Category: Class 9

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 4 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 4 Geometry Additional Questions

I. Multiple Choice Questions.

Question 1.
The angle sum of a convex polygon with number of sides 7 is ………
(a) 900°
(b) 1080°
(c) 1444°
(d) 720°
Solution:
(a) 900°

Question 2.
What is the name of a regular polygon of six sides?
(a) Square
(b) Equilateral triangle
(c) Regular hexagon
(d) Regular octagon
Solution:
(c) Regular hexagon

Question 3.
One angle of a parallelogram is a right angle. The name of the quadrilateral is ………
(a) square
(b) rectangle
(c) rhombus
(d) kite
Solution:
(b) rectangle

Question 4.
If all the four sides of a parallelogram are equal and the adjacent angles are of 120° and 60°, then the name of the quadrilateral is ………
(a) rectangle
(b) square
(c) rhombus
(d) kite
Solution:
(c) rhombus

Question 5.
In a parallelogram ∠A : ∠B = 1 : 2. Then ∠A ………
(a) 30°
(b) 60°
(c) 45°
(d) 90°
Solution:
(b) 60°

Question 6.
Which of the following is a formula to find the sum of interior angles of a quadrilateral of w-sides?
(a) \(\frac{n}{2}\) × 180°
(b) (\(\frac{n+1}{2}\)) 180°
(c) (\(\frac{n-1}{2}\)) 180°
(d) (n – 2) 180°
Solution:
(d) (n – 2) 180°

Question 7.
Diagonal of which of the following quadrilaterals do not bisect it into two congruent triangles?
(a) rhombus
(b) trapezium
(c) square
(d) rectangle
Solution:
(b) trapezium

Question 8.
The point of concurrency of the medians of a triangle is known as ………
(a) circumcentre
(b) incentre
(c) orthocentre
(d) centroid
Solution:
(d) centroid

Question 9.
Orthocentre of a triangle is the point of concurrency of ………
(a) medians
(b) altitudes
(c) angle bisectors
(d) perpendicular bisectors of side
Solution:
(b) altitudes

Question 10.
ABCD is a parallelogram as shown. Find x and y.

(a) 1, 7
(b) 2, 6
(c) 3, 5
(d) 4, 4
Solution:
(c) 3, 5

Question 11.
A circle divides the plane into part ………
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(c) 3

Question 12.
The longest chord of a circle is a of the circle……….
(a) radius
(b) diameter
(c) chord
(d) secant
Solution:
(b) diameter

Question 13.
Opposite angles of a cyclic quadrilateral are ………
(a) supplementary
(b) complementary
(c) equal
(d) none of these
Solution:
(a) supplementary

Question 14.
The value of x from figure is if ‘O’ is the centre of the circle ………

(a) 20 cm
(b) 15 cm
(c) 12 cm
(d) 5 cm
Solution:
(d) 5 cm

Question 15.
If PQ = x and ‘O’ is the centre of the circle, then x= ………

(a) 7 cm
(b) 14 cm
(c) 8 cm
(d) 13 cm
Solution:
(b) 14 cm

Question 16.
In figure OM = ON = 8cm and AB = 30 cm, then CD = ………

(a) 15 cm
(b) 30 cm
(c) 40 cm
(d) 10 cm
Solution:
(b) 30 cm

Question 17.
O is the centre of a circle, ∠AOB = 100°. Then angle ∠ ACB = ………

(a) 80°
(b) 40°
(c) 50°
(d) 60°
Solution:
(c) 50°

Question 18.
In,a circle, with center O, ∠AOB = 20°, ∠BOC = 40°, arc BC = 4 Then length of arc AB will be ………

(a) 8 cm
(b) 6 cm
(c) 2 cm
(d) 1 cm
Solution:
(c) 2 cm

Question 19.
In the figure , OC = 3cm and radius of circle is 5 cm. Then AB = ………

(a) 4 cm
(b) 5 cm
(c) 6 cm
(d) 8 cm
Solution:
(d) 8 cm

Question 20.
O is the centre of the circle. The value of x in the given diagram is ………

(a) 100°
(b) 160°
(c) 200°
(d) 80°
Solution:
(d) 80°

II. Answer the following questions.

Question 1.
In the figure find x° and y°.
Solution:
∠ACD = ∠A + ∠B
(An exterior angle of a triangle is sum of its interior opposite angles)
120° = 50° + x°
x° = 120° – 50°
= 70°
In the triangle ABC

∠A + ∠B + ∠ACB = 180° (Sum of the angles of a Δ)
50° + x + ∠ACB = 180°
50° + 70° + ∠ACB = 180°
∠ACB = 180° – 120°
y = 60°
(OR)
∠ACD + ∠ACB = 180° (Angles of a linear pair)
∠ACB = 180° – 120°
= 60°
The value of x = 70° and y = 60°.

Question 2.
The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.
Solution:
Let the angles of the quadrilateral be 3x, 5x, 9x and 13x.
Sum of all the angles of quadrilateral = 360°.
3x + 5x + 9x + 13x = 360°
30x = 360°
x = \(\frac{360°}{30}\)
= 12°
3x = 3 × 12 = 36°
5x = 5 × 12 = 60°
9x = 9 × 12 = 108°
13x = 13 × 12 = 156°
The required angles of quadrilateral are 36°, 60°, 108° and 156°.

Question 3.
Diagonal AC of a parallelogram ABCD bisects ∠A. Show that

(i) it bisects ∠C also
(ii) ABCD is a rhombus.
Solution:
We have a parallelogram ABCD in which diagonals AC bisect ∠A.
∠DAC = ∠BAC
(i) To prove that AC bisects ∠C
∴ ABCD is a parallelogram

∴ AB || DC and AC is a transversal
∴ ∠1 = ∠3 (Alternate interior angle) …….(1)
Also BC || AD and AC is a transversal
∴ ∠2 = ∠4 (Alternate interior angle) …….(2)
But AC bisects ∠A
∴ ∠1 = ∠2 ……… (3)
From (1), (2) and (3) we get
∠3 = ∠4
∴ AC bisects ∠C.

(ii) To prove that ABCD is a rhombus.
In ΔABC, we have ∠1 = ∠4 [∴ ∠ 1 = ∠2 = ∠4]
∴ BC = AB (side opposite to equal angles are equal) ……..(4)
Similarly AD = DC …….. (5)
But ABCD is a parallelogram AB = DC (Opposite sides of a parallelogram) ………(6)
From (4), (5) and (6) we have AB = BC = CD = DA.
Thus ABCD is a rhombus.

Question 4.
ABCD is a parallelogram and AP and CQ are perpendiculars from vertex A and C on diagonal BD.
Show that

(i) ΔAPB ≅ ΔCQD
(ii) AP = CQ.
Solution:
(i) In ΔAPB and ΔCQD we have
∠APB = ∠CQD (90° each)
AB = CD (opposite sides of parallelogram ABCD)
∠ABP = ∠CDQ (AB || CD and AD is a transversal)
Using ASA congruency we have,
ΔAPB ≅ ΔCQD

(ii) Since ΔAPB ≅ ΔCQD
∴ Their corresponding parts are equal.
∴ AP = CQ.

Question 5.
ABCD is a rectangle and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

Solution:
In rectangle ABCD, P is the mid-point of AB.
Q is the mid-point of BC. R is the mid-point of CD.
S is the mid-point of DA. AC is the diagonal.
Now in ΔABC,
PQ = \(\frac{1}{2}\) AC and PQ || AC ……. (1)
Similarly in ΔACD,
SR = \(\frac{1}{2}\) AC and SR || AC ……. (2)
From (1) and (2) we get,
PQ = SR and PQ || SR
Similarly by joining BD, we have
PS = QR and PS || QR
i.e. Both pairs of opposite sides of quadrilateral PQRS are equal and parallel.
∴ PQRS is a parallelogram.

Question 6.
In the figure A, B and C are three points on a circle with centre O such that ∠BOC = 30° and ∠AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC.
Solution:

O is the centre of the circle. ∠AOB = 60° and ∠BOC = 30°
Sum of all the angles of quadrilateral = 360°.
∠AOB + ∠BOC = ∠AOC
∴ ∠AOC = 30° + 60°
= 90°
∠ADC = \(\frac{1}{2}\) × 90°
= 45°

Question 7.
In the given figure A, B, C and D are four points on a circle, AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC.
Solution:
In ΔCDE,

Exterior ∠BEC = Sum of interior opposite angle
130° = ∠EDC + ∠ECD
130° = ∠EDC + 20°
130° – 20° = ∠EDC
110° = ∠EDC
∴ ∠BAC = ∠BDC = 110° (Both the triangles are standing on the same base)
∠BAC = 110°

Question 8.
In the given figure KLMN is a cyclic quadrilateral. KD is the tangent at K. If ∠N is a diameter ∠NLK = 40° and ∠LNM = 50°. Find ∠MLN and ∠DKL.

Solution:
LN is a diameter
∠LMN = ∠LKN = 90° (Angle in a semi-circle)
∴ ∠MLN = 90° – 50°
= 40°
∠LNK = 90° – 40°
= 50°
∠DKL = ∠LKN = 50° (angles in the alternate segment)
∴ ∠DKL = 50°
∠MLN = 40° and ∠DKL = 50°

Question 9.
In the given figure ∠PQR = 100°, where P, Q and R are points on a circle with centre ‘O’. Find ∠OPR.

Solution:
∠PQR is the angle subtended by the chord PR in the minor segment.
reflex ∠POR = 2∠PQR
= 2 × 100°
= 200°
Now ∠POR + reflex ∠POR = 360°
∠POR + 200° = 360°
∠POR = 360° – 200°
= 160°
From the given diagram, POR is an isosceles triangle (∴ PO = OR = radii)
∴ ∠OPR = ∠ORP (angles opposite to equal sides)
In ΔOPR,
∠OPR + ∠ORP + ∠POR = 180°
∠OPR + ∠OPR + 160° = 180°
2∠OPR = 180° – 160°
2∠OPR = 20°
∠OPR = \(\frac{20°}{2}\)
= 10°
∴ ∠OPR = 10°

Question 10.
AB and CD are two parallel chords of a circle which are on either sides of the centre such that AB = 10 cm and CD = 24 cm. Find the radius if the distance between AB and CD is 17 cm.
Solution:

Given AB = 10 cm, CD = 24 cm and PQ = 17 cm.
PC = PD = \(\frac{24}{2}\)
= 12 cm
AQ = QB = \(\frac{10}{2}\)
= 5 cm
Let OP be x; OQ = (17 – x)
In the ΔOPC,
OC² = OP² + PC²
= x² + 12²
In the ΔOAQ,
OA² = AQ² + QO²
= 5² + (17 – x)²
= 25 + 289 + x² – 34x
= 314 + x² – 34x
But OA² = OC²
314 + x² – 34x = x² + 144
-34x = 144 – 314
-34x = -170
34x = 170
x = \(\frac{170}{34}\)
= \(\frac{10}{2}\)
= 5
We know,
OC² = x²+ 144
= 5² + 144
= 25 + 144
OC² = 169
But OC = \(\sqrt{169}\)
= 13
Radius of the circle = 13 cm
= x² + 144

Students can download Maths Chapter 4 Geometry Ex 4.7 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 4.7

Multiple Choice Questions

Question 1.
The exterior angle of a triangle is equal to the sum of two ……….
(a) Exterior angles
(b) Interior opposite angles
(c) Alternate angles
(d) Interior angles
Solution:
(b) Interior opposite angles

Question 2.
In the quadrilateral ABCD, AB = BC and AD = DC Measure of ∠BCD is …………..

(a) 150°
(b) 30°
(c) 105°
(d) 72°
Solution:
(c) 105°
Hint:
Join BD
∠DBC = 54°; ∠BDC = 21°
∴ ∠BCD = 180° – (54° + 21°)
= 180° – 75°
= 105°

Question 3.
ABCD is a square, diagonals AC and BD meet at O. The number of pairs of congruent triangles with vertex O are ……….

(a) 6
(b) 8
(c) 4
(d) 12
Solution:
(a) 6

Question 4.
In the given figure CE || DB then the value of x° is ………

(a) 45°
(b) 30°
(c) 75°
(d) 85°
∠B = 360° – (∠A + ∠D + ∠BCD)
= 360° – (110° + 75° + 60°)
= 360° – 245°
= 115°
∠DBC = 115° – 30°
= 85°
∴ ∠x = 85° (BD || CE) Alternate angles are equal

Question 5.
The correct statement out of the following is ……..

(a) ΔABC ≅ ΔDEF
(b) ΔABC ≅ ΔEDF
(c) ΔABC ≅ ΔFDE
(d) ΔABC ≅ ΔFED
Solution:
(d) ΔABC ≅ ΔFED

Question 6.
If the diagonal of a rhombus are equal, then the rhombus is a ……..
(a) Parallelogram but not a rectangle
(b) Rectangle but not a square
(c) Square
(d) Parallelogram but not a square
Solution:
(c) Square

Question 7.
If bisectors of ∠A and ∠B of a quadrilateral ABCD meet at O, then ∠AOB is ……..
(a) ∠C + ∠D
(b) \(\frac{1}{2}\) (∠C + ∠D)
(c) \(\frac{1}{2}\) ∠C + \(\frac{1}{3}\) ∠D
(d) \(\frac{1}{3}\) ∠C + \(\frac{1}{2}\) ∠D
Solution:
(b) \(\frac{1}{2}\) (∠C + ∠D)

Question 8.
The interior angle made by the side in a parallelogram is 90° then the parallelogram is a …….
(a) rhombus
(b) rectangle
(c) trapezium
(d) kite
Solution:
(b) rectangle
Hint:
Opposite sides are equal and each angle is 90° then it is a rectangle.

Question 9.
Which of the following statement is correct?
(a) Opposite angles of a parallelogram are not equal
(b) Adjacent angles of a parallelogram are complementary.
(c) Diagonals of a parallelogram are always equal.
(d) Both pairs of opposite sides of a parallelogram are always equal.
Solution:
(d) Both pairs of opposite sides of a parallelogram are always equal.

Question 10.
The angles of the triangle are 3x – 40, x + 20 and 2x – 10 then the value of x is ………
(a) 40°
(b) 35°
(c) 50°
(d) 45°
Solution:
(b) 35°
Hint:
3x – 40 + x + 20 + 2x – 10 = 180° (Sum of the angles of a triangle is 180°)
6x – 30 = 180°
6x = 180°+ 30°
x = \(\frac{210°}{6°}\)
= 35°

Question 11.
PQ and RS are two equal chords of a circle with centre O such that ∠POQ = 70°, then ∠ORS = ……….
(a) 60°
(b) 70°
(c) 55°
(d) 80°
Solution:
(c) 55°
Hint:

∠POQ = 70° (Vertically opposite Angle)
∠ORS and ∠OSR are equal. (OR = OS radius of the circle)
∠ORS + ∠OSR + ∠ROS = 180°
x° + x° + 70° = 180°
2x + 70° = 180°
2x = 180° – 70° = 110°
x = \(\frac{110°}{2}\) = 55°
∴ ∠ORS = 55°

Question 12.
A chord is at a distance of 15cm from the centre of the circle of radius 25cm. The length of the chord is ……….
(a) 25cm
(b) 20cm
(c) 40cm
(d) 18cm
Solution:
(c) 40cm
Hint:

In the right triangle OAC,
AC² = OA² – OC²
= 25² – 15²
= (25 + 15) (25 – 15)
= 40 × 10
AC² = 400
AC = \(\sqrt{400}\)
= 20
Length of the chord AB = 20 + 20 = 40 cm.

Question 13.
In the figure, O is the centre of the circle and ∠ACB = 40° then ∠AOB = ………

(a) 80°
(b) 85°
(c) 70°
(d) 65°
Solution:
(a) 80°
Hint:
∠AOB = 2∠ACB (The angle subtended by an arc of the circle at the centre is double the angle subtended at the remaining part of the circle.)
= 2 × 40° = 80°

Question 14.
In a cyclic quadrilaterals ABCD, ∠A = 4x, ∠C = 2x the value of x is ……..
(a) 30°
(b) 20°
(c) 15°
(d) 25
Solution:
(a) 30°
Hint:

∠A + ∠C = 180° (Sum of the opposite angle of cyclic quadrilateral is 180°)
4x + 2x = 180°
x = \(\frac{180°}{6}\)
= 30°

Question 15.
In the figure, O is the centre of a circle and diameter AB bisects the chord CD at a point E such that CE = ED = 8 cm and EB = 4cm. The radius of the circle is ………

(a) 8cm
(b) 4cm
(c) 6cm
(d) 10cm
Solution:
(i) 10cm
Hint:

Let the radius OD be x.
OE = OB – BE
= x – 4 (OB radius of the circle)
In the ΔOED,
OD² = OE² + ED²
x² = (x – 4)² + 8²
x² = x² + 16 – 8x + 64
8x = 80
x = \(\frac{80}{8}\)
= 10 cm
Radius of the circle = 10 cm.

Question 16.
In the figure, PQRS and PTVS are two cyclic quadrilaterals, If ∠QRS = 80°, then ∠TVS = ………

(a) 80°
(b) 100°
(c) 70°
(d) 90°
Solution:
(a) 80°
Hint:
∠SPQ = 180° – 80° (Opposite angles of a cyclic quadrilateral PQRS)
= 100°
In the cyclic quadrilateral PVTS,
∠V = 180° – 100° = 80° (Opposite angles of a cyclic quadrilateral)

Question 17.
If one angle of a cyclic quadrilateral is 75°, then the opposite angle is ………
(a) 100°
(b) 105°
(c) 85°
(d) 90°
Solution:
(b) 105°
Hint:
Opposite angles = 180° – 75°
= 105° (Sum of the opposite angle is 180°)

Question 18.
In the figure, ABCD is a cyclic quadrilateral in which DC produced to E and CF is drawn parallel to AB such that ∠ADC = 80° and ∠ECF = 20°, then BAD = ?

(a) 100°
(b) 20°
(c) 120°
(d) 110°
Solution:
(c) 120°
Hint:
∠BAD = ∠ABC + ∠ECF (By exterior angle property)
= 100° + 20°
= 120°

Question 19.
AD is a diameter of a circle and AB is a chord If AD = 30cm and AB = 24cm then the distance of AB from the centre of the circle is ………
(a) 10cm
(b) 9cm
(c) 8cm
(d) 6cm
Solution:
(b) 9cm
Hint:

In ΔAOC,
AO = 15 cm
AC = \(\frac{1}{2}\) AB
= \(\frac{1}{2}\) × 24
= 12 cm
In ΔAOC,
OC² = AO² – AC²
= 15² – 12²
= 225 – 144
= 81
OC = \(\sqrt{81}\)
= 9 cm

Question 20.
In the given figure, If OP = 17cm, PQ = 30cm and OS is perpendicular to PQ, then RS is ……….

(a) 10cm
(b) 6cm
(c) 7cm
(d) 9cm
Solution:
(d) 9cm
Hint:
In ΔOPR, OP = 17cm, PR = 30/2 = 15 cm.
OR² = OP² – PR²
= 17² – 15²
= (17 + 15) (17 – 15)
= 32 × 2
OR² = \(\sqrt{64}\)
= 8 cm
RS = OS – OR
= 17 – 8
= 9 cm

Students can download Maths Chapter 4 Geometry Ex 4.6 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 4 Geometry Ex 4.6

Question 1.
Draw a triangle ABC, where AB = 8 cm, BC = 6 cm and ∠B = 70° and locate its circumcentre and draw the circumcircle.
Solution:

Steps for construction:
Step 1: Draw the ΔABC with the given measures.
Step 2: Construct the perpendicular bisector of (AB and BC) any two sides and let them, meet at S which is the circumcenter.
Step 3: With S as centre and SA = SB = SC as radius draw the circumcircle to passes through A, B and C.
Circum radius = 4.3 cm.

Question 2.
Construct the right triangle PQR whose perpendicular sides are 4.5 cm and 6 cm. Also locate its circumcentre and draw circumcircle.
Solution:

Steps for construction:
Step 1: Draw the ΔPQR with the given measures.
Step 2: Construct the perpendicular bisector of (PQ and PR) any two sides and let them meet at S which is the circumcenter.
Step 3: With S as centre and SP = SQ = SR as radius draw the circumcircle to passes through P, Q and R.
Circum radius = 3.8 cm.

Question 3.
Construct ΔABC with AB = 5 cm ∠B = 100° and BC = 6 cm. Also locate its circumcentre draw circumcircle.
Solution:

Steps for construction:
Step 1: Draw the ΔABC with the given measures.
Step 2: Construct the perpendicular bisector of any two sides (AB and BC) and let them meet at S which is the circumcenter.
Step 3: With S as centre and SA = SB = SC as radius draw the circumcircle to passes through A, B and C.
Circum radius = 4.3 cm.

Question 4.
Construct an isosceles triangle PQR where PQ = PR and ∠Q = 50°, QR = 7cm. Also draw its circumcircle.
Solution:

Given PQ = PR
∴ ∠R = 50° (opposite angles are equal)
Steps for construction:
Step 1: Draw the ΔABC with the given measures.
Step 2: Construct the perpendicular bisector of any two sides (QR and PR) and let them meet at S. S is the circumcenter of ΔPQR.
Step 3: With S as centre SP = SQ = SR as radius. Draw the circumcircle.
Circum radius = 3.5 cm.

Question 5.
Draw an equilateral triangle of side 6.5 cm and locate its incentre. Also draw the incircle.
Solution:

Steps for construction:
Step 1: Draw the ΔABC with the each side measure 6.5 cm.
Step 2: Construct the angles bisectors of any two angles (A and B) and let them meet at I. Then I is the incentre of ΔABC.
Step 3: Draw perpendicular from I to any one of the side (AB) to meet AB at D.
Step 4: With I as centre and ID as radius draw the circle. This circle touches all the sides of the triangle internally.
Step 5: Measure of In-radius = 1.5 cm.

Question 6.
Draw a right triangle whose hypotenuse is 10 cm and one of the legs is 8 cm. Locate its incentre and also draw the incircle
Solution:

Steps for construction:
Step 1: Draw the ΔABC with AB = 8cm, BC = 10 cm and ∠A = 90°.
Step 2: Construct the angle bisectors of any two angles (∠B and ∠C) and let them meet at I. Then I is the incentre of ΔABC.
Step 3: Draw perpendicular from I to any one of the side to meet AB at D.
Step 4: With I as centre and ID as radius draw the circle. This circle touches all the sides of the triangle internally.
Step 5: Measure of In-radius = 1.8 cm.

Question 7.
Draw ΔABC given AB = 9 cm, ∠CAB = 115° and ∠ABC = 40°. Locate its incentre and also draw the incircle. (Note: You can check from the above examples that the incentre of any triangle is always in its interior).
Solution:

Step 1: Draw the ΔABC with AB = 9cm, ∠B = 40° and ∠A = 115°.
Step 2: Construct the angle bisectors of any two angles (A and B) and let them meet at I. Then I is the incentre of ΔABC.
Step 3: Draw perpendicular from I to any one of the side (AB) to meet AB at D.
Step 4: With I as centre and ID as radius draw the circle. This circle touches all the sides of the triangle internally.
Step 5: Measure of In-radius = 2.7 cm.

Question 8.
Construct ΔABC in which AB = BC = 6cm and ∠B = 80°. Locate its incentre and draw the incircle.
Solution:

Steps for construction:
Step 1: Draw the ΔABC with AB = 6 cm, ∠B = 80° and BC = 6 cm.
Step 2: Construct the angle bisectors of any two angles (A and B) and let them meet at I. Then I is the incentre of ΔABC.
Step 3: Draw perpendicular from I to any one of the side (AB) to meet AB at D.
Step 4: With I as centre and ID as radius draw the circle. This circle touches all the sides of the triangle internally.
Step 5: Measure of In-radius = 1.9 cm.

Students can download Maths Chapter 4 Geometry Ex 4.5 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 4 Geometry Ex 4.5

Question 1.
Construct the ΔLMN such that LM = 7.5 cm, MN = 5cm and LN = 8cm. Locate its centroid.
Solution:

Steps for construction:
Step 1: Draw the ΔLMN using the given measurement LM = 7.5 cm, MN = 5cm and LN = 8cm.
Step 2: Construct the perpendicular bisectors of any two sides LM and MN intersect LM at P and MN at Q respectively.
Step 3: Draw the median LQ and PN meet at G.
The point G is the centroid of the given ΔLMN.

Question 2.
Draw and locate the centroid of the triangle ABC where right angle at A, AB = 4cm and AC = 3 cm.
Solution:

Steps for construction:
Step 1: Draw the ΔABC using the given measurement AB = 4cm and AC = 3 cm and ZA = 90°.
Step 2: Construct the perpendicular bisectors of any two sides AB and AC to find the mid-points P and Q of AB and AC.
Step 3: Draw the medians PC and BQ intersect at G.
The point G is the centroid of the given ΔABC.

Question 3.
Draw the ΔABC, where AB = 6cm, ∠B = 110° and AC = 9cm and construct the centroid.
Solution:

Steps for construction:
Step 1: Draw the ΔABC using the given measurement AB = 6cm, AC = 9cm and ∠B =110°.
Step 2: Construct the perpendicular bisectors of any two sides AB and BC to find the mid-points P and Q of AB and BC.
Step 3: Draw the medians PC and AQ intersect at G.
The point G is the centroid of the given ΔABC.

Question 4.
Construct the ΔPQR such that PQ = 5cm, PR = 6cm and ∠QPR = 60° and locate its centroid.
Solution:

Steps for construction:
Step 1 : Draw ΔPQR using the given measurements PQ = 5cm, PR = 6cm and ∠P = 60°.
Step 2 : Construct the perpendicular bisectors of any two sides PQ and QR to find the mid-points of M and N respectively.
Step 3 : Draw the median PN and MR and let them meet at G.
The point G is the centroid of the given ΔPQR.

Question 5.
Draw ΔPQR with sides PQ = 7 cm, QR = 8 cm and PR = 5 cm and construct its Orthocentre.
Solution:

Steps for construction:
Step 1: Draw the ΔPQR with the given measurements.
Step 2: Construct altitudes from any two vertices P and Q to their opposite sides QR and PR respectively.
Step 3: The point of intersection of the altitude H is the orthocentre of the given ΔPQR.

Question 6.
Draw an equilateral triangle of sides 6.5 cm and locate its Orthocentre.
Solution:

Steps for construction:
Step 1: Draw the ΔABC with the given measurements.
Step 2: Construct altitudes from any two vertices A and C to their opposite sides BC and AB respectively.
Step 3: The point of intersection of the altitude H is the orthocentre of the given ΔABC.

Question 7.
Draw ΔABC, where AB = 6 cm, ∠B = 110° and BC = 5 cm and construct its Orthocentre.
Solution:

Steps for construction:
Step 1: Draw the ΔABC with the given measurements.
Step 2: Construct altitudes from any two vertices B and C to their opposite sides AC and BC respectively.
Step 3: The point of intersection of the altitude H is the orthocentre of the given ΔABC.

Question 8.
Draw and locate the Orthocentre of a right triangle PQR where PQ = 4.5 cm, QR = 6 cm and PR = 7.5 cm.
Solution:

Steps for construction:
Step 1: Draw the ΔPQR with the given measures.
Step 2: Construct altitude from any two vertices Q and R to their opposite side PR and PQ respectively.
Step 3: The point of intersection of the altitude H is the orthocentre of the given ΔPQR.

Students can download Maths Chapter 4 Geometry Ex 4.4 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 4 Geometry Ex 4.4

Question 1.
Find the value of x in the given figure.

Solution:
∠B = 180° – 120°
(Sum of the opposite angles of a quadrilateral are supplementary)
∠B = 60°
∠BCA = 90° (Angle in a semicircle)
∠BAC + ∠B + ∠BCA = 180°
x + 60° + 90° = 180°
x + 150° = 180°
x = 180° – 150°
= 30°
The value of x = 30°

Question 2.
In the given figure, AC is the diameter of the circle with centre O.

If ∠ADE = 30°; ∠DAC = 35° and ∠CAB = 40°.
Find (i) ∠ACD
(ii) ∠ACB
(iii) ∠DAE
Solution:
(i) ∠ADC = 90° (Angle in a semi-circle)
∠ADC + ∠ACD + ∠DAC = 180° (Sum of the angles of a triangle is 180°)
90° + ∠ACD + 35° = 180°
∠ACD = 180° – 125°
∠ACD = 55°

(ii) ∠ABC = 90° (Angle in a semi-circle)
∠ACB + ∠CBA + ∠BAC = 180° (Sum of the angles of a triangle is 180°)
∠ACB + 90 °+ 40° = 180°
∠ACB = 180° – 130°
= 50°

(iii) In the cyclic quadrilateral ACDE,
∠AED = 180° – 55°
= 125°
In ΔAED,
∠DAE + ∠AED + ∠EDA = 180° (Sum of the angles of a triangle is 180°)
∠DAE + 125° + 30° = 180°
∠DAE = 180°- 155°
= 25°
∴ ∠DAE = 25°

Question 3.
Find all the angles of the given cyclic quadrilateral ABCD in the figure.

Solution:
In a cyclic quadrilateral ABCD,
∠B + ∠D = 180° (Sum of the opposite angles of a cyclic quadrilateral is 180°)
6x – 4° + 7x + 2° = 180°
13x – 2° = 180°
13x = 182°
x = 182°
x = \(\frac{182°}{13}\)
x = 14°
∠B = 6x – 4°
= 6(14) – 4°
= 84 – 4
= 80°
∠D = 7x + 2°
7(14) + 2°
98 + 2
= 100°
180°
2y + 4° + 4y – 4° = (Sum of the opposite angles of a cyclic quadrilateral is 180°)
6y = 180°
y = \(\frac{180°}{6}\)
= 30°
∠A = 2y + 4°
= 2(30) + 4°
= 64°
∠C = 4y – 4°
= 4(30) – 4°
= 120° – 4°
= 116°
∴ ∠A = 64°, ∠B = 80°, ∠C = 116°, ∠D = 100°.

Question 4.
In the given figure, ABCD is a cyclic quadrilateral where diagonals intersect at P such that ∠DBC = 40° and ∠BAC = 60° find

(i) ∠CAD
(ii) ∠BCD
Solution:
(i) ∠CAD = ∠DBC = 40° [Angles at the circumference to the same segment]
(ii) ∠BAC – ∠BDC = 60° [Angles at the circumference to the same segment]
∠BCD + ∠BDC + ∠CBD = 180° (Sum of the three angles of A is 180°)
∠BCD + 60° + 40° = 180°
∠BCD = 180° – 100°
= 80°

Question 5.
In the given figure, AB and CD are the parallel chords of a circle with centre O. Such that AB = 8 cm and CD = 6 cm. If OM ⊥ AB and OL ⊥ CD distance between LM is 7cm. Find the radius of the circle?

Solution:
Let OM be x.
∴ OL = 7 – x
In the right ΔAOM,
OA² = AM² + OM²
= 4² + x²
OA² = 16 + x²
r² = 16 + x² ……… (1) [r is the radius]
In the right ΔOCL,
OC² = OL² + CL²
r² = (7 – x)² + 3²
= 49 + x² – 14x + 9
= 58 + x² – 14x …….. (2)
From (1) and (2) we get,
16 + x² = 58 + x² – 14x
14x = 58 – 16
14x = 42
x = \(\frac{42}{14}\)
x = 3 cm
r² = 16 + x²
= 16 + 9
= 25
∴ r = \(\sqrt{25}\)
= 5
∴ radius of the circle = 5 cm.

Question 6.
The arch of a bridge has dimensions as shown, where the arch measure 2 m at its highest point and its width is 6 cm. What is the radius of the circle that contains the arch?

Solution:
AD = \(\frac{6}{2}\)
= 3 m
In the right ΔADC,

AC² = AD² + DC²
= 32 + 22
= 9 + 4
= 13
AC = \(\sqrt{13}\)
= 3.6m
radius = 3.6 m

Question 7.
In figure ∠ABC = 120°, where A, B and C are points on the circle with centre O. Find ∠OAC?

Solution:
Reflex ∠AOC = 2∠ABC
= 20 × 120°
= 240°
∴ ∠AOC = 360° – 240°
= 120°
∠OCA + ∠OAC = 180° – 120°
= 60°
∴ ∠OAC = \(\frac{60}{2}\)
= 30° (Since OA = OC)

Question 8.
A school wants to conduct tree plantation programme. For this a teacher allotted a circle of radius 6m ground to nineth standard students for planting sapplings. Four students plant trees at the points A, B, C and D as shown in figure. Here AB = 8m, CD = 10m and AB ⊥ CD. If another student places a flower pot at the point P, the intersection of AB and CD, then find the distance from the centre to P.

Solution:
OA = OD = 6 m
AB = 8 m (chord)
CD = 10 m (chord)
In Δ AOM, OM = \(\sqrt{6² – 4²}\) (∴ OM bisects the chord and ⊥ to the chord)
= \(\sqrt{36 – 16}\)
= \(\sqrt{20}\)m
In Δ CON, ON = \(\sqrt{6² – 5²}\)
= \(\sqrt{36 – 25}\)
= \(\sqrt{11}\)m (ON bisects the chord and ⊥ to the chord)
ONPM is a rectangle with all the angles 90° and with length \(\sqrt{20}\) m, breadth \(\sqrt{11}\)l m.
We need to find OP which is the diagonal of the rectangle ONPM.
∴ OP = \(\sqrt{ON² + NP²}\) = \(\sqrt{(\sqrt{11})² + (\sqrt{20})²}\)
(∴ OM = NP, opposite sides of the rectangle)
= \(\sqrt{11 + 20}\)
= \(\sqrt{31}\)
= 5.56 m

Question 9.
In the given figure, ∠POQ = 100° and ∠PQR° = 30°, then find ∠RPO.

Solution:
∠PRQ = \(\frac{1}{2}\) ∠POQ (Angles at the circumference)
= \(\frac{1}{2}\) × 100°
= 50°
∠OPQ + ∠OQP = 180° – 100° (Total angles of A is 180°)
= 80°
∴ ∠OPQ = \(\frac{80}{2}\)
= 40° (Since OP = OQ, radius of the circle)
∠RPQ = 180°- (30 + 50)°
= 100°
∴ ∠RPO = ∠RPQ – ∠OPQ
= 100° – 40°
= 60°

Students can download Maths Chapter 4 Geometry Ex 4.3 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 4 Geometry Ex 4.3

Question 1.
The diameter of the circle is 52 cm and the length of one of its chord is 20 cm. Find the distance of the chord from the centre.
Radius of a circle = \(\frac{56}{2}\) = 26 cm
Solution:

Length of the chord = 20 cm
AC = \(\frac{20}{2}\)
= 10 cm
In ΔOAC, OC² = OA² – AC²
= 26² – 10²
= (26 + 10) (26 – 10)
= 36 × 16
OC = \(\sqrt{30×16}\)
= 6 × 4 cm
= 24 cm
Distance of the chord from the centre = 24 cm.

Question 2.
The chord of length 30 cm is drawn at the distance of 8 cm from the centre of the circle. Find the radius of the circle.
Solution:

Distance AC = \(\frac{1}{2}\) × Length of chord
= \(\frac{1}{2}\) × 30
= 15 cm
Distance from the centre = 8 cm
In ΔOAC Radius (OA) = \(\sqrt{AC² + OC²}\)
= \(\sqrt{15² + 8²}\)
= \(\sqrt{224 + 64}\)
= \(\sqrt{289}\)
= 17
Radius of the circle = 17 cm.

Question 3.
Find the length of the chord AC where AB and CD are the two diameters perpendicular to each other of a circle with radius 4√2 cm and also find ∠OAC and ∠OCA.
Solution:
Radius of a circle = 4√2 cm
In the right ΔAOC,

AC² = OA² + OC²
AC² = (4√2)² + (4√2)²
= 32 + 32 = 64
AC = \(\sqrt{64}\)
= 8
Length of the chord = 8 cm,
∠OAC = ∠OCA = 45°
Since OAC is an isosceles right angle triangle.

Question 4.
A chord is 12 cm away from the centre of the circle of radius 15cm. Find the length of the chord.
Solution:
Radius of a circle (OA) = 15 cm
Distance from centre to the chord (OC) = 12 cm
In the right ΔOAC,

AC² = OA² – OC²
= 15² – 12²
= 225 – 144
= 81
AC = \(\sqrt{81}\)
= 9
Length of the chord (AB) = AC + CB
= 9 + 9 = 18 cm.

Question 5.
In a circle, AB and CD are two parallel chords with centre O and radius 10 cm such that AB = 16 cm and CD = 12 cm determine the distance between the two chords?
Solution:
Length of the chord (AB) = 16 cm
∴ AF = \(\frac{1}{2}\) × 16
= 8 cm
Length of the chord (CD) = 12 cm
∴ CE = \(\frac{1}{2}\) × 12
= 6 cm

In the right ΔOCE,
OE² = OC² – CE²
= 10² – 6²
= 100 – 36
= 64
OE = \(\sqrt{64}\)
= 8 cm
In the right ΔOAF,
OF² = OA² – AF²
= 10² – 8²
= 100 – 64
= 36
OE = \(\sqrt{36}\)
= 6 cm
Distance between the two chords = OE + OF
= 8 + 6
= 14 cm.

Question 6.
Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
Ans

Two circules intersect at C and D
CD is the common chord.
CD = AC + AD
= 3 + 3
= 6 cm
Length of the common chord = 6 cm.

Question 7.
Find the value of x° in the following figures:

Solution:
(i) ∠BOC = 30° + 60°
= 90°
∠BAC (x) = \(\frac{1}{2}\) ∠BOC (By theorem)
= \(\frac{1}{2}\) × 90°
x = 45°

(ii) Join OP
∠QPR = \(\frac{80}{2}\) = 40° (Angle of the circumference is \(\frac{1}{2}\) the angle at the centre)
∠RPO = 30° (OP and OR are equal sides)
∠OPQ = 40° – 30°
= 10°
∠OQP = 10° (OQ and OP are equal sides)
∴ x = 10°

(iii) ∠OPN = ∠ONP (ON and OP are the radius of the circle)
= 90°
In ΔOPN
∠PON + ∠ONP + ∠NPO = 180° (Sum of the angles of a triangle)
70° + x° + x° = 180°
2x° + 70° = 180°
2x = 180° – 70°
2x = 110°
X = \(\frac{110°}{2}\)
= 55°
The value of x = 55°

(iv) Reflex ∠YOZ = 2∠YXZ
= 2(120°)
= 240°
∠YOZ = 360° – reflex ∠YOZ
= 360° – 240°
= 120°
∴ x = 120°

(v) ∠OAC + ∠OCA = 180° – 100°
= 80°
∠OAC = \(\frac{1}{2}\) × 80° [Since OA = OC, (∴ ∠OAC = ∠OCA)]
= 40°
∠OBA + ∠OAB = 180° – 140°
= 40° [Since ∠OBA = ∠OAB, since OB = OA]
∴∠OAB = \(\frac{40°}{2}\)
= 20°
∠BAC = ∠OAB + ∠OAC
= 20° + 40°
x = 60°

Question 8.
In the given figure, ∠CAB = 25°, find ∠BDC, ∠DBA and ∠COB.

Solution:
In ΔACP,
∠ACP = 180° – (25° + 90°)
= 180° – 115°
= 65°
∠CBA = ∠CAB = 25° [Both the angles are standing in the same base]
∠DBA = 65° [∠DBA and ∠BCA standing in the same base]
∠COB = 50°

Students can download Maths Chapter 3 Algebra Ex 3.12 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 3 Algebra Ex 3.12

Question 1.
Solve by the method of elimination
(i) 2x – y = 3; 3x + y = 7
Solution:
2x – y = 3 → (1)
3x + y = 7 → (2)
By adding (1) and (2)
5x + 0 = 10
x = \(\frac{10}{5}\)
x = 2
Substitute the value of x = 2 in (1)
2(2) – y = 3
4 – y = 3
-y = 3 – 4
-y = -1
y = 1
The value of x = 2 and y = 1

(ii) x – y = 5; 3x + 2y = 25
Solution:
x – y = 5 → (1)
3x + 2y = 25 → (2)
(1) × 2 ⇒ 2x – 2y = 10 → (3)
(2) × 1 ⇒ 3x + 2y = 25 → (2)
(3) + (2) ⇒ 5x + 0 = 35
x = \(\frac{35}{5}\)
= 7
Substitute the value of x = 7 in (1)
x – y = 5
7 – y = 5
-y = 5 – 7
-y = -2
y = 2
∴ The value of x = 7 and y = 2

(iii) \(\frac{x}{10}\) + \(\frac{y}{5}\) = 14; \(\frac{x}{8}\) + \(\frac{y}{6}\) = 15
Solution:
\(\frac{x}{10}\) + \(\frac{y}{5}\) = 14
LCM of 10 and 5 is 10
Multiply by 10
x + 2y = 140 → (1)
\(\frac{x}{8}\) + \(\frac{y}{6}\) = 15
LCM of 8 and 6 is 24
3x + 4y = 360 → (2)
(1) × 2 ⇒ 2x + 4y = 280 → (3)
(2) × 1 ⇒ 3x + 4y = 360 → (2)
(3) – (2) ⇒ -x + 0 = -80
∴ x = 80
Substitute the value of x = 80 in (1)
x + 2y = 140
80 + 2y = 140
2y = 140 – 80
2y = 60
y = \(\frac{60}{2}\)
y = 30
∴ The value of x = 80 and y = 30

(iv) 3(2x + y) = 7xy; 3(x + 3y) = 11xy
Solution:
3(2x + y) = 7xy
6x + 3y = 7xy
Divided by xy

3a + 6b = 7 → (1)
3(x + 3y) = 11xy
3x + 9y = 11xy
Divided by xy

9a + 3b = 11 → (2)
(1) × 3 ⇒ 9a + 18b = 21 → (3)
(2) × 1 ⇒ 9a + 3b = -11 → (2)
(3) – (2) ⇒ 15b = 10
b = \(\frac{10}{15}\) = \(\frac{2}{3}\)
Substitute the value of b = \(\frac{2}{3}\) in (1)
3a + 6 × \(\frac{2}{3}\) = 7
3a + 4 = 7
3a = 7 – 4
3a = 3
a = \(\frac{3}{3}\)
= 1
But \(\frac{1}{x}\) = a
\(\frac{1}{x}\) = 1
x = 1
But \(\frac{1}{y}\) = b
\(\frac{1}{y}\) = \(\frac{2}{3}\)
2y = 3
y = \(\frac{3}{2}\)
∴ The value of x = 1 and y = \(\frac{3}{2}\)

(v) \(\frac{4}{x}\) + 5y = 7; \(\frac{3}{x}\) + 4y = 5
Solution:
Let \(\frac{1}{x}\) = a
4a + 5y = 7 → (1)
3a + 4y = 5 → (2)
(1) × 4 ⇒ 16a + 20y = 28 →(3)
(2) × 5 ⇒ 15a + 20y = 25 → (4)
(3) – (4) ⇒ a + 0 = 3
a = 3
Substitute the value of a = 3 in (1)
4(3) + 5y = 7
5y = 7 – 12
5y = -5
5y = \(\frac{-5}{5}\) = -1
But \(\frac{1}{x}\) = a
\(\frac{1}{x}\) = 3
3x = 1 ⇒ x = \(\frac{1}{3}\)
The value of x = \(\frac{1}{3}\) and y = -1

(vi) 13x + 11y = 70; 11x + 13y = 74
Solution:
13x + 11y = 70 → (1)
11x + 13y = 74 → (2)
(1) + (2) ⇒ 24x + 24y = 144
x + y = 6 → (3) (Divided by 24)
(1) – (2) ⇒ 2x – 2y = -4
x – y = -2 → (4) (Divided by 2)
(4) + (3) ⇒ 2x = 4
x = \(\frac{4}{2}\)
= 2
Substitute the value x = 2 in (3)
2 + y = 6
y = 6 – 2
= 4
∴ The value of x = 2 and y = 4

Question 2.
The monthly income of A and B are in the ratio 3 : 4 and their monthly expenditures are in the ratio 5 : 7. If each saves Rs 5,000 per month, find the monthly income of each.
Solution:
Let the income of “A” be “x” and the income of “B” be “y”.
By the given first condition
x : y = 3 : 4
4x = 3y (Product of the extreme is equal to the product of the means)
4x – 3y = 0 → (1)
Expenditure of A = x – 5000
Expenditure of B = y – 5000
By the given second condition
x – 5000 : y – 5000 = 5 : 7
7(x – 5000) = 5(y – 5000)
7x – 35000 = 5y – 25000
7x – 5y = -25000 + 35000
7x – 5y = 10000 → (2)
(1) × 5 ⇒ 20x – 15y = 0 → (3)
(2) × 3 ⇒ 21x – 15y = 30000 → (4)
(3) – (4) ⇒ x + 0 = 30000
x = 30000
Substitute the value of x in (1)
4 (30000) – 3y = 0
120000 = 3y
y = \(\frac{120000}{3}\) = 40000
∴ Monthly income of A is Rs 30,000
Monthly income of B is Rs 40,000

Question 3.
Five years ago, a man was seven times as old as his son, while five year hence, the man will be four times as old as his son. Find their present age.
Solution:
Let the age of a man be “x” and the age of a son be “y”
5 years ago
Age of a man = x – 5 years
Age of his son = y – 5 years
By the given first condition
x – 5 = 7(y – 5)
x – 5 = 7y – 35
x – 7y = -35 + 5
x – 7y = -30 → (1)
Five years hence
Age of a man = x + 5 years
Age of his son = y + 5 years
By the given second condition
x + 5 = 4 (y + 5)
x + 5 = 4y + 20
x – 4y = 20 – 5
x – 4y = 15 → (2)
(1) – (2) ⇒ -3y = -45
3y = 45
y = \(\frac{45}{3}\)
= 15
Substitute the value of y = 15 in (1)
x – 7(15) = -30
x – 105 = -30
x = -30 + 105
= 75
Age of the man is 75 years
Age of his son is 15 years

Students can download Maths Chapter 3 Algebra Ex 3.11 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 3 Algebra Ex 3.11

Question 1.
Solve, using the method of substitution.
(i) 2x – 3y – 7; 5x + y = 9
Solution:
2x – 3y = 7 → (1)
5x + y = 9 → (2)
Equation (2) becomes
y = 9 – 5x
Substitute the value of y in (1)
2x – 3 (9 – 5x) = 7
2x – 27 + 15x = 7
17x = 7 + 27
17x = 34
x = \( \frac{34}{17}\)
= 2
Substitute the value of x = 2 (in) (2)
y = 9 – 5 (2) = 9 – 10 = -1
∴ The value of x = 2 and y = -1

(ii) 1.5x + 0.1y = 6.2; 3x – 0.4y = 11.2
Solution:
1.5x + 0.1y = 6.2
Multiply by 10
15x + y = 62
y = 62 – 15x → (1)
3x – 0.4y = 11.2
Multiply by 10
30x – 4y = 112
divided by (2) we get
15x – 2y = 56 → (2)
Substitute the value of y in (2)
15x – 2(62 – 15x) = 56
15x – 124 + 30x = 56
45x = 56 + 124
45x = 180
x = \( \frac{180}{45}\)
= 4
Substitute the value of x = 4 in (1)
y = 62 – 15(4)
= 62 – 60
y = 2
∴ The value of x = 4 and y = 2

(iii) 10% of x + 20% of y = 24; 3x – y = 20
Solution:
\( \frac{10}{100}\) × x + \( \frac{20}{100}\) × y = 24
\( \frac{x}{10}\) + \( \frac{y}{5}\) = 24
Multiply by 10
x + 2y = 240 → (1)
3x – y = 20
– y = 20 – 3x
y = 3x – 20 → (2)
Substitute the value of y in (1)
x + 2 (3x – 20) = 240
x + 6x – 40 = 240
7x – 40 = 240
7x = 240 + 40
7x = 280
x = \( \frac{280}{7}\)
x = 40
Substitute the value of x = 40 in (2)
y = 3 (40) – 20 = 120 – 20
y = 100
∴ The value of x = 40 and y = 100

(iv) √2x – √3y = 1; √3x – √8y = 0
Solution:
√2x – √3y = 1
– √3y = 1 – √2x
√3y = √2x – 1
y = \(\frac{√2x-1}{√3}\) → (1)
√3x – √8y = 0 → (2)
Substitute the value of y in (2)
\(\sqrt{3x}-\frac{√8(√2x-1)}{√3}\)
multiply by √3
⇒ \(\frac{3x-√8(√2x-1)}{√3}\)
3x – √8(√2x – 1) = 0
3x – 4x + √8 = 0
-x = √8
Substitute the value of x in (1)

The value of x = √8 and y = √3

Question 2.
Raman’s age is three times the sum of the ages of his two sons. After 5 years his age will be twice the sum of the ages of his two sons. Find the age of Raman.
Solution:
Let Raman’s age be “x” years and the sum of the ages of two sons be “y” years.
By the given first condition
x = 3y
x – 3y = 0 → (1)
After 5 years Raman’s age is x + 5 years
Sum of sons age is (y + 10) years
(each son age increases by 5 years)
By the given second condition
x + 5 = 2 (y + 10)
x + 5 = 2y + 20
x – 2y = 20 – 5
x – 2y = 15 → (2)
Equation (1) becomes
x = 3y
Substitute the value of x in (2)
3y – 2y = 15
y = 15
Substitute the value of y = 15 in x = 3y
x = 3(15)
x = 45
∴ Raman’s age = 45 years

Question 3.
The middle digit of a number between 100 and 1000 is zero and the sum of the other digit is 13. If the digits are reversed, the number so formed exceeds the original number by 495. Find the number.
Solution:
Let the unit digit be and the 100 is digit be X. The number is XOY (100x + y)
By the given first condition
x + y = 13 ….(1)
If the digits are reversed the number is 100 y + x.
By the given second condition.
100y + x = 100x + y + 495
-99x + 99y = 495
-x + y = 5 …. (2)
x + y = 13 ….(1)
Add (1) and (2)
2y = 18
y = 9
Substitute the value of y = 9 in (1)
x + 9 = 13
x = 13 – 9
x = 4
∴ The number is 409

Students can download Maths Chapter 3 Algebra Ex 3.10 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 3 Algebra Ex 3.10

Question 1.
Draw the graph for the following:
(i) y = 2x
Solution:
When x = -2, y = 2 (-2) = -4
When x = 0, y = 2 (0) = 0
When x = 2, y = 2 (2) = 4
When x = 3, y = 2 (3) = 6

Plot the points (-2, -4) (0, 0) (2, 4) and (3, 6) in the graph sheet we get a straight line.

(ii) y = 4x – 1
Solution:
When x = – 1; y = 4 (-1) -1 ⇒ y = -5
When x = 0; y = 4 (0) – 1 = 0 – 1 ⇒ y = -1
When x = 2; y = 4 (2) -1 = 8 – 1 ⇒ y = l

Plot the points (-1, -5) (0, -1) and (2, 7) in the graph sheet we get a straight line. At the time of printing change the direction.

(iii) y = (\(\frac{3}{2}\))x + 3
Solution:
When x = -2;
y = \(\frac{3}{2}\)(-2) + 3
y = -3 + 3 = 0
when x = 0;
y = \(\frac{3}{2}\)(0) + 3
y = 3
when x = 2;
y = \(\frac{3}{2}\)(2) + 3
y = 3 + 3
= 6


Plot the points (-2, 0) (0, 3) and (2, 6) in the graph sheet we get a straight line.

(iv) 3x + 2y = 14
Solution:
y = \(\frac{-3x+14}{2}\)
y = – \(\frac{3}{2}\)x + 7
when x = -2
y = –\(\frac{3}{2}\)(-2) + 7 = 10
when x = 0
y = –\(\frac{3}{2}\)(0) + 7 = 7
when x = 2
y = –\(\frac{3}{2}\)(2) + 7 = 4


Plot the points (-2, 10) (0, 7) and (2, 4) in the graph sheet we get a straight line.

Question 2.
Solve graphically (i) x + y = 7, x – y = 3
Solution:
x + y = 7
y = 7 – x

Plot the points (-2, 9), (0, 7) and (3, 4) in the graph sheet
x – y = 3
-y = -x + 3
y = x – 3

Plot the points (-2, -5), (0, -3) and (4, 1) in the same graph sheet.

The point of intersection is (5, 2) of lines (1) and (2).
The solution set is (5,2).

(ii) 3x + 2y = 4; 9x + 6y – 12 = 0
Solution:
2y = -3x + 4
y = \(\frac{-3x+4}{2}\)
= \(\frac{-3}{2}\)x + 2

Plot the points (-2, 5), (0, 2) and (2, -1) in the graph sheet
9x + 6y= 12 (÷3)
3x + 2y = 4
2y = \(\frac{-3x+4}{2}\)
= \(\frac{-3}{2}\)x + 2

Plot the points (-2, 5), (0, 2) and (2, -1) the same graph sheet

Here both the equations are identical, but in different form.
Their solution is same.
This equations have an infinite number of solution.

(iii) \(\frac{x}{2}\) + \(\frac{y}{4}\) = 1: \(\frac{x}{2}\) + \(\frac{y}{4}\) = 2
Solution:
\(\frac{x}{2}\) + \(\frac{y}{4}\) = 1
multiply by 4
2x + y = 4
y = -2x + 4

Plot the points (-3, 10), (-1, 6), (0, 4) and (2, 0) in the graph sheet
\(\frac{x}{2}\) + \(\frac{y}{4}\) = 2
multiply by 4
2x + y = 8
y = -2x + 8

Plot the points (-2, 12), (-1, 10), (0, 8) and (2, 4) in the same graph sheet.

The given two lines are parallel.
∴ They do not intersect a point.
∴ There is no solution.

(iv) x – y = 0; y + 3 = 0
Solution:
x – y = 0
-y = -x
y = x

Plot the points (-2, -2), (0, 0), (1, 1) and (3, 3) in the same graph sheet.
y + 3 = 0
y = -3

Plot the points (-2, -3), (0, -3), (1, -3) and (2, -3) in the same graph sheet.

The two lines l₁ and l₂ intersect at (-3, -3). The solution set is (-3, -3).

(v) y = 2x + 1; 3x – 6 = 0
Solution:
y = 2x + 1

Plot the points (-3, -5), (-1, -1), (0, 1) and (2, 5) in the graph sheet
3x – 6 = 0
y = -3x + 6

Plot the points (-2, 12), (-1, 9), (0, 6) and (2, 0) in the same graph sheet

The two lines l₁ and l₂ intersect at (1, 3).
∴ The solution set is (1, 3).

(vi) x = -3; y = 3
Solution:
x = -3

Plot the points (-3, -3), (-3, -2), (-3, 2) and (-3, 3) in the graph sheet
y = 3

Plot the points (-3, 3), (-1, 3), (0, 3) and (2, 3) in the same graph sheet

The two lines l₁ and l₂ intersect at (-3, 3)
∴ The solution set is (-3, 3)

Question 3.
Two cars are 100 miles apart. If they drive towards each other they will meet in 1 hour. If they drive in the same direction they will meet in 2 hours. Find their speed by using graphical method.
Solution:
Let the speed of the two cars be “x” and “y”.
By the given first condition

x+ y = 100 → (1)
(They travel in opposite direction)
By the given second condition.
\(\)\frac{100}{x-y}= 2 [time taken in 2 hours in the same direction]
2x – 2y = 100
x – y = 50 → (2)
x + y = 100
y = 100 – x

Plot the points (30, 70), (50, 50), (60, 40) and (70,30) in the graph sheet
x – y = 50
-y = -x + 50
y = x – 50

Plot the points (40, -10), (50, 0), (60, 10) and (70, 20) in the same graph sheet

The two cars intersect at (75, 25)
The speed of the first car 75 km/hr
The speed of the second car 25 km/hr