Bhagya

Students can download Maths Chapter 6 Trigonometry Additional Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 6 Trigonometry Additional Questions

I. Choose the Correct Answer

Question 1.
The value of cosec² 60 – 1 is equal to ……..
(a) cos² 60
(b) cot² 60
(c) sec² 60
(d) tan² 60
Solution:
(b) cot² 60

Question 2.
The value of cos 60° cos 30° – sin 60° sin 30° is equal is ……..
(a) cosec 90°
(b) tan 90°
(c) sin 30° + cos 30°
(d) cos 90°
Solution:
(d) cos 90°

Question 3.
The value of \(\frac{sin 57°}{cos 33°}\) is …….
(a) cot 63°
(b) tan 27°
(c) 1
(d) 0
Solution:
(c) 1

Question 4.
If 3 cosec 36° = sec 54° then the value of x is ……..
(a) 0
(b) 1
(c) \(\frac{1}{3}\)
(d) \(\frac{3}{4}\)
Solution:
(c) \(\frac{1}{3}\)

Question 5.
If cos A cos 30° = \(\frac{√3}{4}\), then the measures of A is ……..
(a) 90°
(b) 60°
(c) 45°
(d) 30°
Solution:
(b) 60°

II. Answer the Following Question

Question 1.
Given Sec θ = \(\frac{13}{12}\). Calculate all other trigonometric ratios.
Solution:
In the right triangle ABC

BC² = AC² – AB²
= 13² – 12²
= 169 – 144
= 25
∴ BC = \(\sqrt{25}\)
= 5

Question 2.
If 3 cot A = 4 check weather \(\frac{1- tan²A}{1+ tan²A}\) = cos² A – sin² A or not?
Solution:
3 cot A = 4
cot A = \(\frac{4}{3}\)
In the right ΔABC

AC² = AB² + BC²
= 4² + 3²
= 16 + 9
= 25
= \(\sqrt{25}\)
= 5

Hence \(\frac{1- tan²A}{1+ tan²A}\) = cos² A – sin² A
R.H.S = cos² A – sin² A

L.H.S = R.H.S

Question 3.
Evaluate \(\frac{sin 30° + tan 45° – cosec 60°}{sec 30° + cos 60° + cot 45°}\)
Solution:
sin 30° = \(\frac{1}{2}\); tan 45° = 1; cosec 60° = \(\frac{2}{√3}\); sec 30° = \(\frac{2}{√3}\); cos 60° = \(\frac{1}{2}\); cot 45° = 1
\(\frac{sin 30° + tan 45° – cosec 60°}{sec 30° + cos 60° + cot 45°}\)

The value is \(\frac{43-24√3}{11}\)

Question 4.
Find A if sin 20° tan A sec 70° = √3
Solution:
sin 20° . tan A . sec 70° = √3
sin 20° . sec 70° . tan A = √3
sin (90° – 70°). sec 70° . tan A = √3
cos 70° × latex]\frac{1}{cos 70°}[/latex] tan A = √3
tan A = √3
tan A = tan 60°
∴ ∠A = 60°

Question 5.
Find the area of the right triangle with hypotenuse 8 cm and one of the acute angles is 57°
Solution:
In the ΔABC

sin C = \(\frac{AB}{AC}\)
Sin 57° = \(\frac{AB}{8}\)
0.8387 = \(\frac{AB}{8}\)
∴ AB = 0.8387 × 8
= 0.71 cm
In the ΔABC
cos C = \(\frac{BC}{AC}\)
cos 57° = \(\frac{BC}{8}\)
0.5446 = \(\frac{BC}{8}\)
BC = 0.5446 × 8
= 4.36
Area of the right ΔABC
= \(\frac{1}{2}\) × AB × BC sq. units
= \(\frac{1}{2}\) × 6.71 × 4.36 cm²
= 14.62 cm²
Area of the Δ = 14.62 cm²

Students can download Maths Chapter 7 Mensuration Ex 7.4 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 7 Mensuration Ex 7.4

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

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

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

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

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

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

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

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

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

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

= 10 × 10 × 10
= 1000 bricks

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 7 Mensuration Ex 7.3 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 7 Mensuration Ex 7.3

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

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

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

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

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

Question 5.
The dimensions of a brick are 24 cm × 12 cm × 8 cm. How many such bricks will be required to build a wall of 20 m length, 48 cm breadth and 6 m height?
Solution:
Length of a brick (l) = 24 cm
Breadth of a brick (b) = 12 cm
Depth of a brick (h) = 8 cm
Volume of a brick = lbh cu.units
Volume of one brick = 24 × 12 × 8 cm³
Length of a wall (l) = 20 m = 2000 cm
Breadth of a wall (b) = 48 cm
Height of a wall (h) = 6 m = 600 cm
Volume of a wall = l × b × h cu. units
= 2000 × 48 × 600 cm³
Number of bricks

= 500 × 50 ( ÷ by 4)
= 25000 bricks
∴ Number of bricks = 25000

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

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

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

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

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

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

Students can download Maths Chapter 7 Mensuration Ex 7.2 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 7 Mensuration Ex 7.2

Question 1.
Find the Total Surface Area and the Lateral Surface Area of a cuboid whose dimensions are length = 20 cm, breadth = 15 cm, height = 8 cm.
Solution:
Here l = 20 cm, b = 15 cm, h = 8 cm
L.S.A. of the cuboid = 2(1 + b)h sq.m
= 2(20 + 15) × 8
= 2 × 35 × 8
= 560 sq.m
Total surface area of the cuboid = 2(lb + bh + lh) sq.units
= 2(20 × 15 + 15 × 8 + 8 × 20) sq. cm
= 2(300 + 120 + 160) sq. cm
= 2 × 580 sq. cm
= 1160 sq. cm

Question 2.
The dimensions of a cuboidal box are 6 m x 400 cm x 1.5 m. Find the cost of painting its entire outer surface at the rate of Rs 22 per m².
Solution:
Length of the cuboid box (l) = 6 m
Breadth of the cuboid box (b) = 400 cm = 4m
Height of the cuboid box (h) = 1.5 m
T.S.A. of the cuboid = 2(lb + bh + lh) sq.units
= 2(6 × 4 + 4 × 1.5 + 1.5 × 6) sq.units
= 2(24 + 6 + 9)
= 2 × 39 sq.m
= 78 sq.m
Cost of painting for one sq.m = Rs 22
Total cost of painting = Rs 78 × 22
= Rs 1716

Question 3.
The dimensions of a hall is 10 m × 9 m × 8 m. Find the cost of white washing the walls and ceiling at the rate of Rs 8.50 per m².
Solution:
Length of the hall (l) = 10 m
Breath of the hall (b) = 9 m
Height of the hall (h) = 8 m
Area to be white wash = L.S.A. + Ceiling of the building
= 2(l + b)h + lb sq.units
= 2(10 + 9)8 + 10 × 9 sq.m
= 2 × 19 × 8 + 10 × 9 sq. m
= (304 + 90) sq.m
= 394 sq.m
Cost of white washing one sq.m = Rs 8.50
Cost of white washing for 394 sq.m = Rs 394 × 8.50
= Rs 3349
Total cost of white washing = Rs 3349

Question 4.
Find the TSA and LSA of the cube whose side is
(i) 8 m
(ii) 21 cm
(iii) 7.5 cm
Solution:
(i) 8m
Side of a cube (a) = 8m
T.S.A. of the cube = 6a² sq.units
= 6 × 8 × 8 sq. m
= 384 sq.m
L.S.A. of the cube = 4a² sq.units
= 4 × 8 × 8 sq.m
= 256 sq.m

(ii) 21 cm
Solution:
Side of a cube (a) = 21 cm
T.S.A. of the cube = 6a² sq. units
= 6 × 21 × 21 cm²
= 2646 cm²
L.S.A. of the cube = 4a² sq.units
= 4 × 21 × 21 sq.cm
= 4 × 441 cm²
= 1764 cm²

(iii) 7.5 cm
Solution:
Side of a cube (a) = 7.5 cm
T.S.A. of the cube = 6a² sq.units
= 6 × 7.5 × 7.5 cm²
= 337.5 cm²
L.S.A. of the cube = 4a² sq.units
= 4 × 7.5 × 7.5 sq.cm
= 225 cm²

Question 5.
If the total surface area of a cube is 2400 cm² then, find its lateral surface area.
Solution:
T.S.A. of the cube = 2400 cm²
6a² = 2400
a² = \(\frac{2400}{6}\)
= 400 cm²
L.S.A. of the cube = 4a² sq.units
= 4 × 400 cm²
= 1600 cm²
(OR)
T.S.A. of the cube = 2400 cm²
6a² = 2400
a² = \(\frac{2400}{6}\)
= 400
a = \(\sqrt{400}\)
= 20 cm
Side of a cube (a) = 20 cm
L.S.A. of the cube = 4a² sq.units
= 4 × 20 × 20 cm²
= 1600 cm²

Question 6.
A cubical container of side 6.5 m is to be painted on the entire outer surface. Find the area to be painted and the total cost of painting it at the rate of Rs 24 per m².
Solution:
Side of a cube (a) = 6.5 m
Total surface area of the cube = 6a² sq.units
= 6 × 6.5 × 6.5 sq.m
= 253.50 sq.m
Cost of painting for 1 sq.m = Rs 24
Cost of painting for 253.5 sq.m = 253.5 × 24
= Rs 6084
∴ Area to be painted = 253.50 m²
Total cost of painting = Rs 6084

Question 7.
Three identical cubes of side 4 cm are joined end to end. Find the total surface area and lateral surface area of the new resulting cuboid.
Solution:
Joint the three identical cubes we get a new cuboid
Length of the cuboid (l) = (4 + 4 + 4) cm
l = 12 cm
Breadth of the cuboid (b) = 4 cm
Height of the cuboid (h) = 4 cm
Total surface area of the new cuboid = 2(lb + bh + lh) sq.units
= 2(12 × 4 + 4 × 4 + 4 × 12)
= 2(48 + 16 + 48) cm
= 2(112) cm²
= 224 cm²
Lateral surface area of the new cuboid = 2(l + b)h sq.units
= 2(12 + 4)4 cm²
= 2 × 16 × 4 cm²
= 128 cm²
∴ T.S.A of the new cuboid = 224 cm²
L.S.A of the new cuboid = 128 cm²

Students can download Maths Chapter 7 Mensuration Ex 7.1 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 7 Mensuration Ex 7.1

Question 1.
Using Heron’s formula, find the area of a triangle whose sides are
(i) 10 cm, 24 cm, 26 cm
Solution:
Let a = 10 cm, b = 24 cm and c = 26 cm
s = \(\frac{a + b + c}{2}\)
= \(\frac{10 + 24 + 26}{2}\)
s = \(\frac{60}{2}\)
= 30 cm
s – a = 30 – 10 = 20 cm
s – b = 30 – 24 = 6 cm
s – c = 30 – 26 = 4 cm
Area of a triangle

= 2³ × 3 × 5
= 8 × 3 × 5
= 120 cm²
Area of a triangle = 120 cm²

(ii) 1.8 m, 8 m, 8.2 m
Solution:
Here a = 1.8 m, b = 8 m, c = 8.2 m
s = \(\frac{a+b+c}{2}\)
= \(\frac{(1.8+8+8.2)m}{2}\)
= \(\frac{18}{2}\)
= 9 m
s – a = 9 – 1.8 = 7.2 m
s – b = 9 – 8 = 1 m
s – c = 9 – 8.2 m = 0.8 m
Area of triangle

= 3 × 2.4
= 7.2 m²
∴ Area of the triangle = 7.2 sq. m

Question 2.
The sides of the triangular ground are 22 m, 120 m and 122 m. Find the area and cost of levelling the ground at the rate of Rs 20 per m².
Solution:
The sides of the triangular ground are 22m, 120m and 122 m
a = 22 m, b = 120 m, c = 122 m
s = \(\frac{a+b+c}{2}\)
\(\frac{22+120+122}{2}\)m
= 132
s – a = 132 – 22 = 110 m
s – b = 132 – 120 = 12 m
s – c = 132 – 122 = 10 m

= 4 × 3 × 10 × 11
= 1320 sq.m
Cost of levelling for one sq.m = Rs 20
Cost of levelling the ground = Rs 1320 × 20
= Rs 26400
Area of the ground = Rs 1320 sq.m
Cost of levelling the ground = Rs 26400

Question 3.
The perimeter of a triangular plot is 600 m. If the sides are in the ratio 5 : 12 : 13, then find the area of the plot.
Solution:
Let the side of the triangle a, b and c be 5x, 12x and 13x
Perimeter of a triangular plot = 600 m
5x + 12x + 13x = 600
30x = 600 ⇒ x = \(\frac{600}{30}\)
x = 20
a = 5x = 5 × 20 = 100 m
b = 12x = 12 × 20 = 240 m
c = 13x = 13 × 20 = 260 m
s = \(\frac{600}{2}\)
= 300 m
s – a = 300 – 100 = 200 m
s – b = 300 – 240 = 60 m
s – c = 300 – 260 = 40 m
Area of triangle

= 10³ × 3 × 2 × 2 m²
= 1000 × 12 m²
= 12000 m²
Area of the triangular Plot = 12000 sq.m

Question 4.
Find the area of an equilateral triangle whose perimeter is 180 cm.
Solution:
Perimeter of an equilateral triangle = 180 cm
3a = 180
a = \(\frac{180}{3}\)
= 60 m
Area of an equilateral triangle
= \(\frac{√3}{4}\) a² sq.unit
= \(\frac{√3}{4}\) × 60 × 60 sq.m
= √3 × 15 × 60 sq.m
= 1.732 × 15 × 60 sq.m
= 1558.8 sq.m
Area of an equilateral triangle = 1558.8 sq.m

Question 5.
An advertisement board is in the form of an isosceles triangle with perimeter 36 m and each of the equal sides are 13 m. Find the cost of painting it at Rs 17.50 per square metre.
Solution:
Equal sides of a triangle = 13m
Perimeter of an isosceles triangle = 36 m
Length of the third side = 36 – (13 + 13) m
= 36 – 26
= 10 m
Here a = 13m, b = 13m and c = 10m
s = \(\frac{a+b+c}{2}\)
= \(\frac{36}{2}\)
= 18 m
s – a = 18 – 13 = 5 m
s – b = 18 – 13 = 5 m
s – c = 18 – 10 = 8 m

= 2² × 3 × 5
= 60 sq.m
Cost of painting for one sq. m = Rs 17.50
Cost of painting for 60 sq. m = Rs 60 × 17.50
= Rs 1050

Question 6.
Find the area of the unshaded region.

Solution:
Since ABD is a right angle triangle
AB² = AD² + BD²
= 12² + 16²
= 144 + 256
= 400
AB = \(\sqrt{400}\)
= 20 cm
Area of the right angle triangle = \(\frac{1}{2}\) bh sq.unit
= \(\frac{1}{2}\) × 12 × 16 cm²
= 6 × 16 cm²
= 96 cm²
To find the Area of the triangle ABC
Here a = 42 cm, b = 34 cm, c = 20 cm
s = \(\frac{a+b+c}{2}\)
= \(\frac{42+34+20}{2}\) cm
= \(\frac{96}{2}\)
= 48 cm
s – a = 48 – 42 = 6 cm
s – b = 48 – 34 = 14 m
s – c = 48 – 20 = 28 m
Area of triangle

= 16 × 3 × 7 cm²
= 336 cm²
Area of the unshaded region = Area of the ΔABC – Area of the ΔABD
= (336 – 96) cm²
= 240 cm²

Question 7.
Find the area of a quadrilateral ABCD whose sides are AB = 13 cm, BC = 12 cm, CD = 9 cm, AD = 14 cm and diagonal BD = 15 cm.
Solution:
In the triangle ABD,

Let a = 15 cm, b = 14 cm c = 13 cm
s = \(\frac{a+b+c}{2}\)
= \(\frac{15+14+13}{2}\) cm
= \(\frac{42}{2}\)
= 21 cm
s – a = 21 – 15 = 6 cm
s – b = 21 – 14 = 7 cm
s – c = 21 – 13 = 8 cm
Area of ΔABD

= 2² × 3 × 7 3
= 84 cm²
In the ΔBCD,
Let a = 15 cm, b = 9 cm, c = 12 cm
s = \(\frac{a+b+c}{2}\)
= \(\frac{15+9+12}{2}\) cm
= \(\frac{36}{2}\)
= 18 cm
s – a = 18 – 15 = 3 cm
s – b = 18 – 9 = 9 cm
s – c = 18 – 12 = 6 cm
Area of the ΔBCD

= 2 × 3³
= 2 × 27 sq.cm
= 54 sq. cm
Area of the quadrilateral ABCD = Area of ΔABD + Area of ΔBCD
= (84 + 54) sq.cm
= 138 sq.cm

Question 8.
A park is in the shape of a quadrilateral. The sides of the park are 15 m, 20 m, 26 m and 17 m and the angle between the first two sides is a right angle. Find the area of the park.
Solution:

In the right angle triangle ABC (Given ⌊B= 90°)
AC² = AB² + BC²
= 15² + 20²
= 225 + 400
AC² = 625
AC = \(\sqrt{225}\)
= 25 m
Area of the right ΔABC = \(\frac{1}{2}\) × AB × BC
= \(\frac{1}{2}\) × 15 × 20 sq.m
= 150 sq.m
In the triangle ACD
a = 25 m b = 17 m, c = 26 m
s = \(\frac{a+b+c}{2}\)
= \(\frac{25+17+26}{2}\) cm
= \(\frac{62}{2}\)
= 34 m
s – a = 34 – 25 = 9 m
s – b = 34 – 17 = 17 m
s – c = 34 – 26 = 8 m
Area of the triangle ACD

4 × 3 × 17
= 204 sq.m
Area of the quadrilateral = Area of the ΔABC + Area of the ΔACD
= (150 + 204) sq.m
= 354 sq.m
Area of the quadrilateral = 354 sq.m

Question 9.
A land is in the shape of rhombus. The perimeter of the land is 160 m and one of the diagonal is 48 m. Find the area of the land.
Solution:

Perimeter of the rhombus = 160 m
4 × side = 160
Side of a rhombus = \(\frac{160}{4}\)
= 40 m
In ΔABC, a = 40 m, b = 40 m, c = 48 m
s = \(\frac{a+b+c}{2}\)
= \(\frac{40+40+48}{2}\) cm
= \(\frac{128}{2}\)
= 64 m
s – a = 64 – 40 = 24 m
s – b = 64 – 40 = 24 m
s – c = 64 – 48 = 16m
Area of the ΔABC = \(\sqrt{64×24×24×16}\)
= 8 × 24 × 4
= 768 sq.m
Since ABCD is a rhombus Area of two triangles are equal.
Area of the rhombus ABCD = (768 + 768) sq.m
= 1536 sq.m
∴ Area of the land = 1536 sq.m

Question 10.
The adjacent sides of a parallelogram measures 34 m, 20 m and the measure of the diagonal is 42 m. Find the area of parallelogram.
Solution:
Since ABCD is a parallelogram opposite sides are equal.

In the ΔABC
a = 20 m, b = 42 m and c = 34 m
s = \(\frac{a+b+c}{2}\)
= \(\frac{20+42+34}{2}\) cm
= \(\frac{96}{2}\)
= 48 m
s – a = 48 – 20 = 28 m
s – b = 48 – 42 = 6 m
s – c = 48 – 34 = 14 m
Area of the ΔABC

= 2⁴ × 3 × 7 sq.m
= 16 × 3 × 7 sq.m
= 336 sq.m
Since ABCD is a parallelogram
Area of ΔABC and Area of ΔACD are equal
Area of the parallelogram ABCD = (336 + 336) sq.m
= 672 sq.m
∴ Area of the parallelogram = 672 sq.m

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

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

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

Question 2.
In a quadrilateral ABCD, ∠A = 72° and ∠C is the supplementary of ∠A. The other two angles are 2x – 10 and x + 4. Find the value of x and the measure of all the angles.
Solution:

∠A = 72°
∠C = 180° – 12° (∠A and ∠C are supplementary)
= 108°
∠A + ∠B + ∠C + ∠D = 360° (Total angles of quadrilateral)
72° + 2x – 10 + 108° + x + 4 = 360°
3x + 174° = 360°
x = \(\frac{186°}{3}\)
= 62°
The value of x is 62°
∠B = 2x – 10
= 2(62°) – 10
= 124° – 10°
= 114°
∠D = x + 4
= 62° + 4
= 66°
The other angles are 114°, 62° and 66°.

Question 3.
ABCD is a rectangle whose diagonals AC and BD intersect at O. If ∠OAB = 46°, find ∠OBC.
Solution:
Since the diagonals of a rectangle AC and BD are equal and bisect each other

∴ OA = OB
∠OAB = ∠OBA = 46°
Each angle of a rectangle measures 90°
∠ABC = 90°
∠ABO + ∠OBC = 90°
46° + ∠OBC = 90°
∠OBC = 90°-46°
∴ ∠OBC = 44°

Question 4.
The lengths of the diagonals of a Rhombus are 12 cm and 16 cm. Find the side of the rhombus.
Solution:
Since the diagonals of a rhombus bisect each other at right angles

AO = \(\frac{1}{2}\)AC = \(\frac{1}{2}\) × 12 = 6 cm
BO = \(\frac{1}{2}\)BD = \(\frac{1}{2}\) × 16 = 8 cm
In the right triangle AOD
AD² = AO² + DO²
= 6² + 8²
= 36 + 64
= 100
∴ AD = \(\sqrt{100}\)
= 10
∴ AB = BC = CD = AD = 10 cm.

Question 5.
Show that the bisectors of angles of a parallelogram form a rectangle.
Solution:
Given: A parallelogram in which bisector of angle A, B, C, D intersect at P, Q, R, S to form a quadrilateral PQRS.
To prove: Quadrilateral PQRS is a rectangle.
Proof: Since ABCD is a parallelogram. Therefore, AB || DC.

Now, AB || DC, and transversal AD cuts them, so we have
∠A + ∠D = 180°
\(\frac{1}{2}\)∠A + \(\frac{1}{2}\)∠D = \(\frac{180°}{2}\)
∠DAS + ∠ADS = 90°
But in ΔASD, we have
∠ADS + ∠DAS + ∠ASD = 180°
90° + ∠ASD = 180°
∠ASD = 90°
∠RSP = ∠ASD (vertically opposite angle)
∠RSP = 90°
Similarly, we can prove that
∠SRQ = 90°, ∠RQP = 90° and ∠QPS = 90°
Thus, PQRS is a quadrilateral each of whose angle is 90°.
Hence, PQRS is a rectangle.

Question 6.
If a triangle and a parallelogram lie on the same base and between the same parallels then prove that the area of the triangle is equal to half of the area of parallelogram.
Solution:
Let ΔAPB and parallelogram ABCD lie on base AB and between parallels AB and PC.
To show area ΔAPB = \(\frac{1}{2}\) Area (ABCD)
Now, draw BQ || AP. Then ABQP is a parallelogram.
Now area ABQP = Area ABCD
(They are on same base AB and between same parallels AB and PC)

⇒ ΔPAB ≅ ΔBQP
Area PAB = Area BQP
= \(\frac{1}{2}\) Area ABQP
= \(\frac{1}{2}\) Area ABCD

Question 7.
Iron rods a, b, c, d, e, and f are making a design in a bridge as shown in the figure.

If a || b, c || d, e || f, find the marked angles between
(i) b and c
(ii) d and e
(iii) d and f
(iv) c and f
Solution:
(i) Angle between b and c = 30°
(vertically opposite angles)

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

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

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

Question 8.
In the given figure, ∠A = 64°, ∠ABC = 58°. If BO and CO are the bisectors of ∠ABC and ∠ACB respectively of ΔABC, find x° and y°.

Solution:
In the given ΔABC
∠A = 64° and ∠B = 58°
∠C = 180°- (64° + 58°)
= 180° – 122°
= 58°
Since OC is the bisector of ∠C
y = \(\frac{58°}{2}\)
= 29°
Given ΔOBC
∠OCB = \(\frac{58°}{2}\) = 29°
∠OCB = 29°
∴ ∠BOC = 180°- (29° + 29°)
x = 180° – 58°
x = 122°
∠x = 122° and ∠y = 29°.

Question 9.
In the given figure, if AB = 2, BC = 6, AE = 6, BF = 8, CE = 7 and CF = 7, compute the ratio of the area of quadrilateral ABDE to the area of ΔCDF.

Solution:
Given: AB = 2 cm, BC = 6 cm, AE = 6 cm, BF = 8 cm, CE = 7 cm and CF = 7 cm
Consider ΔAEC and ΔBCF.
In ΔAEC, AE = 6 cm, EC = 7 cm and AC = 8 cm (2 + 6 = 8)
In ΔBCF, BC = 6 cm, CF = 7 cm and BF = 8 cm
∴ ΔAEC s ΔBCF
∴ Area of ΔAEC = Area of ΔBCF (Two triangles are similar areas are equal)
Subtract area of ΔBDC on both sides we get,
Area of ΔAEC – Area of ΔBDC = Area of ΔBCF – Area of ΔBDC
Area of quadrilateral ABDE = Area of ΔCDF

Question 10.
In the given figure, ABCD is a rectangle and EFGH is a parallelogram. Using the measurements given in the figure, what is the length ”d” of the segment that is perpendicular to \(\overline { HE }\) and \(\overline { FG }\)?

Solution:
In the given figure ABCD is a rectangle and EFGH is a parallelogram.
In the right triangle AEH
HE = \(\sqrt{AH^{2} + AE^{2}}\)
= \(\sqrt{3^{2} + 4^{2}}\)
= \(\sqrt{9 + 16}\)
= \(\sqrt{25}\)
HE = 5
∴ GF = 5 (HE and Gf are opposite sides of a parallelogram)
In the right triangle
GC = \(\sqrt{GF^{2} – FC^{2}}\)
= \(\sqrt{5^{2} – 3^{2}}\)
= \(\sqrt{25 – 9}\)
= \(\sqrt{16}\)
∴ DG = 10 – 6 = 4
Area of ΔAEH + Area of ΔBEF + Area of ΔFCG + Area of ΔHDG
= \(\frac{1}{2}\) × 3 × 4 + \(\frac{1}{2}\) × 6 × 5 + \(\frac{1}{2}\) × 3 × 4 + \(\frac{1}{2}\) × 5 × 6
= (6 + 15 + 6 + 15)
= 42
∴ Area of 4 triangles = 42
Area of the parallelogram = Area of the rectangle ABCD – Area of 4 triangles.
= 10 × 8 – 42
= 80 – 42
= 38
b × h = 38
5 × d = 38
d = \(\frac{38}{5}\)
= 7\(\frac{3}{5}\)
Length of d = 7\(\frac{3}{5}\) or 7.6

Question 11.
In parallelogram ABCD of the accompanying diagram, line DP is drawn bisecting BC at N and meeting AB (extended) at P. From vertex C, line CQ is drawn bisecting side AD at M and meeting AB (extended) at Q. Lines DP and CQ meet at O. Show that the area of triangle QPO is \(\frac{9}{8}\) of the area of the parallelogram ABCD.

Solution:
Draw OX perpendicular to QP.
In ΔADP, MN = \(\frac{1}{2}\) AP,
In ΔBCQ, MN = \(\frac{1}{2}\) QB
So, AP = BQ (or) AB + BP = AB + QA
∴ PB = QA
∴ QA = AB = BP (or) QP = QA + AB + BP = 3 AB
Area of ΔOQP = \(\frac{1}{2}\) × QP × OX
= \(\frac{1}{2}\) × 3 AB × OX
= \(\frac{3}{2}\) × AB × OX
= \(\frac{3}{2}\) AB (OY + YX)
= \(\frac{3}{2}\) × AB × OY + \(\frac{3}{2}\) × AB × YX (AB = MN)
= \(\frac{3}{2}\) × MN × OY + \(\frac{3}{2}\) × AB × YX
= 3 Area ΔOMN + \(\frac{3}{2}\) + Area ΔBNM
= 3[\(\frac{1}{4}\) area of MNCD] + \(\frac{3}{2}\)[\(\frac{1}{2}\) area of ABCD]
= \(\frac{3}{4}\)[\(\frac{1}{2}\) area of ABCD] + \(\frac{3}{4}\)[area of ABCD]
= \(\frac{3}{8}\) area of ABCD + \(\frac{3}{4}\) area of ABCD
= area of ABCD [\(\frac{3}{8}\) + \(\frac{3}{4}\)]
= area of ABCD (\(\frac{3+6}{8}\))
= \(\frac{9}{8}\) area of ABCD.
Hence it is proved.

Students can download Maths Chapter 3 Algebra Ex 3.9 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 3.9

Question 1.
Determine the quadratic equations, whose sum and product of roots are
(i) -9, 20
Answer:
Sum of the roots = -9 and Product of the roots = 20
The Quadratic equation is
x² – (sum of the roots) x + product of the roots = 0
x² – (-9) x + 20 = 0 ⇒ x² + 9x + 20 = 0

(ii) \(\frac { 5 }{ 3 } \), 4
Answer:
Sum of the roots = \(\frac { 5 }{ 3 } \); Product of the roots = 4
The Quadratic equation is
x² – (sum of the roots) x + product of the roots = 0
x² – (\(\frac { 5 }{ 3 } \)) x + 4 = 0 ⇒ x² – \(\frac { 5 }{ 3 } \) x + 4 = 0
3x² – 5x + 12 = 0

(iii) \(\frac { -3 }{ 2 } \), -1
Answer:
Sum of the roots = \(\frac { -3 }{ 2 } \); Product of the roots = -1
The Quadratic equation is
x² – (sum of the roots) x + product of the roots = 0
x² – (-\(\frac { 3 }{ 2 } \)) x + (-1) = 0 ⇒ x² + \(\frac { 3 }{ 2 } \) x – 1 = 0
2x² + 3x – 2 = 0

(iv) – (2 – a)², (a + 5)²
Answer:
Sum of the roots = – (2 – a)²; Product of the roots = (a + 5)²
x² – (sum of the roots) x + product of the roots = 0
x² – [-(2 – a)²] x + (a + 5)² = 0
x² + (2 – a)² x + (a + 5)² = 0

Question 2.
Find the sum and product of the roots for each of the following quadratic equations
(i) x² + 3x – 28 = 0
(ii) x² + 3x = 0
(iii) 3 + \(\frac{1}{a}=\frac{10}{a^{2}}\)
(iv) 3y² – y – 4 = 0
Solution:
(i) x² – (-3)x + (-28) = 0.
Comparing this with x² – (α + β)x + αβ = 0.
(α + β) = Sum of the roots = -3
αβ = product of the roots = -28

(ii) x² + 3x = 0 = x² – (-3)x + 0 = 0
x² – (α + β)x + αβ = 0
Sum of the roots α + β = -3
Products of the roots αβ =0

(iii) 3 + \(\frac { 1 }{ a } \) = \(\frac{10}{a^{2}}\)
Answer:
Multiply by a²
3a² + a = 10
3a² + a – 10 = 0
Sum of the roots (α + β) = \(\frac { -1 }{ 3 } \)
Product of the roots (α β) = \(\frac { -10 }{ 3 } \)

(iv) 3y² – y – 4 = 0
Answer:
Sum of the roots (α + β) = \(\frac { -(-1) }{ 3 } \) = \(\frac { 1 }{ 3 } \)
Product of the roots (α β) = \(\frac { -4 }{ 3 } \)

Students can download Maths Chapter 3 Algebra Ex 3.8 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 3.8

Question 1.
Find the square root of the following polynomials by division method
(i) x⁴ – 12x³ + 42x² – 36x + 9
Answer:

(ii) 31 x² – 28x³ + 4x⁴ + 42x + 9
Answer:
Rearrange the order we get
4x⁴ – 28x³ + 37x² + 42x + 9

(iii) 16x⁴ + 8x² + 1
Answer:

(iv) 121 x⁴ – 198x³ – 183x² + 216x + 144
Answer:

Question 2.
Find the square root of the expression
\(\frac{x^{2}}{y^{2}}\) – \(\frac { 10x }{ y } \) + 27 – \(\frac { 10y }{ x } \) + \(\frac{y^{2}}{x^{2}}\)
Answer:

Question 3.
Find the values of a and b if the following polynomials are perfect squares.
(i) 4x⁴ – 12x³ + 37x² + bx + a
Answer:

Since it is a perfect square
b + 42 = 0
b = – 42
a – 49 = 0
a = 49
∴ The value of a = 49 and b = – 42

(ii) ax⁴ + bx³ + 361x² + 220x + 100
Answer:
Re-arrange the order we get
100 + 220x + 361x² + bx³ + ax⁴

Since it is a perfect square
b – 264 = 0
b = 264
a – 144 = 0
a = 144
∴ The value of a = 144 and b = 264

Question 4.
Find the values of m and n if the following expressions are perfect sqaures.
(i) \(\frac{1}{x^{4}}\) – \(\frac{6}{x^{3}}\) + \(\frac{13}{x^{2}}\) + \(\frac { m }{ x } \) + n
Answer:

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

(ii) x⁴ – 8x³ + mx² + nx + 16
Answer:

Since it is a perfect square
m – 16 – 8 = 0
m – 24 = 0
m = 24
n + 32 = 0
n = -32
∴ The value of m = 24 and n = -32

Students can download Maths Chapter 4 Geometry Ex 4.1 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 4 Geometry Ex 4.1

Question 1.
In the figure, AB is parallel to CD, find x.

Solution:
(i) Through T draw TE || AB.

∴ ∠BAT + ∠ATE = 180° (AB || TE)
140° + ∠ATE = 180°
∠ATE = 180°- 140° = 40°
Similarly ∠ETC + ∠TCD = 180° (TE || CD)
∠ETC+150° = 180°
∠ETC = 180°- 150° = 30°
x = ∠ATE + ∠ETC
= 40°+ 30° = 70°
x = 70°

(ii) Draw TE || AB.

∠ABT + ∠ETB = 180° (AB || TE)
48° + ∠ETB = 180°
∠ETB = 180° – 48° = 132°
Similarly ∠CDT + ∠DTE = 180°
24° + ∠DTE = 180°
∴ ∠DTE = 180° – 24°
= 156°
∴ ∠BTE + ∠ETD = 132° + 156°
= 288°
x = 288°

(iii) In the given figure AB || CD, AD is the transversal.
∠CDA = ∠BAD
= 53° (alternate angles are equal)
In ΔECD, ∠D = ∠A = 53° (Alternate angles are equal)
∠E + ∠C + ∠D = 180° (sum of the angles of a triangle)
x° + 38° + 53° = 180°
x° = 180°- 91°
= 89°
x = 89°

Question 2.
The angles of a triangle are in the ratio 1 : 2 : 3, find the measure of each angle of the triangle.
Solution:
The ratio of the angles of a triangle = 1 : 2 : 3.
Let the angles of a triangle be x, 2x and 3x.
x + 2x + 3x = 180° (Total angle of a triangle is 180°)
6x = 180°
x = \(\frac{180°}{6}\)
= 30°
x = 30°; 2x = 2 × 30° = 60°; 3x = 3 × 30° = 90°
Measures of the angles of a triangle = 30°, 60° and 90°.

Question 3.
Consider the given pairs of triangles and say whether each pair is that of congruent triangles. If the triangles are congruent, say ‘how’; if they are not congruent say ‘why’ and also say if a small modification would make them congruent:

(i) In ΔPQR and ΔABC
PQ = AB (Given)
RQ = BC (Given)
ΔABC is not congruent to ΔPQR.
If PR = AC then ΔABC ≅ ΔPQR

(ii) In ΔABD and ΔCDB
AB = CD (Given)
AD = BC (Given)
BD is common
By SSS congruency
ΔABD ≅ ΔCDB

(iii) In ΔPXY and ΔPXZ
PX is common.
XY = XZ (Given)
PY = PZ (Given)
By SSS congruency
ΔPXY ≅ ΔPXZ

(iv) In the given figure BD bisect AC
In ΔAOB and ΔOCD
OA = OC (Given)
∠AOB = ∠DOC (vertically opposite angles)
∠B = ∠D (Given)
By ASA congruency ΔAOB ≅ ΔOCD

(v) In the given figure AC and BD bisect each other at O.
∴ OA = OC (Given); OB = OD (Given)
∠AOB = ∠COD (vertically opposite angles)
By SAS congruency
ΔAOB ≅ ΔOCD

(vi) In the given figure
AB = AC (Given)
BM = MC (AM is the median of the ΔABC)
AM is common (By SSS congruency)
∴ ΔABM ≅ ΔACM

Question 4.
ΔABC and ΔDEF are two triangles in which AB = DF, ∠ACB = 70°, ∠ABC = 60°; ∠DEF = 70° and ∠EDF = 60°. Prove that the triangles are congruent.
Solution:

In ΔABC ∠B = 60° and ∠C = 70°
∴ ∠A = 180° – (60° + 70°)
= 180° – 130°
= 50°
In ΔDEF ∠E = 70° and ∠D = 60°
∠F = 180° – (70° + 60°)
= 180° – 130°
= 50°
∠A = ∠F = 50°
∠B = ∠D = 60°
∠C = ∠E = 70°
By AAA congruency
ΔABC ≅ ΔFDE
(or)
∠B = ∠D = 60°
∠C = ∠E = 70°
AB = FE
By ASA congruency
ΔABC ≅ ΔFDE

Question 5.
Find all the three angles of the ΔABC.

Solution:
∠A + ∠B = ∠ACD (An exterior angle of a triangle is sum of its interior opposite angles)
x + 35 + 2x – 5 = 4x – 15
3x + 30 = 4x – 15
30 + 15 = 4x – 3x
45° = x
∠A = x + 35°
= 45° + 35°
= 80°
∠B = 2x – 5
= 2(45°) – 5°
= 90° – 5°
= 85°
∠ACD = 4x – 15
= 4 (45°) – 15°
= 180° – 15°
= 165°
∠ACB = 180° – ∠ACD
= 180° – 165°
= 15°
∠A = 80°, ∠B = 85° and ∠C = 15°.