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Karnataka Board Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9

Question 1.
A wooden bookshelf has external dimensions as follows:
Height = 110 cm, Depth = 25 cm, Breadth = 85 cm.

The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is 20 paise per cm² and the rate of painting is 10 paise per cm², find the total expenses required for polishing and painting the surface of the bookshelf.
Solution:
Outer length, l = 25 cm.
outer breadth, b = 85 cm.
outer height, h = 110 cm.
Outer surface area = lh + 2lb + 2bh = lh + 2(lb + bh)
= 85 × 110 + 2(85 × 25 + 25 × 110)
= 9350 + 9750
= 19100 cm².
Area of front face,
= [85 × 110 – 75 × 100 + 2(75 × 5)]
= (9350 – 7500 + 2(375)]
= 9350 – 7500 + 750
= 11000 – 7500
= 3500 cm².
Area to be polished,
= 19100 + 3500
= 22600 cm².
Cost of polishing 1 cm³ is Rs. 0.20.
Cost of polishing 22600 cm3 … ? …
= 22600 × 0.20 = Rs. 4520.
Length of horizontal shelf, l = 75 cm.
breadth, b = 20 cm.
height, h = 30 cm.
Area of horizontal shelf
= 2(l + h)b + lh
= [2(75 + 30) × 20 + 75 × 30]
= (4200 + 2250] cm².
= 6450 cm².
∴ Area of painting 3 horizontal rows
= 3 × 6450
= 19350 cm².
Cost of painting for 1 cm³ is Rs. 0.10.
∴ Cost of painting 19350 cm3 … ?
= 19350 × 0.10 = Rs. 1935.
∴ Total cost of polish and painting
= Rs. 4520 + Rs. 1935
= Rs. 6455.

Question 2.
The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in Fig.

Eight such spheres are used for this purpose and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm² and black paint costs 5 paise per cm².
Solution:
Diameter of a wooden frame, d = 21 cm.
Radius r = \(\frac{21}{2}\)
Outer surface area of wooden spheres,
A = 4πr²

∴ A = 1386cm²
Radius of cylinder, r₁ = 1.5 cm
height, h = 7 cm
The curved surface area of cylinder support,
A = 2πrh

Area of silver painted
= [8 × (1386 – 7.07)] cm²
= 8 × 1378.93
= 11031.44 cm²
Cost of silver paint = 11031.44 × 0.25
= Rs. 2757.86
Area of black paint = (8 × 66) cm²
= 528 cm²
Cost of black paint = 528 × 0.05
= Rs. 26.40
Total cost of silver and black paint.
= Rs. 2757.76 + Rs. 26.40
= Rs. 2784.26.

Question 3.
The diameter of a sphere is decreased by 25%. By what percent does its curved surface area decrease?
Solution:
Let the diametrer of sphere be ‘d’
Radius of sphere, r₁ = \(\frac{\mathrm{d}}{2}\)
Radius of outer sphere, r₂ = \(\frac{\mathrm{d}}{2}\left(1-\frac{25}{100}\right)\)
∴ r² = \(\frac{3}{8}\)d
Outer Area of Sphere, S₁ = 4πr₁²
= 4π\(\left(\frac{\mathrm{d}}{2}\right)^{2}\)
S₁ = πd²
The diameter of sphere is decreased by 25%. Then its outer surface area,

KSEEB Solutions for Class 9 Maths

 

KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.1 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 6 Constructions Exercise 6.1.

Karnataka Board Class 9 Maths Chapter 6 Constructions Ex 6.1

Question 1.
Construct an angle of 90° at the initial point of a given ray and justify the construction.
Solution:
Data: Constructing an angle of 90° at the initial point of a given ray.

Steps of Construction:

1. Draw AB straight line.
2. Taking A as centre and any radius draw an arc and intersect AB at C.
3. Taking C as centre with the same radius draw an arc which intersects at D.
4. With the same radius taking D as centre, it intersects at E.
5. With centres E and D with the same radius draw two arcs which meet both at F.
6. If AF is joined, AB straight line is the line in point A with 90°, i.e., AF.
   In the above construction ∠DAB = 60° and ∠DAE = 60°.
   Angular bisector of ∠DAE is AF.
   ∴ ∠DAF = ∠EAF = 30°
   ∴∠BAF = ∠DAB + ∠DAE = 60°+ 30°
   ∴∠BAF = 90°.

Question 2.
Construct an angle of 45° at the initial point of a given ray and justify the construction.
Solution:
Data: To construct an angle of 45° at the initial point of a given ray.

Steps of Construction :

1. Draw a straight line with any measurement.
2. Taking A as centre with any radius draw an arc, which meets AB at C.
3. With C centre, with the same radius draw an arc which meets at D.
4. With D centre, with the same radius, draw an arc which meets at D.
5. With centres E and D, with the previous radius draw two arcs which meet at F.
6. Now AF is joined. Now AP1AF at A. Hence ∠BAF = 90°.
7. Now, AG which is the angular bisector of ∠BAF is drawn.
8. ∠BAG = 45° is constructed.
   ∠FAB = 90°
   AG is the angular bisector of ∠FAB
   ∴ ∠FAG = ∠GAB = 45°
   ∴ ∠GAB = 45°.

Question 3.
Construct the angles of the following measurements :
(i) 30°
(ii) \(22 \frac{1}{2}^{\circ}\)
(iii) 15°.
Solution:
(i) To construct 30° angle :

Steps of Construction :

1. Draw PQ straight line.
2. With P as centre with any radius draw \(\frac{1}{2}\) arc- intersects PQ at A.
3. With A as centre with the same radius draw an arc which intersects at B. Join PB and produced.
4. With A and B centres, with radius more than half of AB draw two arcs which intersect at C. Join PC.
5. ∠BPC = ∠CPQ = 30°.

(ii) To draw an angle of \(22 \frac{1}{2}^{\circ}\)
Solution:

Steps of Construction:

1. Draw a straight line AB.
2. With centre ‘A’, taking a convenient radius to draw an arc which intersect AB at P.
3. With P as centre with the same radius draw an arc at Q, with Q as centre with the same radius draw an arc which intersects at R.
4. With R and Q centres with the same radius draw two arcs which intersect at S. Join AS, ∠BAS = 90°.
5. Now construct AT which is the angular bisector of ∠BAS, and joined, ∠TAB= 45°.
6. Now AU which is the angular bisector of ∠TAB, AU is joined. Now, ∠UAB = \(22 \frac{1}{2}^{\circ}\)

(iii) To construct an angle of 15° :
Solution:

Steps of Construction :

1. Draw a straight line AB.
2. With A as centre with a convenient radius draw an arc which intersect AB at C.
3. With C as centre, with the same radius, draw an arc which intersects at D.
   Now. ∠DAB = 60°
4. Construct AE, the bisector of ∠DAB. Join then, ∠EAB = 30°.
5. Construct AF, the bisector of ∠EAB. Join then, ∠FAB = 15°.

Question 4.
Construct the following angles and verify by measuring them by a protractor :
(i) 75°
(ii) 105°
(iii) 135°.
Solution:
(i) To construct angle 75° :

Steps of Construction :

1. Draw AB with any measurement.
2. With A as centre, with convenient scale draw an arc AB which intersects at C.
3. With C as centre with the same radius draw an arc which intersects at D. ∠DAB = 60°.
4. With D as centre with the same radius draw an arc which intersects at E.
5. With Centres E and D, by taking more than half of ED draw two arcs which meet at E. AF is joined.
   Now, ∠FAB = 90°.
6. Draw AG bisector of ∠EAD, AG joined. ∠GAD = 15°.
7. ∠GAD + ∠DAB = 15 + 60 = 75°
   ∴ ∠GAB = 75°.

(ii) To construct angle 105° :
Solution:

Steps of Construction :

1. Draw PQ with any measurement.
2. With P as centre and with any radius, draw an arc which intersects PQ at A.
3. With A as centre, with the same radius draw an arc which intersect at B.
4. With B as centre with same radius draw another arc, it intersects at C.
5. With centres C and B if two arcs are drawn these two meet at D. Join PD. ∠DPQ = 90°. Straight-line PD intersects the arc CB at E.
6. Now taking radius half of CE, if the line is drawn from C to E it intersects at F. FP is joined. Now ∠FPD = 15°.
7. ∠FPD + ∠DPQ = 15 + 90 = 105°.
   ∴ ∠FPQ = 105° is constructed.

(iii) To construct an angle of 135°
Solution:

Steps of Construction :

1. Draw a straight line AB.
2. With A as centre, a semicircle is drawn which meets AB at C and it intersects the produced BA line at G.
3. With C as centre with the same radius and centre CD, draw a radius of arc DE.
4. With centres E and D by taking a radius more than half draw two arcs which meet at F. Join AF. ∠FAB = 90°. AF intersects ED at H.
5. With G and H centres by taking radius more than \(\frac{1}{2}\) of the radius GH draw arcs which meet at I. AI is joined. Now, ∠IAF = 45°.
6. ∠IAF + ∠FAB = 45° + 90° = 135°
   ∴ ∠IAB = 135° is constructed.

Question 5.
Construct an equilateral triangle, given its side, and justify the construction.
Solution:
Data: Construct an equilateral triangle, given its side.

Steps of Construction :

1. Construct AB = 6 cm.
2. With A as centre and convenient scale draw an arc, it meets AB at P.
3. With P as centre, with same radius draw an arc, it intersects the first arc at Q, AQ is joined and produced. ∠QAB = 60°.
4. With A as centre and taking 6 cm radius draw an arc. It intersects at C which is the produced line of AQ.
5. Next, BC is joined, ABC is an equilateral triangle.

In ∆ABC, ∠A = ∠B = ∠C = 60°
AB = BC = CA = 6 cm.
In ∆ABC, AB = AC = 6 cm. ∠A = 60°
∠B = 60°
∠A + ∠B + ∠C = 180°
60 + ∠B + ∠B =180 (∵ ∠B = ∠C)
60 + 2∠B = 180
2∠B = 180 – 60
2∠B = 120°
∴ ∠B = \(\frac{120}{2}\)
∴ ∠B = 60°
∴∠B = ∠C = 60°
∴∠A = ∠B = ∠C = 60°
AC = BC (Opposite sides of equal angles)
But, AB = AC (known)
∴ AB = BC = CA = 6 cm.

We hope the KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.1 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 6 Constructions Exercise 6.1, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.5 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 5 Triangles Exercise 5.5.

Karnataka Board Class 9 Maths Chapter 5 Triangles Ex 5.5

Question 1.
ABC is a triangle. Locate a point in the interior of ∆ABC which is equidistant from all the vertices of ∆ ABC.
Solution:
ABC is a triangle. To locate a point in the interior of ∆ABC which is equidistant from all the vertices of ∆ABC we have to find out circumcentre means the point where three perpendicular bisectors meet.
E.g. Three sides of ∆ABC are,
AB = 5 cm, BC = 4 cm, and AC = 6 cm.

‘S’ is the circumcentre of ∆ABC

Question 2.
In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.
Solution:
Incentre is equidistant from three sides of a ∆. This is at the point where angular bisectors meet. This is called T.
E.g. In ∆ABC, AB = 6 cm, ∠B = 100°, ∠A = 50°.

Point T is equidistant from three sides.

Question 3.
In a huge park, people are concentrated at three points
A: where there are different slides and swings for children,
B: near which a man-made lake is situated,
C: which is near to a large parking and exit.
Where should an icecream parlour be set up so that the maximum number of persons can approach it?
(Hint: The parlour should be equidistant from A, B, and C).
Solution:
A: In a park where there are different slides and swings for children.
B: near which a man-made lake is situated.
C: which is near to a large parking and exit.

In ∆ABC, to locate a point equidistant from three sides, we have to find out perpendicular bisectors, which means where all perpendicular bisectors meet. This is called S’.
AB = BC = CA = 5 cm.
∴ The ice cream shop is at ‘S’.

Question 4.
Complete the hexagonal and star-shaped Rangolies by filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles?

Solution:
Fig. (i): Regular Hexagonal with side 5 cm. is constructed. It has 6 equal sides.
Measure of each side is 5 cm.

Number of equilateral ∆ with 1 cm side

In Fig. (ii) : there are 12 equilateral triangles with side 5 cm

∴ Number of equilateral A with 1 cm side

We hope the KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.5 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 5 Triangles Exercise 5.5, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.4 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 10.4.

Karnataka Board Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.4

Question 1.
Give the geometric representations of y = 3 as an equation.
(i) in one variable
(ii) in two variables.
Solution:
x + 3 = 0
i) In one variable, y = 3.

ii) In two variables, y = 3 straightline passes through (0, 3) parallel to x-axis. For any value of x, value of y is 3.

Question 2.
Give the geometric representations of 2x + 9 = 0 as an equation.
(i) in one variable
(ii) in two variables.
Solution:
2x + 9 = 0, this equation in
i) One variable 2x + 9 =0
2x = -9
x = \(-\frac{9}{2}\) = -4.5

ii) 2x + 9 = 0 in two variables, 2x + 9 = 0. This equation passes through (-4.5, 0) and this is parallel ot y-axis. Coordinate of x = – 4.5

We hope the KSEEB Solutions for Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.4 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 10 Linear Equations in Two Variables Exercise 10.4, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 5 Triangles Exercise 5.4.

Karnataka Board Class 9 Maths Chapter 5 Triangles Ex 5.4

Question 1.
Show that in a right angled triangle, the hypotenuse is the longest side.

Solution:
In ∆ABC, ∠ABC = 90°
∠A + ∠C = 90°
∠ABC > ∠BAC and ∠ABC < ∠BCA
∴ D is the largest side of ∠ABC.
∴ AC is opposite side of larger angle.
∴ AC hypoternuse is largest side of ∆ABC.

Question 2.
In sides AB and AC of ∆ABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.

Solution:
Data: AB and AC are the sides of ∆ABC and AB and AC are produced to P and Q respectively.
∠PBC < ∠QCB.
To Prove: AC > AB
Proof: ∠PBC < ∠QCB
Now, ∠PBC + ∠ABC = 180°
∠ABC = 180 – ∠PBC ………. (i)
Similarly, ∠QCB + ∠ACB = 180°
∠ACB = 180 – ∠QCB …………. (ii)
But, ∠PBC < ∠QCB (Data)
∴ ∠ABC > ∠ACB Comparing (i) and (ii), AC > AB
(∵ Angle opposite to larger side is larger)

Question 3.
In fig ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

Solution:
Data : ∠B < ∠A and ∠C < ∠D.
To Prove: AD < BC
Proof: ∠B < ∠A
∴ OA < OB …………. (i)
Similarly, ∠C < ∠D
∴ OD < OC …………. (ii)
Adding (i) and (ii), we have
OA + OD < OB + OC
∴ AD < BC.

Question 4.
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD. Show that ∠A > ∠C and ∠B > ∠D.

Solution:
Data : AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD.
To Prove :
(i) ∠A > ∠C
(ii) ∠B > ∠D
Construction: Join AC and BD

Proof: In ∆ABC, AB < BC (data)
∴ ∠3 < ∠4 ………. (i)
In ∆ADC, AD < CD (data)
∴ ∠2 <∠1 …………. (ii)
Adding (i) and (ii),
∠3 + ∠2 < ∠4 + ∠1
∠C < ∠A
∠A > ∠C
By adding B,
In ∆ABD, AB < AD
∠5 < ∠8
In ∆BCD, BC < CD
∠6 < ∠7
∴ ∠5 + ∠6 < ∠8 + ∠7
∠D < ∠B
OR ∠B > ∠D.

Question 5.
In Fig. PR > PQ and PS bisect ∠QPR. Prove that ∠PSR > ∠PSQ.

Solution:
Data: PR > PQ and PS bisect ∠QPR.
To Prove: ∠PSR > ∠PSQ
Proof: In ∆PQR, PR > PQ
∴ ∠PQR > ∠PRQ ………. (i)
PS bisects ∠QPR.
∴ ∠QPS = ∠RPS ………… (ii)
By adding (i) and (ii),
∠PQR + ∠QPS > ∠PRQ + ∠RPS ………. (iii)
But, in ∆PQS, ∠PSR is an exterior angle.
∴ ∠PSR = ∠PQR + ∠QPS ………… (iv)
Similarly, in ∆PRS, ∠PSQ is the exterior angle.
∴ ∠PSQ = ∠PRS + ∠RPS ……… (v)
In (iii), subtracting (iv) and (v),
∠PSR > ∠PSQ.

Question 6.
Show that of all line segments drawn from a given point, not on it, the perpendicular line segment is the shortest.

Solution:
P is a point on straight line l.
PR is the segment for l drawn such that PQ ⊥ l.
Now, in ∆PQR,
∠PQR = 90° (Construction)
∴ ∠QPR + ∠QRP = 90°
∴ ∠QRP < ∠PQR
∴ PQ < PR
Any line segment is drawn from P to l they form a small angle.
∴ PQ ⊥ l is smaller.

We hope the KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 5 Triangles Exercise 5.4, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.3 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 10.3.

Karnataka Board Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.3

Question 1.
Draw the graph of each of the following linear equations in two variables :
i) x + y = 4
ii) x – y = 2
iii) y = 3x
iv) 3 = 2x + y
Solution:
i) x + y = 4
This can be written in the form of y = mx + c, then
y = 4 – x.

x	0	1	4	-1
y	4	3	0	5

ii) x – y = 2
-y = 2 – x
y = -2 + x
y = x – 2

x	0	1	2	4
y	-2	-1	0	2

iii) y = 3x

x	0	1	2	-1
y	0	3	6	-3

iv) 3 = 2x + y
2x + y = 3
y = 3 – 2x

x	0	1	2	-1
y	3	1	-1	5

Question 2.
Give the equations of two lines passing through (2, 14). How many more such lines are there, and why ?
Solution:
E.g. i) y = 7x
ii) x = \(\frac{y}{7}\)
iii) x + y = 16 .
We may write infinite equation, because Graph of a linear equation in two variables has many solutions.

Question 3.
If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.
Solution:
3y = ax + 7
If point (3, 4) then a = ?
3y = ax + 7
3(4) = a(3) + 7 ∵ x = 3, y = 4
12 = 3a + 7
3a + 7 = 12
3a = 12 – 7
3a = 5
∴ a = \(\frac{5}{3}\)

Question 4.
The taxi fare in a city is as follows:
For the first kilometre, the fare is Rs. 8 and for the subsequent distance, it is Rs. 5 per km. Taking the distance covered as x km and the total fare is Rs. y. write a linear equation for this information, and draw its graph.
Solution:
Let the distance travelled be x km.
Let the total distance be y km.

For the first kilometre, the fare is Rs. 8 + fare for the remaining distance, i.e. (x – 1) km.
Rs. 5 for 1 km.
if it travels (x – 1) km … ? Rs.
Rs. (x – 1)5
∴ Total cost = Rs. y
Fare for first km + fare for subsequent distance means (x-1) km = Rs. y
8 + (x – 1)5 = y
8 + 5x – 5 = y
8 – 5 + 5x = y
3 + 5x = y
∴ y = 3 + 5x linear equation.

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

Question 5.
From the choices given below, choose the equation whose graphs are given in Fig. 10.6 and Fig. 10. 7.

Solution:
Solution for graph of linear equation of Fig. 10.6 :

Sum of (x + y) = 0.
Equation for this is

(ii) x + y = 0
because x = -y we have OR -x = y
0 = 0
1 = -1
-1 = 1
Solutions for graph for linear equation in Fig. 10.7 :

x + y = 2 is common in all.
∴ y = -x + 2
∴ Ans: (iii) y = -x + 2 equation.

Question 6.
If the work done by a body on application of a constant force is directly proportional to the distance travelled by the body, express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also read from the graph the work done when the distance travelled by the body is
(i) 2 units
(ii) 0 unit.
Solution:
Let the work done be, W
Distance travelled be D.
If constant force is K, then
Force is directly proportional to the distance travelled by the body.
∴ W ∝ D.
W = K × D
Constant force, K = 5 units.
W = 5 × D linear equation.
If D = 2, then W = 5D
W = 5 × 2
W = 10
If D = 0, then W = 5D
W = 5 × 0
W = 0

Question 7.
Yamini and Fatima, two students of Class IX of a school, together contributed Rs. 100 towards the Prime Minister’s Relief Fund to help the earthquake victims. Write a linear equation which satisfies this data. (You may take their contributions as Rs. x and Rs. y.) Draw the graph of the same.
Solution:
Yamini and Fatima together contributed Rs. 100 towards the Prime Minister’s Relief Fund to help the earthquake victims.
In that contribution of Yamini is R. x Contribution of Fatima is Rs. y Together Rs. 100.
∴ x + y = 100
∴ y = 100 – x.

x	20	40	50	60
y	80	60	50	40

Question 8.
In countries like USA and Canada, the temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius :
F = \(\left(\frac{9}{5}\right)\) C + 32.
(i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for the y-axis.
(ii) If the temperature is 30° C, what is the temperature in Fahrenheit?
(iii) If the temperature is 95° F, what is the temperature in Celsius?
(iv) If the temperature is 0° C, what is the temperature in Fahrenheit, and if the temperature is 0° F, what is the temperature in Celsius?
(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.
Solution:
Let Temperature be Celsius C,
The temperature be Fahrenheit F.
F = \(\left(\frac{9}{5}\right)\) C + 32

C	10	30	0	35
F	50	86	32	95

In the x-axis Celsius,
In the y-axis Fahrenheit.
(i) Graph

(ii) If temperature is 30° C, then 86° F.
(iii) If the temperature is 95° F, then 35° C.
(iv) If temperature is 0° C, then 32° F.
If 0°F, then -32°C
(v) Temperature is not equal in Celsius and Fahrenheit.
because C ≠ F

We hope the KSEEB Solutions for Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.3 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 10 Linear Equations in Two Variables Exercise 10.3, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.3 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

(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

(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 :

(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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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 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)

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 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.

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.

Solution:

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.

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

(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.

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.

Solution:
Data: ∆ABC is an isosceles triangle. AB = AC. Side BA is produced to D such that AD = AB

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,

∠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°.

∠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.