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# NCERT Solutions, Exercise 11.2, Constructions, Class 10 (Mathematics) Class 10 Notes | EduRev

## Class 10 : NCERT Solutions, Exercise 11.2, Constructions, Class 10 (Mathematics) Class 10 Notes | EduRev

Page 1

(Class – X)

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1

Exercise 11.2
Question 1:
Draw a circle of radius 6 cm. From a point 10 cm away from its centre,
construct the pair of tangents to the circle and measure their lengths. Give
the justification of the construction.
A pair of tangents to the given circle can be constructed as follows.
Step 1
Taking any point O of the given plane as centre, draw a circle of 6 cm radius.
Locate a point P, 10 cm away from O. Join OP.
Step 2
Bisect OP. Let M be the mid-point of PO.
Step 3
Taking M as centre and MO as radius, draw a circle.
Step 4
Let this circle intersect the previous circle at point Q and R.
Step 5
Join PQ and PR. PQ and PR are the required tangents.

The lengths of tangents PQ and PR are 8 cm each.
Justification
The construction can be justified by proving that PQ and PR are the tangents
to the circle (whose centre is O and radius is 6 cm). For this, join OQ and
OR.
Page 2

(Class – X)

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1

Exercise 11.2
Question 1:
Draw a circle of radius 6 cm. From a point 10 cm away from its centre,
construct the pair of tangents to the circle and measure their lengths. Give
the justification of the construction.
A pair of tangents to the given circle can be constructed as follows.
Step 1
Taking any point O of the given plane as centre, draw a circle of 6 cm radius.
Locate a point P, 10 cm away from O. Join OP.
Step 2
Bisect OP. Let M be the mid-point of PO.
Step 3
Taking M as centre and MO as radius, draw a circle.
Step 4
Let this circle intersect the previous circle at point Q and R.
Step 5
Join PQ and PR. PQ and PR are the required tangents.

The lengths of tangents PQ and PR are 8 cm each.
Justification
The construction can be justified by proving that PQ and PR are the tangents
to the circle (whose centre is O and radius is 6 cm). For this, join OQ and
OR.
(Class – X)

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2

?PQO is an angle in the semi-circle. We know that angle in a semi-circle is
a right angle.
? ?PQO = 90°
? OQ ? PQ
Since OQ is the radius of the circle, PQ has to be a tangent of the circle.
Similarly, PR is a tangent of the circle

Question 2:
Construct a tangent to a circle of radius 4 cm from a point on the concentric
circle of radius 6 cm and measure its length. Also verify the measurement
by actual calculation. Give the justification of the construction.
Tangents on the given circle can be drawn as follows.
Step 1
Draw a circle of 4 cm radius with centre as O on the given plane.
Step 2
Draw a circle of 6 cm radius taking O as its centre. Locate a point P on this
circle and join OP.
Step 3
Bisect OP. Let M be the mid-point of PO.
Step 4
Taking M as its centre and MO as its radius, draw a circle. Let it intersect
the given circle at the points Q and R.
Page 3

(Class – X)

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1

Exercise 11.2
Question 1:
Draw a circle of radius 6 cm. From a point 10 cm away from its centre,
construct the pair of tangents to the circle and measure their lengths. Give
the justification of the construction.
A pair of tangents to the given circle can be constructed as follows.
Step 1
Taking any point O of the given plane as centre, draw a circle of 6 cm radius.
Locate a point P, 10 cm away from O. Join OP.
Step 2
Bisect OP. Let M be the mid-point of PO.
Step 3
Taking M as centre and MO as radius, draw a circle.
Step 4
Let this circle intersect the previous circle at point Q and R.
Step 5
Join PQ and PR. PQ and PR are the required tangents.

The lengths of tangents PQ and PR are 8 cm each.
Justification
The construction can be justified by proving that PQ and PR are the tangents
to the circle (whose centre is O and radius is 6 cm). For this, join OQ and
OR.
(Class – X)

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2

?PQO is an angle in the semi-circle. We know that angle in a semi-circle is
a right angle.
? ?PQO = 90°
? OQ ? PQ
Since OQ is the radius of the circle, PQ has to be a tangent of the circle.
Similarly, PR is a tangent of the circle

Question 2:
Construct a tangent to a circle of radius 4 cm from a point on the concentric
circle of radius 6 cm and measure its length. Also verify the measurement
by actual calculation. Give the justification of the construction.
Tangents on the given circle can be drawn as follows.
Step 1
Draw a circle of 4 cm radius with centre as O on the given plane.
Step 2
Draw a circle of 6 cm radius taking O as its centre. Locate a point P on this
circle and join OP.
Step 3
Bisect OP. Let M be the mid-point of PO.
Step 4
Taking M as its centre and MO as its radius, draw a circle. Let it intersect
the given circle at the points Q and R.
(Class – X)
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Step 5
Join PQ and PR. PQ and PR are the required tangents.
It can be observed that PQ and PR are of length 4.47 cm each.
In ?PQO,
Since PQ is a tangent,
?PQO = 90°
PO = 6 cm
QO = 4 cm
Applying Pythagoras theorem in ?PQO, we obtain
PQ
2
+ QO
2
= PQ
2

PQ
2
+ (4)
2
= (6)
2

PQ
2
+ 16 = 36
PQ
2
= 36 - 16
PQ
2
= 20
PQ
PQ = 4.47 cm
Justification
The construction can be justified by proving that PQ and PR are the tangents
to the circle (whose centre is O and radius is 4 cm). For this, let us join OQ
and OR.
Page 4

(Class – X)

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1

Exercise 11.2
Question 1:
Draw a circle of radius 6 cm. From a point 10 cm away from its centre,
construct the pair of tangents to the circle and measure their lengths. Give
the justification of the construction.
A pair of tangents to the given circle can be constructed as follows.
Step 1
Taking any point O of the given plane as centre, draw a circle of 6 cm radius.
Locate a point P, 10 cm away from O. Join OP.
Step 2
Bisect OP. Let M be the mid-point of PO.
Step 3
Taking M as centre and MO as radius, draw a circle.
Step 4
Let this circle intersect the previous circle at point Q and R.
Step 5
Join PQ and PR. PQ and PR are the required tangents.

The lengths of tangents PQ and PR are 8 cm each.
Justification
The construction can be justified by proving that PQ and PR are the tangents
to the circle (whose centre is O and radius is 6 cm). For this, join OQ and
OR.
(Class – X)

A Free web support in Education

2

?PQO is an angle in the semi-circle. We know that angle in a semi-circle is
a right angle.
? ?PQO = 90°
? OQ ? PQ
Since OQ is the radius of the circle, PQ has to be a tangent of the circle.
Similarly, PR is a tangent of the circle

Question 2:
Construct a tangent to a circle of radius 4 cm from a point on the concentric
circle of radius 6 cm and measure its length. Also verify the measurement
by actual calculation. Give the justification of the construction.
Tangents on the given circle can be drawn as follows.
Step 1
Draw a circle of 4 cm radius with centre as O on the given plane.
Step 2
Draw a circle of 6 cm radius taking O as its centre. Locate a point P on this
circle and join OP.
Step 3
Bisect OP. Let M be the mid-point of PO.
Step 4
Taking M as its centre and MO as its radius, draw a circle. Let it intersect
the given circle at the points Q and R.
(Class – X)
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3
Step 5
Join PQ and PR. PQ and PR are the required tangents.
It can be observed that PQ and PR are of length 4.47 cm each.
In ?PQO,
Since PQ is a tangent,
?PQO = 90°
PO = 6 cm
QO = 4 cm
Applying Pythagoras theorem in ?PQO, we obtain
PQ
2
+ QO
2
= PQ
2

PQ
2
+ (4)
2
= (6)
2

PQ
2
+ 16 = 36
PQ
2
= 36 - 16
PQ
2
= 20
PQ
PQ = 4.47 cm
Justification
The construction can be justified by proving that PQ and PR are the tangents
to the circle (whose centre is O and radius is 4 cm). For this, let us join OQ
and OR.
(Class – X)

A Free web support in Education

4

?PQO is an angle in the semi-circle. We know that angle in a semi-circle is
a right angle.
? ?PQO = 90°
? OQ ? PQ
Since OQ is the radius of the circle, PQ has to be a tangent of the circle.
Similarly, PR is a tangent of the circle

Question 3:
Draw a circle of radius 3 cm. Take two points P and Q on one of its extended
diameter each at a distance of 7 cm from its centre. Draw tangents to the
circle from these two points P and Q. Give the justification of the
construction.
The tangent can be constructed on the given circle as follows.
Step 1
Taking any point O on the given plane as centre, draw a circle of 3 cm radius.
Step 2
Take one of its diameters, PQ, and extend it on both sides. Locate two points
on this diameter such that OR = OS = 7 cm
Step 3
Bisect OR and OS. Let T and U be the mid-points of OR and OS
respectively.
Step 4
Taking T and U as its centre and with TO and UO as radius, draw two
circles. These two circles will intersect the circle at point V, W, X, Y
respectively. Join RV, RW, SX, and SY. These are the required tangents.
Page 5

(Class – X)

A Free web support in Education

1

Exercise 11.2
Question 1:
Draw a circle of radius 6 cm. From a point 10 cm away from its centre,
construct the pair of tangents to the circle and measure their lengths. Give
the justification of the construction.
A pair of tangents to the given circle can be constructed as follows.
Step 1
Taking any point O of the given plane as centre, draw a circle of 6 cm radius.
Locate a point P, 10 cm away from O. Join OP.
Step 2
Bisect OP. Let M be the mid-point of PO.
Step 3
Taking M as centre and MO as radius, draw a circle.
Step 4
Let this circle intersect the previous circle at point Q and R.
Step 5
Join PQ and PR. PQ and PR are the required tangents.

The lengths of tangents PQ and PR are 8 cm each.
Justification
The construction can be justified by proving that PQ and PR are the tangents
to the circle (whose centre is O and radius is 6 cm). For this, join OQ and
OR.
(Class – X)

A Free web support in Education

2

?PQO is an angle in the semi-circle. We know that angle in a semi-circle is
a right angle.
? ?PQO = 90°
? OQ ? PQ
Since OQ is the radius of the circle, PQ has to be a tangent of the circle.
Similarly, PR is a tangent of the circle

Question 2:
Construct a tangent to a circle of radius 4 cm from a point on the concentric
circle of radius 6 cm and measure its length. Also verify the measurement
by actual calculation. Give the justification of the construction.
Tangents on the given circle can be drawn as follows.
Step 1
Draw a circle of 4 cm radius with centre as O on the given plane.
Step 2
Draw a circle of 6 cm radius taking O as its centre. Locate a point P on this
circle and join OP.
Step 3
Bisect OP. Let M be the mid-point of PO.
Step 4
Taking M as its centre and MO as its radius, draw a circle. Let it intersect
the given circle at the points Q and R.
(Class – X)
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3
Step 5
Join PQ and PR. PQ and PR are the required tangents.
It can be observed that PQ and PR are of length 4.47 cm each.
In ?PQO,
Since PQ is a tangent,
?PQO = 90°
PO = 6 cm
QO = 4 cm
Applying Pythagoras theorem in ?PQO, we obtain
PQ
2
+ QO
2
= PQ
2

PQ
2
+ (4)
2
= (6)
2

PQ
2
+ 16 = 36
PQ
2
= 36 - 16
PQ
2
= 20
PQ
PQ = 4.47 cm
Justification
The construction can be justified by proving that PQ and PR are the tangents
to the circle (whose centre is O and radius is 4 cm). For this, let us join OQ
and OR.
(Class – X)

A Free web support in Education

4

?PQO is an angle in the semi-circle. We know that angle in a semi-circle is
a right angle.
? ?PQO = 90°
? OQ ? PQ
Since OQ is the radius of the circle, PQ has to be a tangent of the circle.
Similarly, PR is a tangent of the circle

Question 3:
Draw a circle of radius 3 cm. Take two points P and Q on one of its extended
diameter each at a distance of 7 cm from its centre. Draw tangents to the
circle from these two points P and Q. Give the justification of the
construction.
The tangent can be constructed on the given circle as follows.
Step 1
Taking any point O on the given plane as centre, draw a circle of 3 cm radius.
Step 2
Take one of its diameters, PQ, and extend it on both sides. Locate two points
on this diameter such that OR = OS = 7 cm
Step 3
Bisect OR and OS. Let T and U be the mid-points of OR and OS
respectively.
Step 4
Taking T and U as its centre and with TO and UO as radius, draw two
circles. These two circles will intersect the circle at point V, W, X, Y
respectively. Join RV, RW, SX, and SY. These are the required tangents.
(Class – X)

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5

Justification
The construction can be justified by proving that RV, RW, SY, and SX are
the tangents to the circle (whose centre is O and radius is 3 cm). For this,
join OV, OW, OX, and OY.

?RVO is an angle in the semi-circle. We know that angle in a semi-circle is
a right angle.
? ?RVO = 90°
? OV ? RV
Since OV is the radius of the circle, RV has to be a tangent of the circle.
Similarly,
OW, OX, and OY are the tangents of the circle

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