pair of linear equation s in two variables and circles
TRANSCRIPT
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ALTERNATIVE ACEDEMIC CALENDER [2021 – 22]
STANDARD: 10th MONTH: September SUBJECT: Mathematics
Sl No
Month/ Week
Expected learning
competencies
Suggested Learning Activities Evaluation
01
Sep
tem
ber
firs
t w
eek (
Th
ree p
eri
od
s)
Defining
Equation, Pair
of linear
equations in
two variables
and Pair of
linear equation
in one variable.
Students already know about
Equation, Linear equation in
one variable, when we teach
Linear equation in two
variables, then they are able
to differentiate between
Linear equation in one
variable and Linear equation
in two variables.
Using Activity
sheet – 1.
02 Graphical
representation
of Pair of linear
equations in
two variables
and Pair of
linear equation
in one variable.
Explain how we can draw a
pair of linear equation in one
variable on Number line and
also graphically represent
solution of pair of linear
equations in two variables as
two lines.
Using Activity
sheet – 2.
03 Defining and
comparing the
ratios of linear
equations in
two variables.
Practicing to find out ratios of
coefficients and comparing
the ratios of linear equations
in two variables.
https://www.youtube.com/watch
?v=nRGJruWqXwU
Using Activity
sheet – 3.
Solving
problems in
Work Book –
part1, Page No-
21 & 22, Main I,
II and III
Pair of linear equations in two variables and Circles
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04
Sep
tem
ber
seco
nd
week(
two
peri
od
s)
Graphical
Method of
solution of a
Pair of Linear
Equations.
Practicing to find solution of a
Pair of Linear Equation in two
variables by Graphical
Method and also steps to
follow to find solution
https://www.youtube.com/watch
?v=J88VLDDQKgI
Using Activity
sheet – 4.
Solving
problems in
Work Book –
part1, Page No-
23 and Page
No- 24 - Main
I(a).
05 Algebraic
Methods of
solving a Pair
of Linear
equations:
1st method:
Substitution
method
Practicing the Definition and
step to find solutions of Pair
of Linear Equation in two
variables by using
Substitution method.
Using Activity
sheet – 5.
Solving
problems in
Work Book –
part1, Page No-
23and page no -
24 - Main I(b).
06
Sep
tem
ber
thir
d w
eek(
thre
e p
eri
od
s)
Algebraic
Methods of
solving a Pair
of Linear
equations:
2nd method:
Elimination
method
Practicing the Definition and
step to find solutions of Pair
of Linear Equation in two
variables by using Elimination
method.
https://www.youtube.com/watch
?v=ivlTc6JOhuc
https://www.youtube.com/wat
ch?v=dbf5_jdTsa8
Using Activity
sheet – 6.
Solving
problems in
Work Book –
part1, Page No-
23and page no -
25 - Main I(c).
07 Algebraic
Methods of
solving a Pair
of Linear
equations:
3rd method:
Cross
Multiplication
Method
Practicing the Definition and
step to find solutions of Pair
of Linear Equation in two
variables by using Cross
Multiplication method.
Using Activity
sheet – 7.
Solving
problems in
Work Book –
part1, Page No-
23and page no -
25 - Main I(d).
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08 Equations
reducible to a
pair of linear
equations in
two variables.
Discussion to solve such
pairs of equations which are
not linear but can be reduced
to linear pair form making
some suitable methods like
Graphical method or any one
of Algebraic method to solve
pair of equations.
https://www.youtube.com/wat
ch?v=Kp_oKu5KW_Y
Using Activity
sheet – 8.
09
Sep
tem
ber
4th
week (
Fo
ur
peri
od
s)
Circle and a
line
Defining Circles and
explaining about relationship
between circle and a line.
Using Activity
sheet – 9.
10 Chapter:
Circles
Theorem – 01
(Theorem on
tangents)
Practicing the steps involved
in proving theorem – The
tangent at any point of a
circle is perpendicular to the
radius through the point of
contact.
https://www.youtube.com/wa
tch?v=IYWfUrtDJBo
https://www.youtube.com/wa
tch?v=Lz6i-Wep5wM
Using Activity
sheet – 10.
Solving
problems in
Work Book –
part1, Page No-
38– Practicing
theorem 01
11 Chapter:
Circles
Theorem –
02(External
point theorem)
Practicing the steps involved
in proving theorem – The
lengths of tangents drawn
from an external point to a
circle are equal.
https://www.youtube.com/wat
ch?v=y2_4CoDboJs
Using Activity
sheet – 11.
Solving
problems in
Work Book –
part1, Page No-
39– Practicing
theorem - 02
12 Problems
related to
theorems.
Practicing problems related to
theorems on circles.
Using Activity
sheet – 12.
Equation:An equation is a mathematical sentence/statement that has two
equal sides separated by an equal sign.
Example: 4 + 6 = 10 is an example of an equation. We can see on the l
of the equal sign, 4 + 6 and on the right hand side of the equal sign 10.
Linear equation in one variable:
equation which is expressed in the form of ax + b = 0
integers, and x is a variable and has only one solution.
Example:5x + 10 = 0 is a linear equation having a single variable in it.
x is – 2.
1) If 6y – 15 = 0, then find the value
of y. Solution: 6y – 15 = 0
6y = 15
y = ���
= ��
3) If 5t – 50 = 0, then find the value
of t. Solution:
Linear equations in two variables
An equation in the form of ax+by+c
a, b are not equal to zero is called a linear equation
power of variable is 1 and it has infinitely many solutions.
Example: In x + y – 5 = 0 equation, the solution is
1) Find one value of x and y which satisfies the equation 2x + y = 7.
Pair of Linear Equations in Two
An equation is a mathematical sentence/statement that has two
equal sides separated by an equal sign.
is an example of an equation. We can see on the l
of the equal sign, 4 + 6 and on the right hand side of the equal sign 10.
Linear equation in one variable:The linear equations in one variable is an
is expressed in the form of ax + b = 0, where a and b are two
variable and has only one solution.
5x + 10 = 0 is a linear equation having a single variable in it.
15 = 0, then find the value 2) If 2y + 8 = 0, then find the value
of y. Solution:
50 = 0, then find the value 4) If 3t + 19 = 0, then find the value
of t. Solution:
in two variables:
An equation in the form of ax+by+c = 0, where a,b and c are real num
called a linear equations in two variables.
power of variable is 1 and it has infinitely many solutions.
= 0 equation, the solution is x = 2 and y = 3.
Find one value of x and y satisfies the equation
2) Find one value of x which satisfies the equation x = 4y.
Pair of Linear Equations in Two Variables ACTIVITY
An equation is a mathematical sentence/statement that has two
is an example of an equation. We can see on the left side
of the equal sign, 4 + 6 and on the right hand side of the equal sign 10.
The linear equations in one variable is an
, where a and b are two
5x + 10 = 0 is a linear equation having a single variable in it. Value of
ind the value
If 3t + 19 = 0, then find the value
, where a,b and c are real numbers and
in two variables. Highest
x = 2 and y = 3.
Find one value of x and y which satisfies the equation
ACTIVITY - 1
Graphical representation of linear equation in one variable:
Graphical representation of linear equation in one variable is actually locatina value on a number line.
For example:2k + 10 = 0, then
Graphical representation of linear equation
Graphical representation of linear equation in two variables is actually a straight line. It may be Horizontal or vertical line.
For example:x – 2y = - 5 then graphical representation is given below.
If y = 0 then x = - 5, y = 1 then x =
When 2x + y = 6 then find the solution graphically.
Pair of Linear Equation
If 6y – 18 = 0 then represent y value in graph sheet
Graphical representation of linear equation in one variable:
Graphical representation of linear equation in one variable is actually locatin
2k + 10 = 0, then graphical representation is given below.
Graphical representation of linear equations in two variables:
Graphical representation of linear equation in two variables is actually a line. It may be Horizontal or vertical line.
5 then graphical representation is given below.
5, y = 1 then x = - 3
When 2x + y = 6 then find the solution graphically.
ACTIVITY Pair of Linear Equations in Two Variables
present y value in graph sheet
Graphical representation of linear equation in one variable is actually locating
representation is given below.
Graphical representation of linear equation in two variables is actually a
5 then graphical representation is given below.
ACTIVITY - 2
Pair of Linear Equations in Two Va
The general representation of a pair of linear equation
and y is a1x + b1y + c1 = 0 and
real numbers and a12 + b1
2 ≠ 0, a
For example:10x + 4y = 3 and
variables.
Comparing the ratios of Pair of Linear Equations in Two Variables:
Compare the
ratios
��
���
��
Number of solutions
Exactly one solution
Types of solution
Consistent and Independent
Graphical representation
Intersecting lines
x – 2y = 0 and
3x + 4y = 20
Comparing the ratios
Number of solutions
Types of solution
Pair of Linear Equation
Pair of Linear Equations in Two Variables:
general representation of a pair of linear equations in two variables
= 0 and a2x + b2y + c2 = 0 where a1, b1, c1, a
≠ 0, a22+ b2
2 ≠ 0.
10x + 4y = 3 and – x + 5y = 2 are Pair of linear equations in two
Comparing the ratios of Pair of Linear Equations in Two Variables:
��
��
��
���
��
���
�
�
��
Exactly one solution
Infinitely many solutions
No solution
Consistent and Independent
Consistent and Dependent
Inconsistent
Intersecting lines Coincident lines
Parallel lines
2y = 0 and 3x + 4y = 20
2x + 3y = 9 and 4x + 6y = 18
x + 2y = 4 and2x + 4y = 12
ACTIVITY
Pair of Linear Equations in Two Variables
in two variables say x
, a2, b2, c2 are all
x + 5y = 2 are Pair of linear equations in two
Comparing the ratios of Pair of Linear Equations in Two Variables:
��
���
��
���
�
�
No solution
Inconsistent
Parallel lines
x + 2y = 4 and 2x + 4y = 12
ACTIVITY - 3
Graphical method of solution of a pair of
linear equations in two variables:
To find a solution of a pair of linear equation
in two variables, we substitute assumed value
of one variable to another variable
1) 10 students of class X took part in
four more than the number of boys, then find the number of boys and the number
of girls who took part in the quiz.
Total number of students who took part in
Let, the number of boys isxand, number
According to question,
x + y =10 ...........(1)
x – y = 4 ..........(2) From equation (1)
From equation (2)
Therefore x = 7 and y = 3
2) The cost of 5 pencils and 7 pens is
₹46. Find the cost of each pencil and each pen.
Let the cost of each pencil be x &the cost of each pen be y From equation (1)
From equation (2)
Therefore x = and y =
x 5 4 6
y=10-x 5 6 4
x 5 4 3
y = x - 4 1 0 -1
x
y
x
y
Pair of Linear Equations in Two
Graphical method of solution of a pair of
in two variables:
To find a solution of a pair of linear equations
, we substitute assumed value
of one variable to another variable
took part in a mathematics quiz. If the number of girls is
four more than the number of boys, then find the number of boys and the number
of girls who took part in the quiz.
Total number of students who took part in the quiz =10
and, number of girls who took part in the quiz
cost of 5 pencils and 7 pens is ₹50. The cost of 7 pencils and 5 pens is
46. Find the cost of each pencil and each pen.
Let the cost of each pencil be x &
ACTIVITY Pair of Linear Equations in Two Variables
Graphical
a mathematics quiz. If the number of girls is
four more than the number of boys, then find the number of boys and the number
of girls who took part in the quiz =y
The cost of 7 pencils and 5 pens is
ACTIVITY - 4
Graphical
Method
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Algebraic methods of solving a pair of linear equations in two variables:
Substitution Method:We have to substitute the value of one variable by expressing it in terms of the other variable to solve the pair of linear equations. So this method is known as the substitution method. For example:x + y = 14 and x – y = 4 Step - 1:Find the value of one variable, say y in terms of the other variable. x from either equation, Whichever is convenient. x + y = 14 ……(1) and x – y = 4 …….(2) , from equation (1) y = 14 – x Step - 2:Substitute this value of y in the other equation and reduce it to an equation in one variable that is in terms of x, which can be solved. Substitute the value of y = 14 – x in (2),x – 14 + x = 4 ⇒ 2x = 18 ⇒ x = 9 Step - 3:Substitute the value of x (or y) obtained in step 2 in the equation used in step 1 to obtain the value of the other variable. Substitute x = 9 in equation (1), x + y = 14 ⇒ 9 + y = 14 ⇒ y = 14 – 9 = 5 Solve the following pair of linear equation by the substitution method:
1) 2x + 3y = 11 and 2x - 4y = - 24 Solution:
2) 3x - y = 3 and 9x - 3y = 9 Solution:
3) x + 2y = 4 and 2x + 4y = 12 Solution:
4) 2x + 3y = 9 and 4x + 6y = 18 Solution:
5) s - 7t = - 4 and s - 3t = 6 Solution:
6) 2x + 3y = 13 and 4x + 5y =23 Solution:
ACTIVITY - 5 Pair of Linear Equations in Two Variables
Substitution
method
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Algebraic methods of solving a pair of linear equations in two variables:
Elimination Method:
In the elimination method, we eliminate any
one of the variables by using basic arithmetic
operations and then simplify the equation to
find the value of the other variable. Then we
can put that value in any of the equations to
find the value of the variable eliminated.
For example:x + y = 5 and 2x - 3y = 4.
Step 1:Let x + y = 5 →(1) and 2x - 3y = 4 →(2).
For equating x coefficient, multiply equation (1) by 2
We get, 2x + 2y = 10 → (3) and 2x - 3y = 4 → (2).
Step 2:Subtract equation (3) from (2), Now we get, -5y = - 6 ⟹ 5y = 6 ⟹⟹⟹⟹ y =��
Step 3:Substituting y value in equation (1), x + y = 5 ⟹⟹⟹⟹ x +��= 5 ⟹⟹⟹⟹ x =
���
.
Solve the following pair of linear equation by the elimination method:
1) 2x + 3y = 8 and 4x + 6y = 7 Solution:
2) x + y = 6 and y – x = 2 Solution:
3) x + y = 5 and 2x - 5y = 4 Solution:
4) 3x + 4y = 10 and 2x - 2y = 2 Solution:
5) 3x - 5y = 4 and 9x - 2y = 7 Solution:
6) 3x + 4y = - 6 and 3x – y = 9 Solution:
ACTIVITY - 6 Pair of Linear Equations in Two Variables
Elimination
Method
Algebraic methods of solving a pair of linear equation
The arrows between the two numbers in
and the second product is to be subtracted from the first.
For example:2x + 3y = 46 and 3x + 5y = 74
Step-1:Write the given equation in the form
c2 = 0, then we get 2x + 3y – 46 = 0 a
Step-2:Taking the help of the diagram, write the equation in the form �
������� =
��������
= �
���������
����� � ���
= �
���� � ��� =
�
���� =
Step-3: When x and y, provided
Solve the following pair of linear equation having unique solution by using
cross multiplication method:
1)x - 3y – 3 = 0 and 3x - 9y – Solution:
3) 3x - 5y = 20 and 6x - 10y = 40Solution:
Pair of Linear Equations in Two
Cross
multiplication
method
Algebraic methods of solving a pair of linear equations in two variables:
The arrows between the two numbers indicate that they are to be multiplied
and the second product is to be subtracted from the first.
2x + 3y = 46 and 3x + 5y = 74
Write the given equation in the form a1x + b1y + c1 = 0 and a
46 = 0 and 3x + 5y – 74 = 0.
Taking the help of the diagram, write the equation in the form
= �
� =
�
�� =
�
�
: When x and y, provided a1 b2 - a2 b1≠ 0. Here values of x = 8 and y = 10.
Solve the following pair of linear equation having unique solution by using
2 = 0 2) 2x + y = 5 and 3x + 2y = 8Solution:
10y = 40 4) x - 3y – 7 = 0 and 3x Solution:
ACTIVITY Pair of Linear Equations in Two Variables
in two variables:
dicate that they are to be multiplied
= 0 and a2x + b2y +
Taking the help of the diagram, write the equation in the form
0. Here values of x = 8 and y = 10.
Solve the following pair of linear equation having unique solution by using
2x + y = 5 and 3x + 2y = 8
7 = 0 and 3x - 3y – 15 = 0
ACTIVITY - 7
Multiply equation (1) by 2 and equation (2)
elimination method we get, m = 2 and n = 3
Now �
� = 2 and
��
= 3, OR x =
Solve the following pairs of equations by reducing them to a pair of linear equations:
�� �
√� +
�
�� = 2 and
�
√� � �
�� = -1.
3) �
� � � +
�� � �
= 2 and �
� � � -
�
Pair of Linear Equations in Two
Equations reducible to a pair of linear equations in two variables:
The solution of such pairs of equations which are not linear but can be reduced to linear form by making some suitable methods like graphical or Algebraic methods.
Example: �
��+
���
= 2 and�
��+
���
=���
Let ��= mand
��
= n
Now, ��
+ ��= 2 ⇒⇒⇒⇒ 3m + 2n = 12
and��
+ ��=
���
⇒⇒⇒⇒ 2m + 3n = 13⟶
Multiply equation (1) by 2 and equation (2) by 3. By simplification using
elimination method we get, m = 2 and n = 3
= 3, OR x = ��and y =
��.
Solve the following pairs of equations by reducing them to a pair of linear
1. �� �� + 3y = 14 and
��- 4y = 23.
� � �
= 1 4) ��
� � � +
�� � �
= 4 and ��
� �
ACTIVITY Pair of Linear Equations in Two Variables
Equations reducible to a pair of linear equations
of such pairs of equations which are not linear but can be reduced to linear form by making some suitable methods like
3m + 2n = 12 ⟶ (1)
⟶(2)
By simplification using
Solve the following pairs of equations by reducing them to a pair of linear
4y = 23.
��
� � -
�
� � � = -2
ACTIVITY - 8
Circle and a line:
Circle:
A circle is all points in the same plane
that lie at an equal distance from a
centre point.
Radius:
Radius of a circle is the distance from
thecentre of the circle to any point
onit’scircumference.
Diameter:
The diameter of a circle is a line
segment that passes through the centre
of a circle and has two endpoints at the
circumference.
Chord:
A chord of a circle is a straight line segment whose endpoints both lie on a
circular arc.
Tangent:
A tangent to a circle is a straight line which
point.P - point is called the point of tangency. The tangent
perpendicular to the radius at the point of tangency.
Secant:
A secant of a circle is a line that intersects a circle at two distinct points
Secant of a circle.
Fill in the blanks with appropriate answers:
1) A tangent to a circle intersects it in ______________ points.
2) _____________ number of tangents can be drawn to a circle passing
through a point lying inside the circle.
3) A circle can have ______________
points of diameter.
4) The common point of a tangent to a circle and the circle is called
_____________________.
5) The angle between tangent to a circle and radius drawn at a point of co
is ___________.
CIRCLES
A circle is all points in the same plane
that lie at an equal distance from a
Radius of a circle is the distance from
of the circle to any point
line
segment that passes through the centre
of a circle and has two endpoints at the
is a straight line segment whose endpoints both lie on a
a straight line which touches the circle at only one
point is called the point of tangency. The tangent XY to a circle is
perpendicular to the radius at the point of tangency.
a line that intersects a circle at two distinct points
Fill in the blanks with appropriate answers:
tangent to a circle intersects it in ______________ points.
_____________ number of tangents can be drawn to a circle passing
through a point lying inside the circle.
an have ______________ parallel tangents at the most at the
4) The common point of a tangent to a circle and the circle is called
between tangent to a circle and radius drawn at a point of co
ACTIVITY CIRCLES
is a straight line segment whose endpoints both lie on a
touches the circle at only one
to a circle is
a line that intersects a circle at two distinct points. AB is
_____________ number of tangents can be drawn to a circle passing
parallel tangents at the most at the end
4) The common point of a tangent to a circle and the circle is called
between tangent to a circle and radius drawn at a point of contact
ACTIVITY - 9
Theorem
The tangents at any point of a circle are
the point of contact.
Data:
To prove:OP⊥XY
Construction: Take any point on Q, other
than P on the tangent XY and join OQ.
It meets circle at a point R.
Proof:
Statement
Hence, Q is a point on the tangent XY, other than the point of contact P. Sthe circle.
Let, OQ intersect the circle at R OP=OR
Now, OQ = OR + RQ
OQ >OR
OQ > OP
__________ is the shortest distance to the tangent from the centre ‘O’
OP ⊥⊥⊥⊥XY
A tangent PQ at a point P of a circle of radius
centre O at a point Q so that OQ
According to Pythagoras theorem, OQ2 = PQ2+ OP2
OP2=144-25 =119
OP=√��� cm.
CIRCLES
You Must Know:
There is one and only one tangent to a circle passing
through a point lying on the circle.
The line containing the radius through the point of
contact is also the normal to the circle at that point.
Theorem – 01[Theorem on tangents]
tangents at any point of a circle are perpendicular to the radius through
: Take any point on Q, other
than P on the tangent XY and join OQ.
Statement Reasons
Hence, Q is a point on the tangent XY, other than the point of contact P. So Q lies outside
[There is only one point of a contact to a tangent ]
Let, OQ intersect the circle at R OP=OR ( )
( )
( )
( )
__________ is the shortest distance to the
( )
of a circle of radius 5cm meets a line through the
OQ=12cm. Find the length of PQ.
According to Pythagoras theorem,
CIRCLES ACTIVITY
There is one and only one tangent to a circle passing
through a point lying on the circle.
The line containing the radius through the point of
contact is also the normal to the circle at that point.
perpendicular to the radius through
Reasons
ly one point of a contact to a tangent ]
( )
( )
( )
( )
( )
meets a line through the
ACTIVITY - 10
Theorem – 02
The lengths of tangents drawn from an external point to a circle are equal.
Data:
To prove:PB=BQ
Construction:
Proof:
Statement
In ∆ APB &∆ AQB
∠APB= ∠AQB=900
AB=AB
AP=AQ
∴ ∆ APB ≅ ∆AQB
BP=BQ.
The length of a tangent from a point A at distance
is 4 cm. Find the radius of the circle.
Solution:From Pythagoras theorem,
AO2 = AB2 + OB2
OB2 = 25 – 16 = 9 OB = 3 cm
CIRCLES
You Must Know:
There is no tangent to a c
point lying inside the circle.
There is one and only one tangent to a circle
passing through a point lying on the circle.
There are exactly two tangents to a circle through a
point lying outside the circle.
The lengths of tangents drawn from an external point to a circle are equal.
Reasons
The length of a tangent from a point A at distance 5 cm from the centre of circle
cm. Find the radius of the circle.
From Pythagoras theorem,
16 = 9 OB = 3 cm
CIRCLES
ACTIVITY
Know:
There is no tangent to a circle passing through a
point lying inside the circle.
There is one and only one tangent to a circle
passing through a point lying on the circle.
There are exactly two tangents to a circle through a
point lying outside the circle.
The lengths of tangents drawn from an external point to a circle are equal.
cm from the centre of circle
ACTIVITY - 11
ircle passing through a
There is one and only one tangent to a circle
passing through a point lying on the circle.
There are exactly two tangents to a circle through a
1) From the point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of circle is: A) 7 cm B) 12 cm C) 15 cm D) 24.5 cmSolution:
2) In the given figure, if TP and TQ are tangents to a circle with centre ‘O’
so that
∠∠∠∠POQ = 1100, then ∠∠∠∠PTQ is equal to:
A) 600 B) 700
C) 800D)900
Solution:
3) If tangents PA and PB from a point P to a circle with centre O are inclined to
each other at angle of 800, then A) 500 B) 600
Solution:
4) A tangent PQ at a point P of a circle of radius 5 cm meets a line through the
centre O at a point Q, so that A) 22 cm B) 23 cm Solution:
5) A tangent PQ at a point P of a circle of radius 6 cm meets a line through
centre ‘O’, so that OQ = 10 cm then the length of PQ is . . . . .
A) 9 cm B) 8 cm
Solution:
CIRCLES
m the point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of circle is:
A) 7 cm B) 12 cm C) 15 cm D) 24.5 cm
TP and TQ are tangents to a circle with centre ‘O’
PTQ is equal to:
If tangents PA and PB from a point P to a circle with centre O are inclined to
, then ∠∠∠∠POA is equal to: C) 700 D) 800
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the
OQ=12 cm. Length of PQ is : C) 24 cm D) 25 cm
A tangent PQ at a point P of a circle of radius 6 cm meets a line through
centre ‘O’, so that OQ = 10 cm then the length of PQ is . . . . .
C) 7 cm D) 6 cm
ACTIVITY CIRCLES
m the point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of circle is:
A) 7 cm B) 12 cm C) 15 cm D) 24.5 cm
TP and TQ are tangents to a circle with centre ‘O’
If tangents PA and PB from a point P to a circle with centre O are inclined to
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the
25 cm
A tangent PQ at a point P of a circle of radius 6 cm meets a line through
ACTIVITY - 12
Filename: 10th AAC MATHS EM Sept-21
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Creation Date: 9/2/2021 11:06:00 PM
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