NCERT Solutions for Class 10 Math Chapter 11 - Constructions

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Steps of construction:
1. Construct the triangle ABC with the given measurements.
BC = 7 cm;
$\angle$B = 45$°$,
$\angle$A = 105$°$
By angle sum property of triangle
$\angle$A + $\angle$B + $\angle$C = 180$°$
105$°$ + 45$°$ + $\angle$C = 180$°$
$\angle$C = 180$°$ – 150$°$ = 30$°$
2. Make any acute angle $\angle$CBX at point B, below the side BC.

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Steps of construction:
1. Draw a circle (I) of radius 6 cm.
2. Take a point P at a distance of 10 cm from the centre of the circle. Join OP.
3. Draw perpendicular bisector of OP. Let ‘M’ be the mid-point of OP.
4. With ‘M’ as centre and radius MO, draw a circle (II) which intersects the circle (I) at T and T′. Join PT and PT′.
5. Then PT and P T′ are two required tangents. Justification:
We know that a tangent at a point is always perpendicular to the radius at the point. Now we have to prove that $\angle$PTO = $\angle$PT′O = 90$°$.
OT is joined.
Now, PMO is the diameter of circle (II) and $\angle$PTO is in the semicircle.
∴ $\angle$PTO = 90$°$
[Angle in semicircle is a right angle] Similarly, $\angle$PT′O = 90$°$
∴ PT and PT′ have to be the tangents to the circle at T and T′ .
On measuring, the lengths of tangents i.e., PT = 8.1 cm PT′ = 8.1 cm.

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Steps of construction:
1. Draw a circle with centre O and radius 4 cm. Mark it as I.
2. Draw another circle with same centre O and radius 6 cm and mark it as II.
3. Take any point P on circle II. Join OP.
4. Draw perpendicular bisector of OP. Let it intersects OP at M.
5. With M as centre and radius MO or MP, draw a circle III which intersects the circle I at T and T′.
6. Join PT.
PT is the required tangent.

Justification:
Join OT.
Now OP is the diameter of the circle III.
$\angle$OTP is in the semicircle.
∴ $\angle$OTP = 90$°$ ...(1)
[∵ Angle in a semicircle is a right angle]
Now OT ⊥ PT [using (1)]
ä A line which makes an angle of 90$°$ with radius at any point on the circle, the line is tangent to the circle.
∴ PT is tangent to the circle ‘I’.

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Steps of construction: 1. Draw a circle of radius 3 cm and centre O.

2. Draw its diameter AB and extend it in both directions as OX and OX′.
3. Take points P on OX′ direction and Q on OX direction such that
OP = OQ = 7 cm.
4. Draw perpendicular bisectors of OP and OQ which intersects OP and OQ at M and M′ respectively.
5. With M as centre and radius = MO or MP, draw a circle II which intersects the circle ‘I’ at T and T′.
6. Similarly, with M′ as centre and radius = M′O or MQ, draw a circle (III) which intersects the circle I at S and S′.
7. Join PT, PT′ and QS, QS′.
Justification:
Join OT and OT′ and OS and OS′.
To prove PT and PT′ are tangents to the circle, we will prove that $\angle$PTO = $\angle$PT′O = 90$°$.
Now OP acts as the diameter of circle II and $\angle$OTP is in the semicircle.
∴ $\angle$OTP = 90º ...(1)
[∵ angle in semicircle is 90º]
But OT is the radius of circle I and line PT touches the circle at T.
ä The line which touches the circle at a point and makes an angle of 90º with radius at that point, is tangent to the circle.
∴ PT is tangent to the circle I at point T through point ‘P’.
Similarly, PT′, QS and QS′ are tangents to the circle I.

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Steps of construction:
1. Draw the rough sketch of required figure.

∵ The tangents make an angle of 60º with each other.
$\angle$OTP = $\angle$OQT = 90º
[Tangent is perpendicular to the radius of circle.]
To find inclination of radii with each other
$\angle$TOQ + $\angle$OTP + $\angle$OQP + $\angle$TPQ = 360º
[angle sum property of quad.]
or $\angle$TOQ + 90º + 90º + 60º = 360º
or $\angle$TOQ = 360º – 90º – 90º – 60º = 120º
2. Draw a circle of radius 5 cm.
3. Draw two radii of circle which make an angle of 120º with each other.
4. The radii intersect the circle at A and B.
5. Make an angle of 90º at each point A and B, which intersect each other at P.

6. PA and PB are the required tangents.

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Steps of construction:
1. Draw a line segment AB = 8 cm.
2. With A as centre and radius 4 cm, draw a circle (I).
3. With B as centre and radius 3 cm, draw a circle (II).
4. Draw the perpendicular bisector of line segment AB which intersects AB at M.
5. With M as centre and radius MA or MB, draw a circle (III) which intersects the circle (I) at S and T and circle (II) at P and Q.
6. Join AP and AQ. These are required tangents to the circle with radius 3 cm from point A. 7. Join BS and BT. These are required tangents to the circle with radius 4 cm from point B.
Justification:
In circle (III), AB acts as diameter then $\angle$ASB and $\angle$BPA are in semicircle.
∴ $\angle$ASB = 90º ...(1) [angle in semicircle] and $\angle$BPA = 90º ...(2)
But $\angle$ASB is angle between radius of circle (I) and line segment BS and $\angle$BPA is angle between radius of circle (II) and line segment AP.
ä Line segment which is perpendicular to the radius of circle, is tangent to the circle through that point.
∴ BS is tangent to circle (I) at point S and AP is tangent to circle (II) at point P.
Similarly, AQ and BT are tangents to the circle (II) and (I) respectively.

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Steps of construction:
1. Construct right angled triangle ABC according to given conditions and measurements.
2. Draw BD ⊥ AC.
3. Take mid-point of side BC take it as M.
4. Take M as centre and BC as diameter, draw a circle through B, C, D using property, angle in semicircle is 90º ($\angle$BDC = 90º). Take this circle as I.
5. Now join A and M.
6. Draw perpendicular bisector of AM to cut AM at N. Now with N as centre and NA or NM as radius, draw a circle (II) which intersects the circle (I) at B and P.
7. Join AP.
8. AP and AB are the required tangents.
Justification of construction:
Line segment AM is diameter of circle (II).
$\angle$APM is in semicircle.
∴ $\angle$APM = 90º [angle in semicircle] i.e., MP ⊥ AP
But MP is the radius of circle (I)
∴ AP is tangent to the circle (II)
[∵ any line ⊥ to radius of circle at any point on the circle is tangent to the circle.]
Similarly, AB is tangent to circle (I).

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To draw circle with bangle means the centre of circle is unknown. First find the centre of circle.

Steps of construction:
1. Draw a circle using a bangle (I).
2. Take any two chords AB and CD (nonparallel) on circle.
3. Draw the perpendicular bisectors of chords AB and CD. The perpendicular bisectors intersect each other at point O.
[∵ any point lying on perpendicular bisector of line segment is equidistant from its end points]
[ä O lies on ⊥ bisector of AB and CD]
∴ OA = OB
and OC = OD
∴ OA = OB
OC = OD (radii of circle)
∴ O is the centre of circle.
4. Take any point P outside the circle.
5. Join OP.
6. Draw the perpendicular bisector of OP. Let M is the mid-point of OP.
7. With M as centre and radius ‘MP’ or ‘MO’, draw a circle II which intersects the circle (I) at T and T′.
8. Join PT and PT′, which is required pair of tangents.
Justification:
Tangent at a point is always perpendicular to the radius at the point. Now, we have to prove that $\angle$PTO = $\angle$PT′O = 90$°$.
Join OT.
Now $\angle$PTO is in the semicircle I.
∴ $\angle$PTO = 90$°$
[angle in semicircle is a right angle]
Similarly, $\angle$PT′O = 90$°$
∴ PT and PT′ has to be the tangents to the circle at T and T′.