Category: Class 10

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 4.2 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 4.1 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 Unit Exercise 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 Unit Exercise 4

Question 1.
In the figure, if BD ⊥ AC and CE ⊥ AB , prove that
(i) ∆ AEC ~ ∆ADB
(ii) \(\frac { CA }{ AB } \) = \(\frac { CE }{ DB } \)

Solution:
(i) ∠AEC = ∠ADB = 90° ∠A is common By AA – Similarity.
∴ ∆AEC ~ ∆ADB
Since the two triangles are similar

(ii) \(\frac { AE }{ AD } \) = \(\frac { AC }{ AB } \) = \(\frac { EC }{ DB } \)
\(\frac { AC }{ AB } \) = \(\frac { CE }{ DB } \)

Question 2.
In the given figure AB || CD || EF. If AB = 6 cm, CD = x cm, EF = 4 cm, BD = 5 cm and DE = y cm. Find x and y.

Solution:
In the given diagram ∆AEF and ∆ACD
∠AEF = ∠ACD = 90°
∠A is common
By AA – Similarity.
∴ ∆AEF ~ ∆ACD
\(\frac { AE }{ AC } \) = \(\frac { AF }{ AD } \) = \(\frac { EF }{ CD } \)
\(\frac { AE }{ AC } \) = \(\frac { EF }{ CD } \)
\(\frac { AE }{ AC } \) = \(\frac { 4 }{ x } \)
AC = \(\frac{\mathrm{AE} \times x}{4}\) …..(1)
In ∆EAB and ∆ECD,
∠EAB = ∠ECD = 90°
∠E is common
∆ ECD ~ ∆EAB
\(\frac { EC }{ EA } \) = \(\frac { ED }{ EB } \) = \(\frac { CD }{ AB } \)
\(\frac { EC }{ EA } \) = \(\frac { x }{ 6 } \)
EC = \(\frac{\mathrm{EA} \times x}{6}\) …….(2)
In ∆AEB; CD || AB
By Basic Proportionality Theorem
\(\frac { AB }{ CD } \) = \(\frac { EB }{ ED } \)
\(\frac { 6 }{ x } \) = \(\frac { 5+y }{ y } \)
x = \(\frac { 6y }{ y+5 } \) (EC = x) …….(3)
Add (1) and (2) we get
AC + EC = \(\frac{A E \times x}{4}+\frac{x \times E A}{6}\)
>
Substitute the value of x = 2.4 in (3)
2.4 = \(\frac { 6y }{ y+5 } \)
6y = 2.4y + 12
6y – 2.4y = 12 ⇒ 3.6 y = 12
y = \(\frac { 12 }{ 3.6 } \) = \(\frac { 120 }{ 36 } \) = \(\frac { 10 }{ 3 } \) = 3.3 cm
The value of x = \(\frac { 12 }{ 5 } \) (or) 2.4 cm and y = \(\frac { 10 }{ 3 } \) (or) 3.3 cm

Question 3.
O is any point inside a triangle ABC. The bisector of ∠AOB, ∠BOC and ∠COA meet the sides AB, BC and CA in point D, E and F respectively. Show that AD × BE × CF = DB × EC × FA.
Solution:
In ∆ABC the bisector meets AB at D, BC at E and AC at F.

The angle bisector AO, BO and CO intersect at “O”.
By Cevas Theorem
\(\frac { AD }{ DB } \) × \(\frac { BE }{ EC } \) × \(\frac { CF }{ AF } \) = 1
AD × BE × CF = DB × EC × AF
Hence it is proved

Question 4.
In the figure, ABC is a triangle in which AB = AC. Points D and E are points on the side AB and AC respectively such that AD = AE. Show that the points B, C, E and D lie on a same circle.

Solution:
∠B = ∠C (Given AB = AC)
AD + DB = AE + EC
BD = EC (Given AD = AE)
DE parallel BC Since AEC is a straight line.
∠AED + ∠CED = 180°
∠CBD + ∠CED = 180°
Similarly of the opposite angles = 180°
∴ BCED is a cyclic quadrilateral

Question 5.
Two trains leave a railway station at the same time. The first train travels due west and the second train due north. The first train travels at a speed of 20 km/hr and the second train travels at 30 km/hr. After 2 hours, what is the distance between them?
Solution:
A is the position of the 1^st train.
B is the position of the 2^nd train.

Distance Covered in 2 hours
OA = 2 × 20 = 40 km
OB = 2 × 30 = 60 km
Distance between the train after 2 hours

Distance between
the two train = 72.11 km (or) 20\(\sqrt { 13 }\) km

Question 6.
D is the mid point of side BC and AE ⊥ BC. If BC = a, AC = b, AB = c, ED = x, AD = p and AE = h, prove that
(i) b² = p² + ax + \(\frac{a^{2}}{4}\)
(ii) c² = p² – ax + \(\frac{a^{2}}{4}\)
(iii) a² + c² = 2 p² + \(\frac{a^{2}}{2}\)
Solution:
(i) Given ∠AED = 90°

ED = x; DC = \(\frac { a }{ 2 } \)
(D is the mid point of BC)
∴ EC = x + \(\frac { a }{ 2 } \), BE = \(\frac { a }{ 2 } \) – x
∴ In the right ∆ AED
AD² = AE² + ED²
p² = h² + x²
In the right ∆ AEC,
AC² = AE² + EC²

(ii) In the right triangle ABE,
AB² = AE² + BE²
c² = h² + (\(\frac { a }{ 2 } \) – x)²
c² = h² + \(\frac{a^{2}}{4}\) + x² – ax
c² = h² + x² + \(\frac { 1 }{ 4 } \) a² – ax
c² = p² – ax + \(\frac{a^{2}}{4}\) (from 1)

(iii) By adding (2) and (3)
b² + c² = p² + ax + \(\frac{a^{2}}{4}\) + p² – ax + \(\frac{a^{2}}{4}\)
= 2p² + \(\frac{2a^{2}}{4}\)
= 2p² + \(\frac{a^{2}}{2}\)

Question 7.
A man whose eye-level is 2 m above the ground wishes to find the height of a tree. He places a mirror horizontally on the ground 20 m from the tree and finds that if he stands at a point C which is 4 m from the mirror B, he can see the reflection of the top of the tree. How height is the tree?
Solution:
Let the height of the tree AD be “h”.

In ∆ ACD and ∆ BCF,
∠A = ∠B = 90°
∠C is common
∆ ACD ~ ∆ BCF by AA similarity
\(\frac { AD }{ BF } \) = \(\frac { AC }{ BC } \)
\(\frac { h }{ x } \) = \(\frac { 24 }{ 2 } \) = 6
h = 6x ………(1)
In ∆ ACE and ∆ ABF,
∠C = ∠B = 90°
∠A is common
∴ ∆ ACE ~ ∆ ABF
\(\frac { CE }{ BF } \) = \(\frac { AC }{ AB } \)
\(\frac { 2 }{ x } \) = \(\frac { 24 }{ 20 } \)
24x = 20 × 2
x = \(\frac{20 \times 2}{24}=\frac{5 \times 2}{6}=\frac{10}{6}\)
x = \(\frac { 5 }{ 3 } \)
Substitute the value of x in (1)
h = 6 × \(\frac { 5 }{ 3 } \) = 10 m
∴ Height of the tree is 10 m

Question 8.
An emu which is 8 ft tail is standing at the foot of a pillar which is 30 ft high. It walks away from the pillar. The shadow of the emu falls beyond emu. What is the relation between the length of the shadow and the distance from the emu to the pillar?
Solution:
Let the shadow of the emu AE be “x” and BE be “y” ED || BC

By basic proportionality theorem
\(\frac { AE }{ AB } \) = \(\frac { ED }{ BC } \)
\(\frac { x }{ x+y } \) = \(\frac { 8 }{ 30 } \)
30x = 8x + 8y
22x – 8y = 0
(÷ by 2) 11x – 4y = 0
11x = 4y
x = \(\frac { 4 }{ 11 } \) × y
x = \(\frac { 4 }{ 11 } \) × distance from the pillar to emu
Length of = \(\frac { 4 }{ 11 } \) × distance from the shadow the pillar to emu

Question 9.
Two circles intersect at A and B. From a point P on one of the circles lines PAC and PBD are drawn intersecting the second circle at C and D. Prove that CD is parallel to the tangent at P.
Solution:
Proof:

A and B are the points intersecting the circles. Join AB.
∠P’PB = ∠PAB (alternate segment theorem)
∠PAB + ∠BAC = 180° …(1)
(PAC is a straight line)
∠BAC + ∠BDC = 180° …(2)
ABDC is a cyclic quadrilateral.
From (1) and (2) we get
∠P’PB = ∠PAB = ∠BDC
P’P and DC are straight lines.
PD is a transversal alternate angles are equal.
∴ P’P || DC.

Question 10.
Let ABC be a triangle and D, E, F are points on the respective sides AB, BC, AC (or their extensions).
Let AD : DB = 5 : 3, BE : EC = 3 : 2 and AC = 21. Find the length of the line segment CF.
Solution:

\(\frac { AD }{ DB } \) = \(\frac { 5 }{ 3 } \); \(\frac { BE }{ EC } \) = \(\frac { 3 }{ 2 } \); AC = 21
By Ceva’s theorem

Length of the line segment CF = 6 units

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 4.3 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 Ex 3.19 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 3.19

Question 1.
A system of three linear equations in three variables is inconsistent if their planes
(1) intersect only at a point
(2) intersect in a line
(3) coincides with each other
(4) do not intersect
Answer:
(4) do not intersect

Question 2.
The solution of the system x + y -3z = -6, -7y + 7z = 7,3z = 9 is
(1) x = 1, y = 2, z = 3
(2) x = -1, y = 2, z = 3
(3) x = -1, y = -2, z = 3
(4) x = 1, y = 2, z = 3
Solution:
(4) x = 1, y = 2, z = 3
Hint:
x + y – 3z = -6
-7y + 7z = 7
3z = 9
z = 3
-7y + 21 = 7
-7y = -14
y = 2
x + 2 -3 × 3 = -6
x + 2 – 9 = -6 .
x = -6 + 7 = 1

Question 3.
If (x – 6) is the HCF of x² – 2x – 24 and x² – kx – 6 then the value of k is ……………..
(1) 3
(2) 5
(3) 6
(4) 8
Answer:
(2) 5
Hint.
HCF = x – 6
p(x) = x² – 2x – 24
= (x – 6) (x + 4)
g(x) = x² – kx – 6
∴ x – 6 is the common factor.
g (6) = 6² – k(6) – 6
= 36 – 6k – 6
= 30 – 6k
g(6) = 0
30 – 6k = 0
30 = 6k ⇒ k = \(\frac { 30 }{ 6 } \) = 5
The value of k = 5

Question 4.
\(\frac { 3y-3 }{ y } \) + \(\frac { 7y-7 }{ 3y2 } \) is …………

Answer:
(1) \(\frac{9 y}{7}\)

Question 5.
y² + \(\frac{1}{y^{2}}\) is not equal to ………….
(1) \(\frac{y^{4}+1}{y^{2}}\)
(2) (y + \(\frac { 1 }{ y } \))²
(3) (y – \(\frac { 1 }{ y } \))²
(4) (y + \(\frac { 1 }{ y } \))² – 2
Answer:
(2) (y + \(\frac { 1 }{ y } \))²
Hint.

Question 6.
\(\frac{x}{x^{2}-25}-\frac{8}{x^{2}+6 x+5}\) gives …………..

Answer:

Hint.

Question 7.
The square root of \(\frac{256 x^{8} y^{4} z^{10}}{25 x^{6} y^{6} z^{6}}\) is equal to ………….

Answer:

Hint.

Question 8.
Which of the following should be added to make x⁴ + 64 a perfect square
(1) 4x²
(2) 16x²
(3) 8x²
(4) -8x²
Solution:
(2) 16x²
Hint:
x⁴ + 64 = (x²)² + 8² + 2 × 8x²
= (x² + 8)²

Question 9.
The solution of (2x – 1)² = 9 is equal to ………..
(1) -1
(2) 2
(3) -1, 2
(4) None of these
Answer:
(3) -1, 2
Hint.
(2x – 1)² = 9 ⇒ (2x – 1) = \(\sqrt { 9 }\)
2x – 1 = ± 3 ⇒ 2x – 1 = 3 or 2x – 1 = 3
2x – 1 = 3
2x = 4 ⇒ x = \(\frac { 4 }{ 2 } \) = 2
2x – 1 = -3 ⇒ 2x = -3 + 1
x = -1

Question 10.
The values of a and b if 4x⁴ – 24x³ + 76x² + ax + b is a perfect square are ………..
(1) 100,120
(2) 10,12
(3) -120,100
(4) 12,10
Answer:
(3) -120,100
Hint.
Since it is a perfect square
a + 120 = 0
a = -120
b – 100 = 0
b = 100

Question 11.
If the roots of the equation q²x² + p²x + r² = 0 are the squares of the roots of the equation qx² + px + r = 0 , then q,p,r are in ……………..
(1) A.P.
(2) G.P.
(3) Both A.P and G.P
(4) none of these
Answer:
(2) G.P.
Hint.
q²x² + p²x + r² = 0
Let the roots be α² + β²
α² + β² = \(\frac{-p^{2}}{q^{2}}\)
α²β² = \(\frac{r^{2}}{q^{2}}\)
qx² + px + r = 0
Let the root be α and β

Question 12.
Graph of a linear polynomial is a
(1) straight line
(2) circle
(3) parabola
(4) hyperbola
Solution:
(1) straight line

Question 13.
The number of points of intersection of the quadratic polynomial x2 + 4x + 4 with the X axis is …………….
(1) 0
(2) 1
(3) 0 or 1
(4) 2
Answer:
(2) 1
Hint:
(x + 2)² = 0 ⇒ (x + 2) (x + 2) = 0
x + 2 = 0 or x + 2 = 0 ⇒ x = -2 or x = -2
Number of points of intersection is 1 (both the values are same)

Question 14.
For the given matrix

the order of the matrix A^T is ……………
(1) 2 × 3
(2) 3 × 2
(3) 3 × 4
(4) 4 × 3
Answer:
(3) 3 × 4

Question 15.
If A is a 2 × 3 matrix and B is a 3 × 4 matrix, how many columns does AB have ………..
(1) 3
(2) 4
(3) 2
(4) 5
Answer:
(2) 4
Hint.
The order of AB is 2 × 4
∴ Number of columns is 4.

Question 16.
If the number of columns and rows are not equal in a matrix then it is said to be a
(1) diagonal matrix
(2) rectangular matrix
(3) square matrix
(4) identity matrix
Solution:
(2) rectangular matrix

Question 17.
Transpose of a column matrix is …………..
(1) unit matrix
(2) diagonal matrix
(3) column matrix
(4) row matrix
Answer:
(4) row matrix

Question 18.
Find the matrix X if


Answer:

Hint.

Question 19.
Which of the following can be calculated from the given matrices?

(i) A²
(ii) B²
(iii) AB
(iv) BA
(1) (i) and (ii) only
(2) (ii) and (iii) only
(3) (ii) and (iv) only
(4) all of these
Answer:
(3) (ii) and (iv) only
Hint: (i) A² is possible to find
(ii) B² is also possible to find
(iii) not possible: order of A= (3 × 2) order of B is (3 × 3). AB is not possible number of column of matrix A ≠ number of rows of the matrix B.
(iv) Possible number column of the matrix is equal to the number of the matrix A.

Question 20.

Which of the following statements are correct?

(1) (i) and (ii) only
(2) (ii) and (iii) only
(3) (iii) and (iv)
(4) all of these
Answer:
(1) (i) and (ii) only
Hint.
(i) AB + C : order of A = 2 × 3
order of B = 3 × 2
order of AB = 2 × 2
order of C = 2 × 2
It is possible to add AB + C

(ii) BC :
order of B = 3 × 2
order of C = 2 × 2
It is possible to find BC

(iii) BA is not possible
order of B = 3 × 2
order of A = 3 × 2
BA does not exist.
BA + C is not a correct statement

(iv) ABC is not possible
order of A = 2 × 3 order of
B = 3 × 2
order of AB = 2 × 2
order of C = 3 × 2
It is not possible to multiply AB and C.
∴ The statement ABC is not correct.

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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.

Students can download Maths Chapter 3 Algebra Ex 3.17 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 3.17

Question 1.
If then

verify that (i) A + B = B + A
(ii) A + (-A) = (-A) + A = O.
Answer:

From (1) and (2) we get A + B = B + A

From (1) and (2) we get
A + (-A) = (-A) + A = 0

Question 2.
If
then verify that
A + (B + C) = (A + B) + C.
Answer:


From (1) and (2) we get
A + (B + C) = (A + B) + C

Question 3.
Find X and Y if and
Answer:

Question 4.
If
find the value of (i) B – 5A (ii) 3A – 9B
Answer:

Question 5.
Find the values of x, y, z if

Answer:
(i) x – 3 = 1 ⇒ x = 1 + 3 ⇒ x = 4
3x – z = 0 (substitute the value of x)
3(4) – z = 0
12 – z = 0
∴ z = 12
x + y + z = 6
4 + y + 12 = 0
y + 16 = 6
y = 6- 16
∴ y = -10
The value of x = 4, y = -10 and z = 12

(ii) [x y – z z + 3] + [y 4 3] = [4 8 16]
x + y = 4 ….(1)
y – z + 4 = 8
Substitute the value
of z in (2)
(2) ⇒ y – 10 = 4
Substitute the value of y in (1)
z + 3 + 3 = 16
z + 6 = 16
z = 16 – 16 = 10
y = 14
(1) ⇒ x + 14 = 4
x – 4 – 14 = -10
The value of x = -10, y = 14 and z = 10

Question 6.
Find x and y if x
Answer:

4x – 2y = 4
(1) ⇒ 2x – y = 2
(2) ⇒ 3x – y = 2
– x + y = 2
Add (1) and (2)

Substitute the value of x = 4 in (2)
– 4 + y = 2
y = 2 + 4 = 6
The value of x = 4 and y = 6

Question 7.
Find the non-zero values of x satisfying the matrix equation

Answer:

The value of x = 4

Question 8.
Solve for x,y :

Answer:

x² – 4x = -5
x² – 4x + 5 = 0
(x – 5) (x + 1) = 0
x – 5 = 0 or x + 1 = 0
x = 5 or x = – 1

The value of x = -1 and 5
y² – 2y = 8
y² – 2y – 8 = 0
(y – 4) (y + 2) = 0
y – 4 = 0 or y + 2 = 0

y = 4 or y = -2
The value of y = -2 and 4

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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.16

1. In the

write (i) The number of elements
(ii) The order of the matrix
(iii) Write the elements a₂₂, a₂₃, a₂₄, a₃₄, a₄₃, a₄₄.
Answer:
(i) The number of elements is 16
(ii) The order of the matrix is 4 × 4
(iii) Elements corresponds to

Question 2.
If a matrix has 18 elements, what are the possible orders it can have? What if it has 6 elements?
Answer:
The possible orders of the matrix having 18 elements are

The possible orders of the matrix having 6 elements are

Question 3.
Construct a 3 × 3 matrix whose elements are given by
(i) a_ij = |i – 2j|
Answer:
a_ij = |i – 2j|
The general 3 × 3 matrices is

a₁₁ = |1 – 2(1)| = |1 – 2| = | – 1| = 1
a₁₂ = |1 – 2(2)| = |1 – 4| = | – 3| = 3
a₁₃ = |1 – 2(3)| = |1 – 6| = | – 5| = 5
a₂₁ = |2 – 2(1)| = |2 – 2| = 0 = 0
a₂₂ = |2 – 2(2)| = |2 – 4| = | – 2| = 2
a₂₃ = |2 – 2(3)| = |2 – 6| = | – 4| = 4
a₃₁ = |3 – 2(1)| = |3 – 2| = | 1 | = 1
a₃₂ = |3 – 2(2)| = |3 – 4| = | – 1 | = 1
a₃₃ = |3 – 2(3)| = |3 – 6| = | – 3 | = 3
The required matrix

(ii) a_ij = \(\frac{(i+j)^{3}}{3}\)
Answer:
a₁₁ = \(\frac{(1+1)^{3}}{3}\) = \(\frac{2^{3}}{3}\) = \(\frac { 8 }{ 3 } \)
a₁₂ = \(\frac{(1+2)^{3}}{3}\) = \(\frac { 27 }{ 3 } \) = 9
a₁₃ = \(\frac{(1+3)^{3}}{3}\) = \(\frac { 64 }{ 3 } \) = \(\frac { 64 }{ 3 } \)
a₂₁ = \(\frac{(2+1)^{3}}{3}\) = \(\frac { 27 }{ 3 } \) = 9
a₂₂ = \(\frac{(2+2)^{3}}{3}\) = \(\frac { 64 }{ 3 } \) = \(\frac { 64 }{ 3 } \)
a₂₃ = \(\frac{(2+3)^{3}}{3}\) = \(\frac { 125 }{ 3 } \) = \(\frac { 125 }{ 3 } \)
a₃₁ = \(\frac{(3+1)^{3}}{3}\) = \(\frac { 64 }{ 3 } \) = \(\frac { 64 }{ 3 } \)
a₃₂ = \(\frac{(3+2)^{3}}{3}\) = \(\frac { 125 }{ 3 } \) = \(\frac { 125 }{ 3 } \)
a₃₃ = \(\frac{(3+3)^{3}}{3}\) = \(\frac { 216 }{ 3 } \) = 72
The required matrix

Question 4.
If  then find the tranpose of A.
Answer:

transpose of A = (A^T)

Question 5.
If then find the tranpose of – A
Answer:

Transpose of – A = (-A^T) =

Question 6.
If A = then verify (AT)T = A
Answer:

Hence it is verified

Question 7.
Find the values of x, y and z from the following equations
(i)

Answer:
Since the given matrices are equal then all the corresponding elements are equal.
y = 12, z = 3, x = 3
The value of x = 3, y = 12 and z = 3

(ii)

Answer:
x + y = 6 ……(1)
5 + z = 5
z = 5 – 5 = 0
xy = 8
y = \(\frac { 8 }{ x } \)
Substitute the value of y = \(\frac { 8 }{ x } \) in (1)
x + \(\frac { 8 }{ x } \) = 6
x² + 8 = 6x
x² – 6x + 8 = 0
(x – 4) (x – 2) = 0
∴ x – 4 = 0 or x – 2 = 0
x = 4 or x = 2
y = \(\frac { 8 }{ 4 } \) = 2 or y = \(\frac { 8 }{ 2 } \) = 4

∴ The value of x, y and z are 4, 2, 0 (or) 2, 4, 0

(iii)

Solution:
x + y + z = 9 ……….(1)
x + z = 5 ……….(2)
y + z = 7 ……….(3)

Substitute the value of y = 4 in (3)
y + z = 7
4 + z = 7
z = 7 – 4
= 3
Substitute the value of z = 3 in (2)
x + 3 = 5
x = 5 – 3
= 2
∴ The value of x = 2 , y = 4 and z = 3