -1 and 3

Question 2.
Solve the following quadratic equations by formula method
(i) 2x² – 5x + 2 = 0
Answer:
a = 2, b = -5, c = 2

The solution set is \(\frac { 1 }{ 2 } \) and 2

(ii) \(\sqrt { 2 }\) f² – 6 f + 3 \(\sqrt { 2 }\) = 0
Answer:
Here a = \(\sqrt { 2 }\), b = -6 and c = 3\(\sqrt { 2 }\)

The solution set is \(\frac{3+\sqrt{3}}{\sqrt{2}}\) and \(\frac{3-\sqrt{3}}{\sqrt{2}}\)

(iii) 3y² – 20y – 23 = 0
Answer:
a = 3, b = -20, c = -23

The solution set is -1 and \(\frac { 23 }{ 3 } \)

(iv) 36y² – 12ay + (a² – b²) = 0
Answer:
Here a = 36, b = -12a, c = a² – b²

The solution set is \(\frac { (a+b) }{ 6 } \) and \(\frac { (a-b) }{ 6 } \)

Question 3.
A ball rolls down a slope and travels a distance d = t² – 0.75t feet in t seconds. Find the time when the distance travelled by the ball is 11.25 feet.
Answer:
Distance = t² – 0.75t
11.25 = t² – 0.75t
Multiply by 100
1125 = 100t² – 75t
100t² – 75t – 1125 = 0 (Divided by 25)
4t² – 3t – 45 = 0
a = 4,
b = -3,
c = -45

(time will not be negative)
The required time = \(\frac { 15 }{ 4 } \) seconds
= 3 \(\frac { 3 }{ 4 } \) second or 3.75 seconds

Students can download Maths Chapter 3 Algebra Ex 3.10 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 3.10

Question 1.
Solve the following quadratic equations by factorization method
(i) 4x² – 7x – 2 = 0
Answer:
4x² – 7x – 2 = 0
4x² – 8x + x – 2 = 0
4x(x – 2) + 1(x – 2) = 0
(x – 2) + (4x + 1) = 0

x – 2 = 0 or 4x + 1 = 0 (equate the product of factors to zero)
x = 2 or 4x = -1 ⇒ x = \(\frac { -1 }{ 4 } \)
The roots are 2; \(\frac { -1 }{ 4 } \)

(ii) 3(p² – 6) = p(p + 5)
Answer:
3p² – 18 = p² + 5p

2p² – 5p – 18 = 0
2p² – 9p + 4p – 18 = 0
p(2p – 9) + 2(2p – 9) = 0
(2p – 9)(p + 2) = 0
2p – 9 = 0 or p + 2 =
The roots are p = \(\frac { 9 }{ 2 } \), -2

(iii) \(\sqrt{a(a-7)}\) = 3 \(\sqrt{2}\)
Answer:
Squaring on both sides
a(a – 7) = (3\(\sqrt{2}\))²

a² – 7a = 18
a² – 7a – 18 = 0
(a – 9) (a + 2) = 0
a – 9 = 0 or a + 2 = 0
The roots are -2 and 9

(iv) \(\sqrt { 2 }\) x² + 7x + 5\(\sqrt { 2 }\) = 0
Answer:
\(\sqrt { 2 }\) x² + 7x + 5 \(\sqrt { 2 }\) = 0
\(\sqrt { 2 }\) x² + 2x + 5x + 5\(\sqrt { 2 }\) = 0
\(\sqrt { 2 }\) x (x + \(\sqrt { 2 }\)) + 5(x + \(\sqrt { 2 }\)) = 0

(x + \(\sqrt { 2 }\)) + (\(\sqrt { 2 }\)x + 5) = 0 (equate the product of factors to zero)
x + – \(\sqrt { 2 }\) = 0 or \(\sqrt { 2 }\)x + -5
x = \(\frac{-5}{\sqrt{2}}\)
The roots are – \(\sqrt { 2 }\), \(\frac{-5}{\sqrt{2}}\)

(v) 2x² – x + \(\frac { 1 }{ 8 } \) = 0
Answer:
2x² -x + \(\frac { 1 }{ 8 } \) = 0
16x² – 8x + 1 = (multiply by 8)

16x² – 4x – 4x + 1
4x(4x – 1) – 1 (4x – 1) = 0
(4x- 1) (4x- 1) = 0
4x = 1, 4x = 1
x = \(\frac { 1 }{ 4 } \), x = \(\frac { 1 }{ 4 } \)
The roots are \(\frac { 1 }{ 4 } \) and \(\frac { 1 }{ 4 } \)

Question 2.
The number of volleyball games that must be scheduled in a league with n teams is given by G(n) = \(\frac{n^{2}-n}{2}\) where each team plays with every other team exactly once. A league schedules 15 games. How many teams are in the league?
Solution:
G(n) = \(\frac{n^{2}-n}{2}\)
⇒ 15 = \(\frac{n^{2}-n}{2}\) ⇒ 30 = n² – n
n² – n – 30 = 0
⇒ n² – 6n + 5n – 30 = 0
n(n – 6) + 5 (n – 6) = 0
(n – 6)(n + 5) = 0 ⇒ n = 6, -5
As n cannot be (-ve), n = 6.
∴ There are 6 teams in the league.