KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.3

KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.3 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 5 Triangles Exercise 5.3.

Karnataka Board Class 9 Maths Chapter 5 Triangles Ex 5.3

Question 1.
∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at P, show that
KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.3 1
(i) ∆ABD ≅ ∆ACD
(ii) ∆ABP ≅ ∆ACP
(iii) AP bisects ∠A as well as ∠D.
(iv) AP is the perpendicular bisector of BC.
Solution:
Data : ∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. AD is extended to intersect BC at P.
To Prove:
(i) ∆ABD ≅ ∆ACD
(ii) ∆ABP ≅ ∆ACP
(iii) AP bisects ∠A as well as ∠D.
(iv) AP is the perpendicular bisector of BC
(v) AD is the angular bisector of ∠A.
Proof:
(i) In ∆ABD and ∆ACD,
AB = AC (data)
BD = DC (data)
AD is common.
S.S.S. Congruence rule.
∴ ∆ABD ≅ ∆ACD

(ii) In ∆ABP and ∆ACP,
AB = AC (data)
∠ABP = ∠ACP (Opposite angles)
∠BAP = ∠CAP (∵ ∆ABD ≅ ∆ACD proved)
Now ASA postulate.
∆ABP ≅ ∆ACP.

(iii) ∆BAD ≅ ∆CAD proved.
AP bisects ∠A.
In ∆BDP and ∆CDP,
BD = DC (data)
BP = PC (proved)
DP is common.
∴ ∆BDP ≅ ∆CDP (SSS postulate)
∴ ∠BDP = ∠CDP
∴ DP bisects ∠D.
∴ AP bisects ∠D.

(iv) Now, ∠APB + ∠APC = 180° (Linear pair)
∠APB + ∠APB = 180°
2 ∠APB = 180
∴ ∠APB = \(\frac{180}{2}\)
∴∠APB = 90°
∠APB = ∠APC = 90°
BP = PC (proved)
∴ AP is the perpendicular bisector BC.

(v) AP is the angular bisector of ∠A.
Angular bisector of ∠A is aD, because AD, AP is in one line.

Question 2.
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that A
KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.3 2
(i) AD bisects BC
(ii) AD bisects ∠A.
Solution:
Data: AD is an altitude of an isosceles triangle ABC in which AB = AC.
To Prove:
(i) AD bisects BC.
(ii) AD bisects ∠A.
Proof: i) In ∆ABD and ∆ACD,
∠ADB = ∠ADC (∵ AD ⊥ BC)
AB = AC (data)
AD is common.
∴ ∆ABD ≅ ∆ACD
∴ BD = DC
∴ AD bisects BC.

(ii) ∠BAD = ∠CAD (∵ ∆ADB ≅ ∆ADC)
∴ AD bisects ∠A.

Question 3.
Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ∆PQR. Show that :
KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.3 3
(i) ∆ABM ≅ ∆PQN
(ii) ∆ABC ≅ ∆PQR.
Solution:
Data: Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ∆PQR.
To Prove:
(i) ∆ABM ≅ ∆PQN
(ii) ∆ABC ≅ ∆PQR.
Proof: (i) In ∆ABC,
AM is the median drawn to BC.
∴ BM = \(\frac{1}{2} \) BC
Similarly, in ∆PQR,
QN = \(\frac{1}{2}\) QR
But, BC = QR
\(\frac{1}{2} \) BC = \(\frac{1}{2}\) QR
∴ BM = QN
In ∆ABM and ∆PQN,
AB = PQ (data)
BM = QN (data)
AM = PN (proved)
∴ ∆ABM ≅ ∆PQN (SSS postulate)

(ii) In ∆ABC and ∆PQR,
AB = PQ (data)
∠ABC = ∠PQR (proved)
BC = QR (data)
∴ ∆ABC ≅ ∆PQR (SSS postulate)

Question 4.
BE and CF are two equal altitudes of a triangle ABC. Using the RHS congruence rule, prove that the triangle ABC is isosceles.
KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.3 4
Solution:
Data: BE and CF are two equal altitudes of a triangle ABC.
To Prove: ABC is an isosceles triangle.
Proof : BE = CF (data)
In ∆BCF and ∆CBE,
∠BFC = ∠CEB = 90° (data)
BC is a common hypotenuse.
As per Right angle, hypotenuse, side postulate,
∴ ∆BCF ≅ ∆CBE
∴ ∠CBF = ∠BCE
∴ ∠CBA = ∠BCA
∴ AB = AC
∴ ∆ABC is an isosceles triangle.

Question 5.
ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.
KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.3 5
Solution:
Data: ABC is an isosceles triangle with AB = AC.
To Prove : ∠B = ∠C
Construction: Draw AP ⊥ BC.
Proof: In ∆ABC, AP ⊥ BC and AB = BC.
∴ In ∆ABP and ∆ACP
∠APB = ∠APC = 90° ( ∵ AP ⊥ BC)
Hypotenuse AB = Hypotenuse AC
AP is common.
As per RHS Postulate,
∆ABP ≅ ∆ACP
∴ ∠ABP = ∠ACP
∴ ∠ABC = ∠ACB
∴∠B = ∠C.

We hope the KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.3 helps you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 5 Triangles Exercise 5.3, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.5

Karnataka Board Class 9 Maths Chapter 12 Circles Ex 12.5

Question 1.
In Fig., 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.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.5 1
Solution:
∠AOB + ∠BOC = 60° + 30° = 90°
∴ ∠AOC = 90° (Angle subtended at the centre)
∠ADC = ?
The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
∴ Angle subtended at the centre
∠AOC= 2 × ∠ADC
90° = 2 × ∠ADC 90
∴ ∠ADC = \(\frac{90}{2}\) = 45°
∴ ∠ADC = 45°.

Question 2.
A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.5 2
Solution:
(i) Angle subtended in the circumference, ∠BAD =?
(ii) Angle subtended in the circumference, ∠BCD =?
In this figure BD is a chord, OB is the radius, it is equal to OD.
∴ OB = OD = BD
∴ ∆OBD is equilateral triangle.
∴ each angle is equal to 60°.
∴ angle subtended at the centre ∠BOD = 60°.
(i) Angle subtended in the circumference
∠BAD= \(\frac{1}{2}\) × angle subtended at centre ∠BOD
= \(\frac{1}{2}\) × 60°
∴ ∠BAD = 30°.

(ii) The sum of either pair of opposite angles of a cyclic quadrilateral is 180°.
∴ In cyclic quadrilateral ABCD,
∠BAD + ∠ACD = 180
30 + ∠ACD = 180 (∵ ∠BAD = 30°)
∴ ∠ADC = 180 – 30
∴ ∠ACD = 150°.

Question 3.
In Fig., ∠PQR = 100°. where P, Q and R are points on a circle with centre O. Find ∠OPR.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.5 3
Solution:
If ∠PQR = 100°, then ∠OPR = ?
∠PQR = 100°.
∴ ∠POR = ?
∠POR = 2 × ∠PQR = 2 × 100°
∠POR = 200°
∴ ∠POR – Reflex angle ∠POR = 360°
∴ ∠POR = 360 – 200
∴ ∠POR = 160°
In ∆POR, OP = OR radii.
∴ ∠OPR = ∠ORB
∴ ∠OPR + ∠ORP + ∠POR = 180°
∠OPR + ∠OPR + 160 = 180
2∠OPR+ 160 = 180
2∠OPR = 180 – 160
2∠OPR = 20
∠OPR = \(\frac{20}{2}\)
∴ ∠OPR= 10°.

Question 4.
In Fig., ∠ABC = 69°, ∠ACB = 31°, find ∠BDC.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.5 4
Solution:
If ∠ABC = 69°, ∠ACB = 31°, then ∠BDC = ?
In ∆ABC,
∠ABC = ∠ACB + ∠BAC = 180°
69 + 31 + ∠BAC = 180
100 + ∠BAC = 180
∠BAC = 180 – 100
∠BAC = 80°
∠BAC and ∠BDC are angles in same segment. These are equal.
∠BDC = ∠BAC = 80°
∴ ∠BDC = 80°.

Question 5.
In Fig., 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.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.5 5
Solution:
∠BEC = 130°, ∠ECD = 20°, ∠BAC = ?
Angles formed by arc AD are ∠ABD, ∠ACD is equal.
∴ ∠ABD = ∠ACB = 20°
∠ABD = 20°
∠AEB + ∠BEC = 180 (adjacent angles)
∠AEB + 130° = 180
∠AEB = 180 – 130
∴ ∠AEB = 50°
Now, in ∆BAE,
∠BAE = ∠ABE + ∠AEB = 180°
∠BAE + 20 + 50 = 180
∠BAE + 70 = 180
∠BAE = 180 – 70
∴ ∠BAE = 110°
But ∠BAE and ∠BAC are the same.
∴ ∠BAC = 110°

Question 6.
ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC = 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.5 6
Solution:
∠DBC = 70°, ∠BAC = 30°, then ∠BCD =?
AB = BC, then ∠ECD = ?
∠DAC and ∠DBC are angles in the same segment.
∴ ∠DAC = ∠DBC = 70°
∴ ∠DAC = 70°
ABCD is a cyclic quadrilateral.
∴ Sum of opposite angles is 180°.
∠DAB + ∠DCB = 180
100 + ∠DCB = 180
[∵ ∠DAC + ∠BAC = ∠DAB 70 + 30 = 100]
∠DCB = 180 – 100
∴ ∠DCB = 80
∠DCB = ∠BCD = 80
∴ ∠BCD = 80
In ∆ABC, AB = AC,
∴ ∠BAC = ∠BCA = 30°
∠BCA = 30°
∠ECD = ∠BCD – ∠BCA = 80 – 30
∴ ∠ECD = 50°.

Question 7.
If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.5 7
Solution:
Data: In cyclic quadrilateral ABCD, AC and BD are diameters of circle.
To Prove: ABCD is a rectangle.
Proof: AC is a diameter. ∠ABC is angle in semicircle. Angle in semicircle is a right angle.
∴ ∠ABC = 90° ∠ADC = 90°
Similarly, BD is a diamgers, ∠DAB, ∠DCB are angles in semicircle.
∠DAB = 90° ∠DCB = 90°
Now, four angles of quadrilateral ABCD are right angles.
∴ ∠A = ∠B = ∠C = ∠D = 90°
∴ ABCD is a rectangle.

Question 8.
If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.5 8
Solution:
Data: ABCD is a trapezium, DC || AB and AD = BC which are non-parallel sides.
To Prove: ABCD is a cyclic quadrilateral.
Proof: ABCD is a trapezium.
AB || CD and AD = BC.
∴ ∠DAB + ∠CDA = 180° …………. (i)
(sum of interior angles)
Similarly, ∠DCB + ∠ABC = 180° …………. (ii)
As we know, AD = BC,
∴∠DAB = ∠CBA
Substituting Eqn. (i) in Eqn. (ii),
∠CBA + ∠CDA = 180°
∠DAB + ∠DCB = 180°
If sum of angles of a quadrilateral is 180°, then it is cyclic quadrilateral.
∴ ABCD is a cyclic quadrilateral.

Question 9.
Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively. Prove that ∠ACP = ∠QCD.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.5 9
Solution:
Two circles are drawn taking PQ and PR of a triangle as diameter. Let these intersect at P and S.
To Prove: ∠ACP = ∠QCD
Proof: ∠ABP = ∠QBD ………….. (i) (vertically opposite angles)
∠ABP = ∠ACP ……….. (ii) (angles in the same segment)
Similarly, ∠QCD = ∠QBD …………. (iii)(angles in the same segment)
From (i), (ii), and (iii),
∠ACP = ∠QCD.

Question 10.
If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.5 10
Solution:
Data: Two circles are drawn taking PQ and PR of a triangle as diameter. Let these intersect at P and S.
To Prove: The point of intersection ‘S’ is on the third side QR of ∆PQR.
Construction: Join PS.
Proof: QAP is a diameter.
∴ ∠QSP = 90° (angle in the semi-circle) Similarly, ABR is a diameter.
∠PSR – 90° (angle in the semicircle)
∠QSR = ∠QSP + ∠RSP = 90 + 90
∠QSR = 180°
∴ ∠QSR is a straight angle.
∴ QSR is a straight line.
∴ Point ‘S’ is on the third side QR of ∆PQR.

Question 11.
∆ABC and ∆ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.5 11
Solution:
Data: ∆ABC and ∆ADC are right-angled triangles having common hypotenuse AC.
To Prove: ∠CAD = ∠CBD
Proof: In ∆ABC, ∠ABC = 90°
∴ ∠BAC + ∠BCA = 90° …………. (i)
In ∆ADC, ∠ADC = 90°
∴ ∠DAC + ∠DCA = 90° …………… (ii)
Adding (i) and (ii),
∠BAC + ∠BCA + ∠DAC + ∠DCA = 90 + 90
(∠BAC + ∠DAC) + (∠BCA + ∠DCA) = 180°
∠BAD + ∠BCD = 180°
∴ ∠ABC + ∠ADC = 180°
If opposite angles of a quadrilateral are supplementary, then it is a cyclic quadrilateral.
∴ ∠CAD = ∠CBD (∵ Angles in the same segment).

Question 12.
Prove that a cyclic parallelogram is a rectangle.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.5 12
Solution:
ABCD is a cyclic parallelogram in the circle with ‘O’ centre.
To Prove: ABCD is a rectangle.
Proof: ABCD is a cyclic parallelogram.
∴ AB || DC and AD || BC
∠A = ∠C (Opposite angles of parallelogram)
But, ∠A + ∠C = 180° (Opposites angles of cyclic quadrilateral)
∠A + ∠A = 180°
2∠A = 180°
∠A = \(\frac{180}{2}\)
∴ ∠A = 90°
If each angle of the parallelogram is a right angle, it is a rectangle.
∴ ABCD is a rectangle.

KSEEB Solutions for Class 9 Maths

 

KSEEB Solutions for Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.2

KSEEB Solutions for Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.2 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 10 Linear Equations in Two Variables Exercise 10.2.

Karnataka Board Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.2

Question 1.
Which one of the following options is true, and why ?
y = 3x + 5 has
i) a unique solution,
ii) only two solutions,
iii) infinitely many solutions.
Solution:
y = 3x + 5 has many solutions. Because this is a linear equation with two variables.

Question 2.
Write four solutions for each of the following equations :
i) 2x + y = 7
ii) πx + y = 9
iii) x = 4y
Solution:
i) 2x + y = 7
If x = 0, 2x + y = 7
2 × 0 + y = 7
0 + y = 7
y = 7
Solution is (0, 7).
If x = 1, 2x + y = 7
2 × 1 + y = 7
2 + y = 7
y = 7 – 2 = 5
Solution is (1, 5).
If x = 2, 2x + y = 7
2 × 2 + y = 7
4 + y = 7
y = 7 – 4 = 3
Solution is (2, 3).
If x = 3, 2x + y = 7
2 × 3 + y = 7
6 + y = 7
∴ y = 7 – 6 = 1
Solution is (3, 1).

ii) πx + y = 9
If x = 0, πx + y = 9
π × 0 + y = 9
0 + y = 9
y = 9
Solution (x, y) = (0, 0).
If x = 1, πx + y = 9
π × 1 + y = 9
π + y = 9
y = 9 – π
Solution (x, y) = (1, 9 – π).
If y = 0, πx + y = 9
πx + 0 = 9
πx = 9
∴ x = \(\frac{9}{\pi}\)
Solution (x, y) = ( \(\frac{9}{\pi}\), o).
If y = 1, πx + y = 9
πx + 1 = 9
πx = 9 – 1
πx = 8
∴ x = \(\frac{8}{\pi}\)
Solution (x, y) = (\(\frac{8}{\pi}\) ,1).

iii) x = 4y
If x = 0, x = 4y
0 = 4y
4y = 0
∴ y = \(\frac{0}{4}\) = ∞
∴ Solution (x, y) = (0, ∞).
If x = 1, x = 4y
1 = 4y
4y = 1
∴ y = \(\frac{1}{4}\)
∴ Solution (x, y) = (1, \(\frac{1}{4}\)).
If x = 2, x = 4y
2 = 4y
4y = 2
∴ y = \(\frac{4}{2}\)
∴ y = \(\frac{1}{2}\)
Solution (x, y) = (2, \(\frac{1}{2}\)).
If x = 4, x = 4y
4 = 4y
4y = 4
∴ y = \(\frac{4}{4}\)
∴ y = 1
∴ Solution (x, y) = (4, 1).

Question 3.
Check which of the following are solutions of the equation x – 2y = 4 and which are not:
i) (0, 2)
ii) (2, 0)
iii) (4, 0)
iv) (\(\sqrt{2}, 4 \sqrt{2}\))
v) (1, 1)
Solution:
i) (0, 2)
x – 2y = 4
0 – 2(2) = 4
0 – 4 = 4
-4 ≠ 4
∴ (0, 2) is not a solution.

ii) (2, 0)
x – 2y = 4
2 – 2(0) = 4
2 – 0 = 4
2 ≠ 4
∴ (2, 0) is not a solution

iii) (4, 0)
x – 2y = 4
4 – 2(0) = 4
4 – 0 = 4
4 = 4
∴ (4, 0) is a solution.

iv)
KSEEB Solutions for Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.2 1

v) (1, 1)
x – 2y = 4
1 – 2(1) = 4
1 – 2 = 4
-1 ≠ 4
∴ (1, 1) is not a solution.

Question 4.
Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.
Solution:
2x + 3y = k
If x = 2, y = 1, then k = ?
2x + 3y = k
2(2) + 3(1) = k
4 + 3 = k
7 = k
∴ k = 7.

We hope the KSEEB Solutions for Class 9 Maths Chapter Linear Equations in Two Variables Ex 10.2 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter Linear Equations in Two Variables Exercise 10.2, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.2

KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.2 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 5 Triangles Exercise 5.2.

Karnataka Board Class 9 Maths Chapter 5 Triangles Ex 5.2

Question 1.
In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that:
(i) OB = OC
(ii) AO bisects ∠A.
KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.2 1
Solution:
Data: In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O.
To Prove:
(i) OB = OC
(ii) AO bisects ∠A.
Proof:
(i) In ∆ABC, AB = AC
∴ ∠ABC = ∠ACB
\(\frac{1}{2}\) ABC = \(\frac{1}{2}\) ACB
∠OBC = ∠OCB.
In ∆OBC Now, ∠OBC = ∠OCB is proved.
∴ ∆OBC is an isosceles triangle.
∴ OB = OC.

(ii) In ∆AOB and ∆AOC,
AB = AC (Data)
OB = OC (proved)
AO is common.
Side, Side, Side postulate.
∴ ∆AOB ≅ ∆AOC
∴ ∠OAB = ∠OAC
∴ AO bisects ∠A.

Question 2.
In ∆ABC, AD is the perpendicular bisector of BC. Show that ∆ABC is an isosceles triangle in which AB = AC.
KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.2 2
Solution:

KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.2 3
Data: In ∆ABC, AD is the perpendicular bisector of BC.
To Prove: ∆ABC is an isosceles triangle in which AB = AC.
Proof: In ∆ABC, AD is the perpendicular bisector of BC.
∴ BD = DC
∴ ∠ADB = ∠ADC = 90°.
Now, in ⊥∆ADB and ⊥∆ADC,
BD = DC (AD is the perpendicular bisector)
∠ADB = ∠ADC = 90° (Data)
AD is common
∴ ∆ADB ≅ ∆ADC
∴ Angles opposite to equal sides of an isosceles triangle are equal.
∴ AB = AC
∴ In ∆ABC, If AB = AC, then
∆ABC is an isosceles triangle.

Question 3.
ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively. Show that these altitudes are equal.
KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.2 4
Solution:
Data: ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively
To Prove: Altitude BE = Altitude CF.
Proof: In ∆ABC,
AB = AC and CF ⊥ AB, BE ⊥ AC.
∴ ∠BEC = ∠CFB = 90° (Data)
Angles opposite to equal sides of an isosceles triangle are equal.
BC is common.
∴ ∆BEC ≅ ∆CFB (ASA postulate)
∴ BE = CF.

Question 4.
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal. Show that
KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.2 5
(i) ∆ABE ≅ ∆ACF
(ii) AB = AC i.e., ∆ABC is an isosceles triangle.
Solution:
Data: In ∆ABC, altitudes BE and CF to sides AC and AB are equal and BE = CF.
To Prove:
(i) ∆ABE ≅ ∆ACF
(ii) AB = AC i.e., ∆ABC is an isosceles triangle.
Proof: In ∆ABE and ∆ACF,
∠AEB = ∠AFC = 90° (Data)
Altitude BE = Altitude CF (Data)
∠A is Common.
∴ ∆ ABE ≅ ∆ACF (AAS Postulate)
∴ AB = AC
∴ ∆ABC is an isosceles triangle.

Question 5.
ABC and DBC are two isosceles triangles on the same base BC. Show that ∠ABD = ∠ACD.
KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.2 6
Solution:
Data: ABC and DBC are two isosceles triangles on the same base BC.
To Prove: ∠ABD = ∠ACD
Proof: In ∆ABC, AB = AC,
∴ Opposite angle ∠ABC = ∠ACB …………. (i)
Similarly in ∆BDC, BD = DC.
Opposite angle ∠DBC = ∠DCB ………….. (ii)
From (i) and (ii),
∠ABC = ∠ACB
Adding ∠DBC and ∠DCB on both sides,
∠ABC + ∠DBC = ∠ABD
∠ACB + ∠DCB = ∠ACD
Equals are added to equal angles.
∴ ∠ABD = ∠ACD.

Question 6.
∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB. Show that ∠BCD is a right angle.
KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.2 7
Solution:
Data: ∆ABC is an isosceles triangle. AB = AC. Side BA is produced to D such that AD = AB
KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.2 8
To Prove: ∠BCD is a right angle.
Proof: In ∆ABC, AB = AC.
∴ ∠ABC = ∠ACB = x°
Similarly, in ∆ACD,
AB = AC.
AB = AD.
∴ AD = AC
Angles opposite to equal sides of a triangle are equal.
∴ ∠ACD = ∠ADC = x.
Now in ∆DCB,
∠B + ∠C + ∠D = 180°
∠DBC + ∠ACB + ∠ACD + ∠ADC = 180°
x + x + x + x = 180
4x = 180
∴ x = \(\frac{180}{4}\)
∴ x = 45°
Now, ∠DCB = ∠DCA + ∠ACB
= x + x
= 2x
= 2 × 45 (∵ x = 45°)
∴∠DCB = 90°
∴ ∠BCD is an right angle.

Question 7.
ABC is a right-angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.
Solution:
Data: ABC is a right-angled triangle in which ∠A = 90° and AB = AC.
To Prove: ∠B = ? and ∠C = ?
Proof: In ∆ABC, AB = AC, then
∴ ∠B = ∠C.
In ∆ABC, ∠A + ∠B + ∠C = 180°
90 + ∠B + ∠C = 180°
∠B + ∠C = 180 – 90°
∠B + ∠C = 90°
∠B + ∠C = 90°.
But, ∠B = ∠C,
KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.2 9
∠B + ∠C = 90°
∠B + ∠C = 90°
2∠B = 90°
∴ ∠B = \(\frac{90}{2}\) = 45
∴ ∠C = 45° ∵∠ABC = ∠ACB

Question 8.
Show that the angles of an equilateral triangle are 60° each.
Solution:
In ∆ABC, AB = AC = BC.
∴ ∠A = ∠B = ∠C, let this is x°.
But sum of three angles is 180°.
KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.2 10
∠A + ∠B + ∠C = 180°
x + x + x = 180°
3x = 180°
∴ x = \(\frac{180}{3}\) = 60°
∴ ∠A = 60°
∠B = 60°
∠C = 60°

We hope the KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.2 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 5 Triangles Exercise 5.2, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.4

Karnataka Board Class 9 Maths Chapter 12 Circles Ex 12.4

Question 1.
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.
Solution:
R = 5 cm, r = 3 cm. d = 4 cm.
length of the common chord = ?
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.4 1
When measured Chord CD = 6 cm.
Circles with centre A and B, intersect at points C and D.
CD is common chord.
Perpendicular bisector of CD is AB.
AC = 5cm. BC = 3cm.
∴ Chord, CD= 2 × OC = 2 × 3
∴ Chord, CD = 6 cm.

Question 2.
If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
Solution:
Data : Circle with centre ‘O’, AB and CD chords are equal and intersect at E.
To Prove: Segment EAD = Segment ACB.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.4 2
Construction: Draw OP⊥CD and OQ⊥AB.
Join OE.
Proof: AQ = QB (∵ Chord drawn perpendicular CP = PD ,to centre bisects chord.).
In ∆OPE and ∆OQE,
∠OPE = ∠OQE = 90° (Construction)
OP = OQ
OE is common.
∴ ∆OPE ≅ ∆OQE (RHS Theorem)
QE = PE ……….. (i)
But QB = PD
∴ QB – QE = PD – PE
EB = ED ……….. (ii)
From (i) and (ii),
AE = CE.
CBD is minor segment formed by Chord CD.
ADB is minor segment formed by Chord AB.
∴ Minor segment CBD = Minor segment ADB
Major segment CAD = Major segment ACB.

Question 3.
If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.4 3
Solution:
Data: O is the centre of Circle. Chord AB = Chord CD. These chords intersects at E. OE is joined.
To Prove: ∠OEP = ∠OEQ Construction:
Draw OP⊥CD and OQ⊥AB.
Proof: In ∆OPE and ∆OQE,
∠OPE = ∠OQE = 90° (construction)
OE is common.
OP = OQ (Equal chords are equidistant)
∴ ∆OPE ≅ ∆OEQ (RHS postulate)
∴ ∠OEP = ∠OEQ.

Question 4.
If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.4 4
Solution:
Data: A line intersects two concentric circles with centre O at A, B, C, and D.
To Prove: AB = CD.
Construction: Draw OM⊥AD.
Proof: AD is the chord of a bigger circle.
OM⊥AD
∴ AM = MD ………(i) (∵ Perpendicular bisects the chord from centre)
Similarly, BC is the chord of the smaller circle.
OM⊥BC.
∴ BM = MC ………. (ii)
Subtracting (ii) from (i),
AM – BM = MD – MC
∴ AB = CD.

Question 5.
Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip?
Solution:
Data: Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5 cm drawn in a park. Distance between Reshma and Salma and between Salma and Mandip is 6 cm. each.
To Prove: Distance between Reshma and Mandip = ?, RM = ?
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.4 5
Construction: Draw OX⊥RS and OY⊥SM. OR, OS and OM are joined.
Proof: OR = OS = OM = 5 m.
RX = XS = 3 m.
Now, in ⊥∆ORX,
OR2 = OX2 + RX2
(5)2 = (OX)2 + (3)2
25 = OX2 + 9
25 – 9 = OX2
16 = OX2
OX2 = 16
∴ OX = 4 WI …………… (i)
In quadrilateral ORSM,
OR = OM and SR = SM
∴ ORSM is a kite.
Diagonals of a kite bisect at the right angle.
∠RAS = 90°
∴ RA = AM.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.4 6
Now, RM = 2RC = 2 × 4.8 = 9.8 m
∴ RM = 9.8 m.
∴ Distance between Reshma and Mandip is 9.8 m.

Question 6.
A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.4 7
Solution:
Data: A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed, and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk to each other.
To Prove: Length of wire of telephone =?
Proof: In this figure, AS = SD = DA
∴ ASD is an equilateral triangle.
∴ OA = 20 m. (∵ radius)
Median passed through centre of circle.
∴ AB is the median.
AD ⊥ SD
Median of an equilateral triangle divides in the ratio 2 : 1 in the centre?
∴ AO : OB = 2 : 1
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.4 8
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.4 9

KSEEB Solutions for Class 9 Maths

 

KSEEB Solutions for Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.1

KSEEB Solutions for Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.1 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 10 Linear Equations in Two Variables Exercise 10.1.

Karnataka Board Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.1

Question 1.
The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement. (Take the cost of a notebook to be Rs. x and that of a pen to be Rs. y).
Solution:
Let the cost of pen b Rs. x
If the cost of the Notebook is Rs. y.
Cost of 1 notebook is twice the cost of a pen.
∴ y = 2x
-2x + y = 0
If this is written in the form of
ax + by + c = 0 it is
-2x + y + 0 = 0
Here, a = -2, b = 1, c = 0.

Question 2.
Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:
i) 2x + 3y = \(9 . \overline{35}\)
ii) x – \(\frac{y}{5}\) – 10 = 0
iii) -2x + 3y = 6
iv) x = 3y
v) 2x = -5y
vi) 3x + 2 = 0
vii) y – 2 = 0
viii) 5 = 2x
Solution:
i) 2x + 3y = \(9 . \overline{35}\)
OR 2x + 3y – \(9 . \overline{35}\) = 0
If we compare this to ax + by + c = 0,
Here, a = 2, b = 3, c = \(9 . \overline{35}\).

ii) x – \(\frac{y}{5}\) – 10 = 0
If we compare this to ax + by + c = 0,
Here, a = 1, b = \(\frac{-1}{5}\) , c=10.

iii) -2x + 3y = 6
OR -2x + 3y – 6 = 0
If we compare this to ax + by + c = 0,
Here, a = -2, b = 3, c = -6.

iv) x = 3y
x – 3y = 0
x – 3y + 0 = 0
If we compare this to ax + by + c = 0,
Here, a = 1, b = -3, c = 0.

v) 2x = -5y
2x + 5y = 0
2x + 5y + 0 = 0
If we compare this to ax + by + c = 0,
Here, a = 2, b = 5, c = 0.

vi) 3x + 2 = 0
3x + 0y + 2 = 0
If we compare this to ax + by + c = 0,
Here, a = 3, b = 0, c = 2.

vii) y – 2 = 0
0x + 1y – 2 = 0
If we compare this to ax + by + c = 0,
Here, a = 0, b = 1, c = -2.

viii) 5 = 2x
-2x + 5 = 0
-2x + 0y + 5 = 0
If we compare this to ax + by + c = 0,
Here, a = -2, b = 0, c = 5.

We hope the KSEEB Solutions for Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.1 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 10 Linear Equations in Two Variables Exercise 10.1, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 9 Coordinate Geometry Ex 9.3

KSEEB Solutions for Class 9 Maths Chapter 9 Coordinate Geometry Ex 9.3 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 9 Coordinate Geometry Exercise 9.3.

Karnataka Board Class 9 Maths Chapter 9 Coordinate Geometry Ex 9.3

Question 1.
In which quadrant or on which axis do each of the points (-2, 4), (3, -1), (-1. 0), (1, 2) and (-3, -5) lie? Verify your answer by locating them on the Cartesian plane.
Solution:
KSEEB Solutions for Class 9 Maths Chapter 9 Coordinate Geometry Ex 9.3 1
A (-2, 4) II Quadrant
B (3, -1) IV Quadrant
C (-1, 0) Negative x-axis
D(1, 2) I Quadrant
E (-3, -5) III Quadrant

Question 2.
Plot the points (x, y) given in the following table on the plane, choosing suitable units of distance on the axes.

x -2 -1 0 1 3
y 8 7 -1.25 3 -1

Solution:
A                            B                              C                                D                                 E

x -2 -1 0 1 3
y 8 7 -1.25 3 -1

KSEEB Solutions for Class 9 Maths Chapter 9 Coordinate Geometry Ex 9.3 2
A (-2, 8)
B (-1, 7)
C (0,-1.25)
D (1, 3)
E (3, -1)

We hope the KSEEB Solutions for Class 9 Maths Chapter 9 Coordinate Geometry Ex 9.3 helps you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 9 Coordinate Geometry Exercise 9.3, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.3

Karnataka Board Class 9 Maths Chapter 12 Circles Ex 12.3

Question 1.
Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?
Solution:
(1) Circles which do not intersect each other :
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.3 1
There is no common point for these circles.

(2) Circles which touch externally :
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.3 2
For these pairs of circles, there is only one common point ‘x’.

(3) Circles which touch internally :
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.3 3
These pair of circles touch internally. These have common end point ‘P’.

(4) Intersecting Circles :
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.3 4
For these pair of circles, M and N are two common points.
If we observe such circles we conclude that only two points are there.

Question 2.
Suppose you are given a circle. Give a construction to find its centre.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.3 5
Solution:
If a circle with any measurement, to find out centre of circle, steps are as follows :

  1. Construct a circle with any radius.
  2. Draw two chords AB and CD with any length.
  3. Draw perpendicular bisector for AB and CD chords.
  4. If these bisectors intersect each other when produced they meet at ’O’. O is the centre of the circle.

Question 3.
If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.3 6
Solution:
Consider two circles centred at point O and O’ intersecting each other at point A and B respectively.
Join AB. AB is the chord of the circle centred at O.
∴ Perpendicular bisector of AB will pass through O.
Again, AB is also the chord of the circle centred at O’.
∴ Perpendicular bisector ofAB will also pass through O’.
Clearly, the centres of these circles lie on the perpendicular bisector of the common chord.

KSEEB Solutions for Class 9 Maths

 

KSEEB Solutions for Class 9 Maths Chapter 9 Coordinate Geometry Ex 9.2

KSEEB Solutions for Class 9 Maths Chapter 9 Coordinate Geometry Ex 9.2 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 9 Coordinate Geometry Exercise 9.2.

Karnataka Board Class 9 Maths Chapter 9 Coordinate Geometry Ex 9.2

Question 1.
Write the answer of each of the following questions :
(i) What is the name of horizontal and the vertical lines drawn to determine the position of any point in the Cartesian plane ?
(ii) What is the name of each part of the plane formed by these two lines?
(iii) Write the name of the point where these two lines intersect.
Solution:
(i) In a Cartesian plane, to determine the position of any point, the horizontal line is called the x-axis. A vertical line is y-
The y-coordinate is also called the ordinate.
Horizontal line → xOx’
Vertical line → yOy’

(ii) The name of each part of the plane formed by these two lines is called Quadrant.
I Quadrant → xOy
II Quadrant → yOx’
III Quadrant → x’Oy’
IV Quadrant → y’Ox

(iii) The name of the point where these two lines intersect is called the origin. Coordinates of origin are (0. 0).

Question 2.
See Fig.9.14, and write the following:
KSEEB Solutions for Class 9 Maths Chapter 9 Coordinate Geometry Ex 9.2 1
Solution:

  1. The coordinates of B : (-5, +2)
  2. The coordinates of C. : (+5, -5)
  3. The point identified by the coordinates (-3, -5) : E.
  4. The point identified by the coordinates (2, -4) : G.
  5. The abscissa of the point D : 6
  6. The ordinate of the point H: -3
  7. The coordinates of the point L : (0, +5)
  8. The coordinates of the point M : (-3, 0).

We hope the KSEEB Solutions for Class 9 Maths Chapter 9 Coordinate Geometry Ex 9.2 helps you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 9 Coordinate Geometry Exercise 9.2, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 9 Coordinate Geometry Ex 9.1

KSEEB Solutions for Class 9 Maths Chapter 9 Coordinate Geometry Ex 9.1 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 9 Coordinate Geometry Exercise 9.1.

Karnataka Board Class 9 Maths Chapter 9 Coordinate Geometry Ex 9.1

Question 1.
How will you describe the position of a table lamp on your study table to another person ?
Solution:
Length of study table = 100 cm.
Breadth = 60 cm.
KSEEB Solutions for Class 9 Maths Chapter 9 Coordinate Geometry Ex 9.1 1
Scale : 10 cm = 1 row
Length of table 100 cm = 10 column.
Breadth 60 cm = 6 rows.

Question 2.
(Street Plan) : A city has two main roads which cross each other at the centre of the city. These two roads are along the North-South direction and East-West direction. All the other streets of the city-run parallel to these roads and are 200 m apart. There are 5 streets in each direction. Using 1 cm = 200 m, draw a model of the city in your notebook. Represent the roads/streets by single lines.
There are many cross-streets in your model. A particular cross-street is made by two streets, one running in the North-South direction and another in the East-West direction. Each cross street is referred to in the following manner:
If the 2nd street running in the North-South direction and 5th in the East-West direction meet at some crossing, then we will call this cross-street (2, 5). Using this convention, find:
(i) how many cross – streets can be referred to as (4, 3).
(ii) How many cross – streets can be referred to as (3, 4).
Solution:
Scale : 200 m = 1 cm.
KSEEB Solutions for Class 9 Maths Chapter 9 Coordinate Geometry Ex 9.1 2

We hope the KSEEB Solutions for Class 9 Maths Chapter 9 Coordinate Geometry Ex 9.1 helps you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 9 Coordinate Geometry Exercise 9.1, drop a comment below and we will get back to you at the earliest.