Category: Class 9

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 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 6 Trigonometry Ex 6.5 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 6 Trigonometry Ex 6.5

Question 1.
If sin 30° = x and cos 60° = y, then x² + y² is …….
(a) \(\frac{1}{2}\)
(b) 0
(c) sin 90°
(d) cos 90°
Solution:
(a) \(\frac{1}{2}\)
Hint:
sin 30° = x = \(\frac{1}{2}\)
cos 60° = y = \(\frac{1}{2}\)
x² + y²

Question 2.
If tan θ = cot 37°, then the value of θ is ………
(a) 37°
(b) 53°
(c) 90°
(d) 1°
Solution:
(b) 53°
Hint:
tan θ = cot 37°
= cot (90° – 53°)
= tan 53°
The value of θ is 53°

Question 3.
The value of tan 72°. tan 18° is ………
(a) 0
(b) 1
(c) 18°
(d) 72°
Solution:
(b) 1
Hint:
tan 72° . tan 18° = tan 72° . tan (90° – 72°)
= tan 72° . cot 72°
= tan 72° × \(\frac{1}{tan 72°}\)
= 1

Question 4.
The value of \(\frac{2 tan 30°}{1 – tan^{2} 30°}\) is equal to ………
(a) cos 60°
(b) sin 60°
(c) tan 60°
(d) sin 30°
Solution:
(c) tan 60°
Hint:

= √3 = tan 60°

Question 5.
If 2 sin 2θ = √3 then the value of θ is ………
(a) 90°
(b) 30°
(c) 45°
(d) 60°
Solution:
(b) 30°
Hint:
2 sin 2θ = √3 ⇒ sin 2θ = \(\frac{√3}{2}\)
sin 2θ = sin 60° ⇒ 2θ = 60°
θ = 30°

Question 6.
The value of 3 sin 70° sec 20° + 2 sin 39° sec 51° is ………
(a) 2
(b) 3
(c) 5
(d) 6
Solution:
(c) 5
Hint:
3 sin 70° sec 20° + 2 sin 39° sec 51°
= 3. sin 70° . sec (90° – 70°) + 2 sin 39° . sec (90° – 39°)
= 3. sin 70° . cosec 70° + 2 sin 39° . cosec 39°
= 3. sin70° × \(\frac{1}{sin 70°}\) + 2 sin 39° × \(\frac{1}{sin 39°}\)
= 3 + 2
= 5

Question 7.
The value of \(\frac{1-tan^{2}45°}{1+tan^{2}45°}\) is ……..
(a) 2
(b) 1
(c) 0
(d) \(\frac{1}{2}\)
Solution:
(c) 0
Hint:
\(\frac{1-tan^{2}45°}{1+tan^{2}45°}\) = \(\frac{1-1}{1+1}\)
= \(\frac{0}{2}\) = 0

Question 8.
The value of cosec (70° + θ) – sec (20° + θ) + tan (65° + θ) – cot (25° + θ) is ……..
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(a) 0
Hint:
cosec (70° + θ) – sec (20° – θ) + tan (65° + θ) – cot (25° – θ)
= sec [90° – (70° + θ)] – sec (20° – θ) + tan (65° + θ) – tan [90° – (25° – θ)]
= sec (20° – θ) – sec (20° – θ) + tan (65° + θ) – tan (65° + θ)
= 0

Question 9.
The value of tan 1° . tan 2° . tan 3°…. tan 89° is ………
(a) 0
(b) 1
(c) 2
(d) \(\frac{√3}{2}\)
Solution:
(b) 1
Hint:
tan 1° . tan 2° . tan 3° …….. tan 89°
= tan (90° – 89°). tan (90° – 88°) .tan (90° – 87°) …….. tan 45° . tan (89°)
= cot 89° . cot 88°. cot 87°. ……. tan 45° …….. tan 87°. tan 88°. tan 89°
= 1

Question 10.
Given that sin α = \(\frac{1}{2}\) and cos β = \(\frac{1}{2}\), then the value of α + β is ………
(a) 0°
(b) 90°
(c) 30°
(d) 60°
Solution:
Hint:
sin α = \(\frac{1}{2}\)
sin 30° = \(\frac{1}{2}\) ∴ α = 30°
cos β = \(\frac{1}{2}\) ∴ β = 60°
∴ α + β = 30° + 60°
= 90°

Students can download Maths Chapter 6 Trigonometry Ex 6.4 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 6 Trigonometry Ex 6.4

Question 1.
Find the value of the following:
(i) sin 49°
(ii) cos 74° 39′
(iii) tan 54° 26′
(iv) sin 21° 21′
(v) cos 33° 53′
(vi) tan 70° 17′
Solution:
(i) sin 49° = 0.7547

(ii) cos 74° 39′ = cos 74° 39′ + 3′
From the table we get,
cos 74° 36′ = 0.2656
Mean difference of 3 = 8
cos 74° 39′ = 0.2656 – 8
= 0.2648

(iii) tan 54° 26′ = tan 54° 24′
tan 54° 24′ = 1.3968
Mean difference of 2 = 17
tan 54° 26’ = 1.3968 + 17
= 1.3985

(iv) sin 21° 21′ = sin 21° 18’+ 3’
sin 21° 18’ = 0.3633
Mean difference of 3 = 8(+)
sin 21° 21′ = 0.3633 + 8
= 0.3641

(v) cos 33° 53′ = cos 33° 48’ + 5′
cos 33° 48′ = 0.8310
Mean difference of 5 = 8
cos 33° 53′ = 0.8310 – 8
= 0.8302

(vi) tan 70° 17′ = tan 70° 12’+ 5’
tan 70° 12′ = 2. 7776
Mean difference of 5 = 131
tan 70° 17′ = 2. 7776 + 131
= 2.7907

Question 2.
Find the value of θ if
(i) sin θ = 0.9975
(ii) cos θ = 0.6763
(iii) tan θ = 0.0720
(iv) cos θ = 0.0410
(v) tan θ = 7.5958
Solution:
(i) sin θ = 0.9975
From the table we get,
= 0.9974 + 0.0001
= 85° 54′ + 1′
= 85° 55′ (or) 85° 56′ (or) 85° 57

(ii) cos θ = 0.6763
= 0. 6769 – 0.0006′
= 47° 24′ + 3′
= 47° 27

(iii) tan θ = 0. 0720
= 0. 0717 + 0.0003
= 4° 6′ + 1′
= 4° 7′

(iv) cos θ = 0.0410
= 0.0419 – 0.0009
= 87° 36′ + 3′
= 87° 39′

(v) tan θ = 7. 5958
= 7. 5958 + 0 (from the natural table)
= 82° 30′ (tangent table)

Question 3.
Find the value of the following:
(i) sin 65° 39′ + cos 24° 57’ + tan 10° 10′
(ii) tan 70° 58′ + cos 15° 26′ – sin 84° 59′
Solution:
(i) sin 65° 39′ = 0.9111
cos 24° 57′ = 0.9066
tan 10° 10′ = 0.1793
sin 65° 39′ + cos 24° 57′ + tan 10° 10′
= 0.9111 + 0.9066 + 0.1793
= 1.9970
sin 65° 39′ + cos 24° 57′ + tan 10° 10′ = 1.9970

(ii) tan 70° 58′ = 2. 8982
cos 15° 26′ = 0. 9639
sin 84° 59′ = 0.9962
tan 70° 58′ + cos 15° 26′ – sin 84° 59′
= 2. 8982 + 0. 9639 – 0.9962
= 3.8621 – 0.9962
= 2.8659

Question 4.
Find the area of a right triangle whose hypotenuse is 10 cm and one of the acute angle is 24°24′.
Solution:
In the ΔABC,

sin 24° 24′ = \(\frac{AB}{AC}\)
0. 4131 = \(\frac{AB}{10}\)
∴ AB = 4.131 cm
In the ΔABC,
cos 24° 24′ = \(\frac{BC}{AC}\)
0.9107 = \(\frac{BC}{10}\)
∴ BC = 9.107 cm
Area of the right angle = \(\frac{1}{2}\) × BC × AB
= \(\frac{1}{2}\) × 9.107 × 4.131
= \(\frac{37.62}{2}\)
Area of the triangle = 18.81 cm²

Question 5.
Find the angle made by a ladder of length 5m with the ground, if one of its end is 4m away from the wall and the other end is on the wall.
Solution:
AC is the length of the ladder.

In the right ΔABC,
cos θ = \(\frac{BC}{AC}\)
cos θ = \(\frac{4}{5}\)
= 0.8
cos θ = 0.8000
θ = 36° 52′
Angle made by a ladder with the ground is 36° 52′

Question 6.
In the given figure, HT shows the height of a tree standing vertically. From a point P, the angle of elevation of the top of the tree (that is ∠P) measures 42° and the distance to the tree is 60 metres. Find the height of the tree.
Solution:
Let the height of the tree HT be “x”

In the ΔHTP,
tan 42° = \(\frac{HT}{PT}\)
0.9004 = \(\frac{x}{60}\)
x = 0.9004 × 60
= 54.024
The height of the tree = 54.02 m

Students can download Maths Chapter 6 Trigonometry Ex 6.3 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 6 Trigonometry Ex 6.3

Question 1.
Find the value of the following:

(iii) tan 15° tan 30° tan 45° tan 60° tan 75°

Solution:
(i) cos 45° = \(\frac{1}{√2}\)

= 1² + 1² – 2(\(\frac{1}{√2}\))²
= 1 + 1 – 2(\(\frac{1}{2}\))
= 2 – 1
= 1

(ii) cos 60° = \(\frac{1}{2}\)

= 1 + 1 + 1 – 8(\(\frac{1}{2}\))²
= 3 – 8 × \(\frac{1}{4}\)
= 3 – 2
= 1

(iii) tan 30° = \(\frac{1}{√3}\), tan 45° = 1, tan 60° = √3
tan 15° . tan 30°. tan 45° . tan 60°. tan 75° = tan 15° . \(\frac{1}{√3}\) . 1 . √3 tan 75°
= tan 15° × tan 75° × \(\frac{1}{√3}\) × 1 × √3
= tan(90° – 75°) × \(\frac{1}{cot 75°}\) × 1 [tan 90° – θ = cot θ]
= cot 75° × \(\frac{1}{cot 75°}\) × 1
= 1

= 1 + 1
= 2

Students can download Maths Chapter 6 Trigonometry Ex 6.2 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 6 Trigonometry Ex 6.2

Question 1.
Verify the following equalities:
(i) sin² 60° + cos² 60° = 1
Solution:
sin 60° = \(\frac{√3}{2}\); cos 60° = \(\frac{1}{2}\)
L.H.S = sin² 60° + cos² 60°

L.H.S = R.H.S
Hence it is proved.

(ii) 1 + tan² 30° = sec² 30°
Solution:
tan 30° = \(\frac{1}{√3}\); sec 30° = \(\frac{2}{√3}\)
L.H.S = 1 + tan² 30°

∴ L.H.S = R.H.S
Hence it is proved.

(iii) cos 90° = 1 – 2sin² 45° = 2cos² 45° – 1
Solution:
cos 90° = 0, sin 45° = \(\frac{1}{√2}\), cos 45° = \(\frac{1}{√2}\)
cos 90° = 0 ……. (1)
1 – 2 sin² 45° = 1 – 2 (\(\frac{1}{√2}\))²
= 1 – 2 × \(\frac{1}{2}\)
= 1 – 1 = 0 → (2)
2 cos² 45° – 1 = 2(\(\frac{1}{√2}\))² – 1
= \(\frac{2}{2}\) – 1
= \(\frac{2 – 2}{2}\) = 0 → (3)
From (1), (2) and (3) we get
cos 90° = 1 – 2 sin² 45° = 2 cos² 45° – 1
Hence it is proved.

(iv) sin 30° cos 60° + cos 30° sin 60° = sin 90°
Solution:
sin 30° = \(\frac{1}{2}\); cos 60° = \(\frac{1}{2}\); cos 30° = \(\frac{√3}{2}\); sin 60° = \(\frac{√3}{2}\); sin 90° = 1
L.H.S = sin 30° cos 60° + cos 30° sin 60°

= 1
R.H.S = sin 90° = 1
L.H.S = R.H.S
Hence it is proved.

Question 2.
Find the value of the following:
(i) \(\frac{tan 45°}{cosec 30°}\) + \(\frac{sec 60°}{cot 45°}\) – \(\frac{5 sin 90°}{2 cos 0°}\)
(ii) (sin 90° + cos 60° + cos 45°) × (sin 30° + cos 0° – cos 45°)
(iii) sin²30° – 2 cos³ 60° + 3 tan⁴ 45°
Solution:
(i) tan 45° = 1, cosec 30° = 2; sec 60° = 2; cot 45° = 1; tan 45°, sin 90° = 1; cos 0° = 1

= 0

(ii) sin 90° = 1; cos 60° = \(\frac{1}{2}\); cos 45° = \(\frac{1}{√2}\); sin 30° = \(\frac{1}{2}\); cos 0° = 1
(sin 90° + cos 60° + cos 45°) × (sin 30° + cos 0° – cos 45°)

= \(\frac{7}{4}\)

(iii) sin 30° = \(\frac{1}{2}\); cos 60° = \(\frac{1}{2}\); tan 45° = 1
sin² 30° – 2 cos³ 60° + 3 tan⁴ 45°

= 3

Question 3.
Verify cos 3 A = 4 cos³ A – 3 cos A, when A = 30°
Solution:
L.H.S = cos 3 A
= cos 3 (30°)
= cos 90°
= 0
R.H.S = 4 cos³ A – 3 cos A
= 4 cos³ 30° – 3 cos 30°

= 0
∴ L.H.S = R.H.S
Hence it is proved.

Question 4.
Find the value of 8 sin2x, cos 4x, sin 6x, when x = 15°.
Solution:
8 sin 2x cos 4x sin 6x = 8 sin 2 (15°) × cos 4 (15°) × sin (6 × 15°)
= 8 sin 30° × cos 60° × sin 90°
= 8 × \(\frac{1}{2}\) × \(\frac{1}{2}\) × 1
= 2

Students can download Maths Chapter 6 Trigonometry Ex 6.1 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 6 Trigonometry Ex 6.1

Question 1.
From the given figure, find all the trigonometric ratios of angle B.

Solution:

Question 2.
From the given figure, find the values of

(i) sin B
(ii) sec B
(iii) cot B
(v) tan C
(vi) cosec C
Solution:
In the right ΔABD,
AD² = AB² – BD²
= 13² – 5²
= 169 – 25
= 144
AD = \(\sqrt{144}\)
= 12
In the right ΔADC,
AC² = AD² + DC²
= 12² + 16²
= 144 + 256
= 400
AC = \(\sqrt{400}\)
= 20

Question 3.
If 2 cos θ = √3, then find all the trigonometric ratios of angle θ.
Solution:

2 cos θ = √3 ⇒ cos θ = \(\frac{√3}{2}\)
AB² = AC² – BC²
= 2² – (√3)² ⇒ = 4 – 3 = 1
AB = √1 = 1

Question 4.
If cos A =\(\frac{3}{5}\), then find the value of \(\frac{sin A-cos A}{2 tan A}\)
Solution:

cos A = \(\frac{3}{5}\)
ΔABC
BC² = AC² – AB²
= 5² – 3²
= 25 – 9
= 16
BC = \(\sqrt{16}\) = 4
sin A = \(\frac{4}{5}\); tan A = \(\frac{4}{3}\)

∴ The value of \(\frac{sin A-cos A}{2 tan A}\) = \(\frac{3}{40}\)

Question 5.
If cos A = \(\frac{2x}{1+x_{2}}\) then find the values of sin A and tan A in terms of x.
Solution:
cos A = \(\frac{2x}{1+x_{2}}\)
In the triangle ABC

BC² = AC² – AB²
= (1 + x²)² – (2x)²
= 1 + x⁴ + 2x² – 4x²
= x⁴ – 2x² + 1
= (x² – 1)² (or) (1 – x²)² (using (a – b)²)
BC = \(\sqrt{(x^{2}-1)^{2}}\) (or) \(\sqrt{(1-x^{2})^{2}}\)
BC = x² – 1
The value of sin A = \(\frac{BC}{AC}\) = \(\frac{x²-1}{x²+1}\)
tan A = \(\frac{BC}{AB}\) = \(\frac{x²-1}{2x}\)
and
BC = 1 – x²
The value of sin A = \(\frac{1-x²}{1+x²}\)
tan A = \(\frac{1-x²}{2x}\)

Question 6.
If sin θ = \(\frac{a}{\sqrt{a²+b²}}\) then show that b sin θ = a cos θ.
Solution:

sin θ = \(\frac{a}{\sqrt{a²+b²}}\)
In the triangle ΔABC
BC² = AC² – AB²

L. H. S = R. H. S

Question 7.
If 3 cot A = 2, then find the value of \(\frac{4sin A-3cos A}{2 sin A+3 Cos A}\)
Solution:

3 cot A = 2 ⇒ cot A = \(\frac{2}{3}\)
AC² = AB² + BC²
= 3² + 2²
= 9 + 4
AC = \(\sqrt{13}\)
cos A = \(\frac{AB}{AC}\) = \(\frac{3}{\sqrt{13}}\)
sin A = \(\frac{BC}{AC}\) = \(\frac{2}{\sqrt{13}}\)
The value of \(\frac{4sin A-3cos A}{2 sin A+3 Cos A}\)

The value is (\(\frac{-1}{13}\))

Question 8.
If cos θ : sin θ = 1 : 2, then find the value of \(\frac{8cos θ-2cos θ}{4 cos θ+2 sin θ}\)
Solution:
cos θ : sin θ = 1 : 2

Aliter:
cos θ : sin θ = 1 : 2
2 cos θ = sin θ ⇒ 2 = \(\frac{sin θ}{cos θ}\) ⇒ 2 = tan θ

∴ The value is \(\frac{1}{2}\)

Question 9.
From the given figure, prove that θ + ∅ = 90°. Also prove that there are two other right angled triangles. Find sin α, cos β and tan ∅.

Solution:
In the ΔABC,
AB = 9 + 16 = 25
AC = 15; BC = 20
AB² = 25²
= 625 ……. (1)
AC² + BC² = 15² + 20²
= 225 + 400
= 625 …….. (2)
From (1) and (2) we get
AB² = AC² + BC²
ABC is a right angle triangle at C (Pythagoras theorem)
∴ ∠C = 90°
θ + ∅ = 90°
Also ADC is a right angle triangle ∠ADC = 90° (Given)
BDC is also a right angle triangle ∠BDC = 90° (since ADB is a straight line sum of the two angle is 180°)
From the given diagram

Question 10.
A boy standing at a point O finds his kite flying at a point P with distance OP = 25 m. It is at a height of 5 m from the ground. When the thread is extended by 10 m from P, it reaches a point Q. What will be the height QN of the kite from the ground? (use trigonometric ratios).

Solution:
Let the angle O be “θ”
In ΔONQ

In ΔOMP
sin θ = \(\frac{PM}{OP}\) ⇒ sin θ = \(\frac{5}{25}\)
sin θ = \(\frac{1}{5}\) ……… (2)
From (1) and (2) we get
\(\frac{h}{35}\) = \(\frac{1}{5}\)
5 h = 35 ⇒ h = \(\frac{35}{5}\) = 7
The height of the kite from the ground is 7m.

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.6 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.6

Question 1.
If the y-coordinate of a point is zero, then the point always lies ……..
(a) in the I quadrant
(b) in the II quadrant
(c) on x-axis
(d) on y-axis
Solution:
(c) on x-axis

Question 2.
The points (-5, 2) and (2, -5) lie in the ……….
(a) same quadrant
(b) II and III quadrant respectively
(c) II and IV quadrant respectively
(d) IV and II quadrant respectively
Solution:
(c) II and IV quadrant respectively

Question 3.
On plotting the points O (0, 0), A (3, -4), B (3, 4) and C (0, 4) and joining OA, AB, BC and CO, which of the following figure is obtained?
(a) Square
(b) Rectangle
(c) Trapezium
(d) Rhombus
Solution:
(c) Trapezium

Question 4.
If P (-1, 1), Q ( 3, -4), R (1, -1), S (-2, -3) and T (- 4, 4) are plotted on a graph paper, then the points in the fourth quadrant are ………
(a) P and T
(b) Q and R
(c) only S
(d) P and Q
Solution:
(b) Q and R

Question 5.
The point whose ordinate is 4 and which lies on the v-axis is ……….
(a) (4, 0)
(b) (0, 4)
(c) (1, 4)
(d) (4, 2)
Solution:
(b) (0, 4)

Question 6.
The distance between the two points (2, 3) and (1, 4) is ……..
(a) 2
(b) \(\sqrt{56}\)
(c) \(\sqrt{10}\)
(d) √2
Solution:
(d) √2
Hint:
\(\sqrt{(1-2)^{2}+(4+3)^{2}}\)
= \(\sqrt{(-1)^{2}+1^{2}}\)
= \(\sqrt{1+1}\)
= √2

Question 7.
If the points A (2, 0), B (-6, 0), C (3, a – 3) lie on the x-axis then the value of a is ……..
(a) 0
(b) 2
(c) 3
(d) -6
Solution:
(c) 3
Hint:
a – 3 = 0 ⇒ a = 3

Question 8.
If (x + 2, 4) = (5, y – 2), then the coordinates (x, y) are ………
(a) (7, 12)
(b) (6, 3)
(c) (3, 6)
(d) (2, 1)
Solution:
(c) (3, 6)
Hint:
x + 2 = 5
∴ x = 5 – 2 = 3
and
4 = y – 2
4 + 2 = y
∴ y = 6

Question 9.
If Q₁, Q₂, Q₃, Q₄ are the quadrants in a Cartesian plane then Q₂ ∩ Q₃ is ……..
(a) Q₁ U Q₂
(b) Q₂ U Q₃
(C) Null set
(d) Negative x-axis
Solution:
(c) Null set

Question 10.
The distance between the point (5, -1 ) and the origin is ………
(a) \(\sqrt{24}\)
(b) \(\sqrt{37}\)
(c) \(\sqrt{26}\)
(d) \(\sqrt{17}\)
Solution:
(c) \(\sqrt{26}\)
Hint:

Question 11.
The coordinates of the point C dividing the line segment joining the points P (2, 4) and Q (5, 7) internally in the ratio 2 : 1 is ……..
(a) (\(\frac{7}{2}, \frac{11}{2}\))
(b) (3, 5)
(c) (4, 4)
(d) (4, 6)
Solution:
(d) (4, 6)
Hint:
A line divides internally in the ratio m : n
m = 2, n = 1
x₁ = 2, x₂ = 5
y₁ = 4, y₂ = 7

= (4, 6)

Question 12.
If p (\(\frac{a}{3}, \frac{b}{2}\)) is the mid-point of the line segment joining A (-4, 3) and B (-2, 4) then (a, b) is ………
(a) (-9, 7)
(b) (-3, \(\frac{7}{2}\))
(c) (9, -7)
(d) (3, –\(\frac{7}{2}\))
Solution:
(a) (-9, 7)
Hint:
Mid point of a line

\(\frac{a}{3}\) = -3 ⇒ a = -9
\(\frac{b}{2}\) = \(\frac{7}{2}\) ⇒ b = 7
(a, b) is (-9, 7)

Question 13.
In what ratio does the point Q (1, 6) divide the line segment joining the points P (2, 7) and R(-2, 3) ………
(a) 1 : 2
(6) 2 : 1
(c) 1 : 3
(d) 3 : 1
Solution:
(c) 1 : 3
Hint:
A line divides internally in the ratio m : n the point P =
(\(\frac{mx_{2}+nx_{1}}{m+n}\), \(\frac{my_{2}+ny_{1}}{m+n}\))

(1, 6) = (\(\frac{-2m+2n}{m+n}\), \(\frac{3m+7n}{m+n}\))
\(\frac{-2m+2n}{m+n}\) = 1
-2m + 2n = m + nx
-3m = n – 2n
3m = n
\(\frac{m}{n}\) = \(\frac{1}{3}\)
∴ m : n = 1 : 3
and
\(\frac{3m+7n}{m+n}\) = 6
3m + 7n = 6m + 6n
6m – 3n = 7n – 6n
3m = n
\(\frac{m}{n}\) = \(\frac{1}{3}\)

Question 14.
If the coordinates of one end of a diameter of a circle is (3, 4) and the coordinates of its centre is (-3, 2), then the coordinate of the other end of the diameter is ……..
(a) (0, -3)
(b) (0, 9)
(c) (3, 0)
(d) (-9, 0)
Solution:
(d) (-9, 0)
Hint:
Let the other end of the diameter be (a, b)

Mid point of a line =
(\(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\))
(-3, 2) = \(\frac{3+a}{2}, \frac{4+b}{2}\)
\(\frac{3+a}{2}\) = -3
3 + a = -6
a = -6 – 3 = -9
and
\(\frac{4+b}{2}\) = 2
4 + b = 4
b = 4 – 4 = 0
The other end is (-9, 0)

Question 15.
The ratio in which the x-axis divides the line segment joining the points A (a₁, b₁) and B (a₂, b₂) is ……..
(a) b₁ : b₂
(b) -b₁ : b₂
(c) a₁ : a₂
(d) -a₁ : a₂
Solution:
(b) -b₁ : b₂
Hint:
A line divides internally in the ratio m : n the point P is,
(\(\frac{mx_{2}+nx_{1}}{m+n}, \frac{my_{2}+ny_{1}}{m+n}\))

The point P is (a, 0) = (\(\frac{ma_{2}+na_{1}}{m+n}, \frac{mb_{2}+nb_{1}}{m+n}\))
∴ \(\frac{mb_{2}+nb_{1}}{m+n}\) = 0
mb₂ + nb₁ = 0 ⇒ mb₂ = -nb₁
\(\frac{m}{n}\) = \(\frac{b_{1}}{b_{2}}\)
∴ m : n = -b₁ : b₂

Question 16.
The ratio in which the x-axis divides the line segment joining the points (6, 4) and (1, -7) is ………
(a) 2 : 3
(b) 3 : 4
(c) 4 : 7
(d) 4 : 3
Solution:
(c) 4 : 7
Hint:
A line divides internally in the ratio m : n the point P

-7m + 4n = 0
4n = 7m
\(\frac{m}{n}\) = \(\frac{4}{7}\)
The ratio is 4 : 7

Question 17.
If the coordinates of the mid-points of the sides AB, BC and CA of a triangle are (3, 4), (1, 1) and (2, -3) respectively, then the vertices A and B of the triangle are ……….
(a) (3, 2), (2, 4)
(b) (4, 0), (2, 8)
(c) (3, 4), (2, 0)
(d) (4, 3), (2, 4)
Solution:
(b) (4, 0), (2, 8)
Hint:

x₁ + x₂ = 6 → (1)
y₁ + y₂ = 8 → (2)
x₂ + x₃ = 2 → (3)
y₂ + y₃ = 2 → (4)
x₁ + x₃ = 4 → (5)
y₁ + y₃ = -6 → (6)
Adding (1) + (3) + (5) we get,
2(x₁ + x₂ + x₃) = 3
x₁ + x₂ + x₃ = 6
x₁ + x₃ = 4
∴ x₂ = 6 – 4 = 2
x₂ + x₃ = 2
x₁ = 6 – 2 = 4
Adding (2) + (4) + (6) we get,
2 (y₁ + y₂ + y₃) = 4
y₁ + y₂ + y₃ = 2
y₂ + y₃ = 2
∴ y₁ = 2 – 2 = 0
y₁ + y₃ = -6
y₂ = 2 + 6 = 8
∴ The vertices A is (4, 0) and B is (2, 8).

Question 18.
The mid-point of the line joining (-a, 2b) and (-3a, -4b) is ……..
(a) (2a, 3b)
(b) (-2a, -b)
(c) (2a, b)
(d) (-2a, -3b)
Solution:
(b) (-2a, -b)
Mid points of line

= (-2a, -b)

Question 19.
In what ratio does the y-axis divides the line joining the points (-5, 1) and (2, 3) internally ………
(a) 1 : 3
(b) 2 : 5
(c) 3 : 1
(d) 5 : 2
Solution:
(d) 5 : 2
Hint:
When it cut the y-axis the point P is (0, a)
A line divides internally in the ratio m : n the point

2m – 5n = 0 ⇒ 2m = 5n
\(\frac{m}{n}\) = \(\frac{5}{2}\) ⇒ m : n = 5 : 2

Question 20.
If (1, -2), (3, 6), (x, 10) and (3, 2) are the vertices of the parallelogram taken in order, then the value of x is ………
(a) 6
(b) 5
(c) 4
(d) 3
Solution:
(b) 5
Hint:
Since ABCD is a parallelogram

Mid-point of AC = Mid-point of BD

\(\frac{1+x}{2}\) = \(\frac{6}{2}\) ⇒ 1 + x = 6 ⇒ x = 6 – 1 = 5
The value of x = 5

Students can download Maths Chapter 5 Coordinate Geometry 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 5 Coordinate Geometry Additional Questions

I. Multiple Choice Questions.

Question 1.
On which quadrant does the point (- 4, 3) lie?
(a) I
(b) II
(c) III
(d) IV
Solution:
(b) II

Question 2.
The point whose abscissa is 5 and lies on the x-axis is …….
(a) (-5, 0)
(b) (5, 5)
(c) (0, 5)
(d) (5, 0)
Solution:
(d) (5, 0)

Question 3.
A point which lies in the III quadrant is ……..
(a) (5, 4)
(b) (5, -4)
(c) (-5, -4)
(d) (-5, 4)
Solution:
(c) (-5, -4)

Question 4.
A point on the y-axis is ……..
(a) (1, 1)
(b) (6, 0)
(c) (0, 6)
(d) (-1, -1)
Solution:
(c) (0, 6)

Question 5.
The distance between the points (4, -1) and the origin is ……..
(a) \(\sqrt{24}\)
(b) \(\sqrt{37}\)
(c) \(\sqrt{26}\)
(d) \(\sqrt{17}\)
Solution:
(d) \(\sqrt{17}\)

Question 6.
The distance between the points (-1, 2) and (3, 2) is ……..
(a) \(\sqrt{14}\)
(b) \(\sqrt{15}\)
(c) 4
(d) 0
Solution:
(c) 4

Question 7.
The centre of a circle is (0, 0). One end point of a diameter is (5, -1), then the radius is …….
(a) \(\sqrt{24}\)
(b) \(\sqrt{37}\)
(c) \(\sqrt{26}\)
(d) \(\sqrt{17}\)
Solution:
(c) \(\sqrt{26}\)

Question 8.
The point (0, -3) lies on
(a) + ve x-axis
(b) + ve y-axis
(c) – ve x-axis
(d) – ve y-axis
Solution:
(d) – ve y-axis

Question 9.
The point which is on y-axis with ordinate -5 is ……..
(a) (0, -5)
(b) (-5, 0)
(c) (5, 0)
(d) (0, 5)
Solution:
(a) (0, -5)

Question 10.
The diagonal of a square formed by the points (1, 0), (0, 1), (-1, 0) and (0, -1) is …….
(a) 2
(b) 4
(c) √2
(d) 8
Solution:
(a) 2

Question 11.
The distance between the points (-2, 2) and (3, 2) is ……..
(a) 10 units
(b) 5 units
(c) 5√3 units
(d) 20 units
Solution:
(b) 5 units

Question 12.
The midpoint of the line joining the points (1, -1) and (-5, 3) is ……..
(a) (2, 1)
(b) (2, -1)
(c) (-2, -1)
(d) (-2, 1)
Solution:
(d) (-2, 1)

Question 13.
If the centroid of a triangle is at (1, 3) and two of its vertices are (-7, 6) and (8, 5) then the third vertex is ……..
(a) (-2, 2)
(b) (2, -2)
(c) (-2, -2)
(d) (2, 2)
Solution:
(b) (2, -2)

Question 14.
The ratio in which the X-axis divides the line segment joining the points (6, 4) and (1, -7) is ……..
(a) 1 : 2
(b) 2 : 3
(c) 4 : 7
(d) 7 : 4
Solution:
(c) 4 : 7

Question 15.
The centroid of a triangle (3, -5), (-7, 4) and (10, -2) is …….
(a) (2, -1)
(b) (2, 1)
(c) (-2, 1)
(d) (1, -2)
Solution:
(a) (2, -1)

II. Answer the Following Questions.

Question 1.
Show that the given points (1, 1), (5, 4), (-2, 5) are the vertices of an isosceles right angled triangle.
Solution:
Let A (1, 1), B (5, 4) and G (-2, 5)

AB = 5, AC = 5
∴ ABC is an isosceles triangle …….. (1)
BC² = AB² + AC²
50 = 25 + 25 ⇒ 50 = 50
∴ ∠A = 90° ……… (2)
From (1) and (2) we get ABC is an isosceles right angle triangle.

Question 2.
Show that the point (3, -2), (3, 2), (-1, 2) and (-1, -2) taken in order are the vertices of a square.
Solution:

= \(\sqrt{16}\)
= 4
AB = BC = CD = DA = 4. All the four sides are equal.
∴ ABCD is a Rhombus ……..(1)

Diagonal AC = Diagonal BD = \(\sqrt{32}\) ……..(2)
From (1) and (2) we get ABCD is a square.

Question 3.
Show that the point A (3, 7) B (6, 5) and C (15, -1) are collinear.
Solution:

AB + BC = AC ⇒ \(\sqrt{13}\) + 3\(\sqrt{13}\) = 4\(\sqrt{13}\)
∴ The points A, B, C are collinear.

Question 4.
Find the type of triangle formed by (-1, -1), (1, 1) and (-√, √3)
Solution:
Let the point A (-1, -1), B (1, 1) and C (-√3, √3)

AB = BC = AC = √8
∴ ABC is an equilateral triangle.

Question 5.
Find x such that PQ = QR where P(6, -1) Q(1, 3) and R(x, 8) respectively.
Solution:

But PQ = QR
\(\sqrt{(x-1)^{2}+25}\) = \(\sqrt{41}\)
Squaring on both sides
(x – 1)² + 25 = 41
(x – 1)² = 41 – 25 = 16
x – 1 = \(\sqrt{16}\) = ± 4
x – 1 = 4 (or) x – 1 = – 4
x = 5 (or) x = -4 + 1 = -3
The value of x = 5 or – 3

Question 6.
Find the coordinate of the point of trisection of the line segment joining (4, -1) and
Solution:
Let A (4, -1) and B (-2, -3) are the given points
Let P (a, b) and Q (c, d) be the points of trisection of AB.
∴ AP = PQ = QB

The required coordinate P is (2, –\(\frac{5}{3}\)) and Q is (0, –\(\frac{7}{3}\))

Question 7.
Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
Solution:
Given points are A(-3, 10), B(6, -8) and P(-1, 6)
divides AB internally in the ratio m : n
By section formula.
A line divides internally in the ratio m : n the point P =

∴ \(\frac{6m-3n}{m+n}\) = -1
6m – 3n = -m – n
6m + m = 3n – n
7m = 2n ⇒ \(\frac{m}{n}\) = \(\frac{2}{7}\)
∴ m : n = 2 : 7
and
\(\frac{-8m+10n}{m+n}\) = 6
-8m + 10n = 6m + 6n
-8m – 6m = 6n – 10n
14m = 4n
∴ \(\frac{m}{n}\) = \(\frac{14}{4}\) = \(\frac{2}{7}\)
Hence P divides AB internally in the ratio 2 : 7

Question 8.
If (1, 2) (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find “x” and “y”.
Solution:
Let A(1, 2), B(4, y), C(x, 6) and D(3, 5)

Since ABCD is a parallelogram the diagonal bisect each other
Mid point of AC = Mid point of BD
(\(\frac{1+x}{2}\), 4) = (\(\frac{7}{2}\), \(\frac{y+5}{2}\))
\(\frac{1+x}{2}\) = \(\frac{7}{2}\)
1 + x = 7
x = 7 – 1
= 6
and
\(\frac{y+5}{2}\) = 4
y + 5 = 8
y = 8 – 5
= 3
∴ The value of x = 6 and y = 3

Students can download Maths Chapter 9 Probability 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 9 Probability Additional Questions

I. Choose the Correct Answer

Question 1.
Which of the following cannot be taken as probability of an event?
(a) 0
(b) 0.35
(c) \(\frac{7}{20}\)
(d) –\(\frac{7}{20}\)
Solution:
(d) –\(\frac{7}{20}\)

Question 2.
A letter is chosen at random from the word “MATHEMATICS” the probability of getting a vowel is ……..
(a) \(\frac{2}{11}\)
(b) \(\frac{3}{11}\)
(c) \(\frac{4}{11}\)
(d) \(\frac{5}{11}\)
Solution:
(c) \(\frac{4}{11}\)

Question 3.
If P(A) =\(\frac{1}{3}\) then P(A)’ is ………
(a) \(\frac{1}{3}\)
(b) \(\frac{2}{3}\)
(c) \(\frac{3}{2}\)
(d) 1
Solution:
(b) \(\frac{2}{3}\)

Question 4.
An integer is chosen from the first twenty natural number, the probability that it is a prime number is ……..
(a) \(\frac{1}{5}\)
(b) \(\frac{2}{5}\)
(c) \(\frac{3}{5}\)
(d) \(\frac{4}{5}\)
Solution:
(b) \(\frac{2}{5}\)

Question 5.
From a well shuffled pack of 52 cards one card is drawn at random. The probability of getting not a king is ………
(a) \(\frac{12}{13}\)
(b) \(\frac{1}{13}\)
(c) \(\frac{4}{13}\)
(d) \(\frac{2}{13}\)
Solution:
(a) \(\frac{12}{13}\)

II. Answer the Following Questions

Question 6.
1500 families with 2 children were selected randomly and the following data were recorded.

Compute the probability of a family chosen at random having one girl.
Solution:
Total number of families = 1500
∴ n(S) = 1500
Let E be the event of getting one girl
n(E) = 814
P(E) = \(\frac{n(E)}{n(S)}\)
= \(\frac{814}{1500}\)
= \(\frac{407}{750}\)

Question 7.
The record of weather station shows that out of the part 250 consecutive days its whether forecast were correct 175 time. What is the probability that it was not correct on a given data?
Solution:
n(S) = 250
Let E be the event of getting whether forecast were correct
n(E) = 175
P(E) = \(\frac{n(E)}{n(S)}\)
= \(\frac{175}{250}\)
= \(\frac{7}{10}\)
Forecast was not correct on a given day = 1 – P(E)
= 1 – \(\frac{7}{10}\)
= \(\frac{10-7}{10}\)
= \(\frac{3}{10}\)

Question 8.
If A coin is tossed 200 times and is found that a tail comes up for 120 times. Find the probability of getting a tail.
Solution:
Number of trials = 200
n(S) = 200
Let E be the event of getting a tail
n(E) = 120
p(E) = \(\frac{n(E)}{n(S)}\)
= \(\frac{120}{200}\)
= \(\frac{12}{20}\)
= \(\frac{3}{5}\)

Question 9.
A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball from the bag is thrice that of drawing a red ball, then find the number of blue balls in the bag.
Solution:
Let the number of blue balls be “x”
Total number of balls = 5 + x
∴ n(S) = 5 + x
Let B be the event of drawing a blue ball and R be the event of drawing a red ball
Given P(B) = 3P(R)

∴ x = 15
Number of blue balls = 15

Question 10.
Find the probability that a non leap year selected at random will have 53 fridays.
Solution:
No. of days in a non leap year = 365 days
This year contain 52 weeks and one day
Sample space = {Sun, Mon, Tue, Wed, Thu, Fri, Sat}
n(S) = 7
Let A be the event of getting a friday
n(A) = 1
P(A) = \(\frac{n(A)}{n(S)}\)
= \(\frac{1}{7}\)