NCERT Solutions for Class 10 Maths Constructions [Old Book]

NCERT
SOLUTIONS
CLASS - 10th

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Class : 10th
Subject : Maths
Chapter : 11
Chapter Name : Constructions

Exercise 11.1

Q1 In the question, give the justi cation of the construction also Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8.
Measure the two parts.

Answer. A line segment of length 7.6 cm can be divided in the ratio of 5 : 8 as follows
Step 1 Draw line segment AB of 7.6 cm and draw a ray AX making an acute angle with line segment AB .
Step 2 Locate L3=(= 5+8) points, A 1, A 2, A 3, A 4, …. . A 13, on AX such that AA 1 = A 1A 2 = A 2A 3 and so on .
Step 3 Join BA 13.
Step 4 Through the point A 5, draw a line parallel to BA 13 (by making an angle equal to ∠AA 13B) at A 5 intersecting AB at point C.
C is the point dividing line segment AB of 7.6 cm in the required ratio of 5 : 8.
The lengths of AC and CB can be measured. It comes out to 2.9 cm and 4.7 cm respectively.

The construction can be justi ed by proving that
AC 5
CB = 8

By construction, we have A 5C‖A 13B . By applying Basic proportionality theorem for
AC AA 5
CB
= AA
5 13
From the figure, it can be observed that AA 5 and A 5A 13 contain 5 and 8 equal
divisions of line segments respectively.
AA f 5
∴ AA = 8
3 13

On comparing equations (1) and (2), we obtain
AC 5
CB = 8

Page : 219 , Block Name : Exercise 11.1

Q2 In the question, give the justi cation of the construction also , construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle
2
similar to it whose sides are 3 of the corresponding sides of the rst triangle.

Answer. Step 1 Draw a line segment AB = 4 cm. Taking point A as centre, draw and arc of 5 cm radius. Similarly, taking point B as its
centre, draw an arc of 6 cm radius. These arcs will intersect each other at point C. Now, AC = 5 cm and BC=6 cm
And △ABC is the required triangle.
Step 2 Draw a ray AX making an acute angle with line AB on the opposite site of vertex C.
Step 3 Locate 3 points (as 3 is greater between 2 and 3) online AX such that AA 1 = A 1A 2 = A 2A 3.

Step 4 JoinBA 3 and draw a line through A 2 parallel to BA 3 to intersect AB at point B ′.
Step 5 Draw a line through B′ parallel to the line BC to intersect AC at C ′ △AB ′C ′is the required triangle.

The construction can be justi ed by proving that
2 2 2
AB ′ = 3 AB, BC ′ = 3 BC, AC ′ = 3 AC
By construction, we have B ′C ′‖BC
∴ ∠ABC ′ = ∠ABC(Corresponding angles)
△AB ′C ′ and △ABC
∠ABC ′ = ∠ABC( Proved above )
∠B ′AC ′ = ∠BAC( Common )
∴ △ABC ′ − ΔABC( AA similarity criterion )
AB ′ B ′C ′ AC ′
⇒ AB = BC = AC
△AA 2B ′ and △AA 3B
∠A 2AB ′ = ∠A 3AB(Corresponding angles )

∠AA 2B ′ = ∠AA 3B(Corresponding angles)
∴ △AA 2B ′ − ΔAA 3B( AA similarity criterion)
AB ′ AA 2
⇒ AB = AA
3
AB ′ 2
⇒ AB = 3
From equations (1) and (2), we obtain
AB ′ BC ′ AC ′ 2
AB
= BC = AC = 3
2 2 2
⇒ AB ′ = 3 AB, B ′C ′ = 3 BC, AC ′ = 3 AC

Page : 220 , Block Name : Exercise 11.1

Q3 In the question, give the justi cation of the construction also ,construct a triangle with sides 5 cm, 6 cm and 7 cm and then
7
another triangle whose sides are 5 of the corresponding sides of the rst triangle.

Answer. Step 1 Draw a line segment AB of 5 cm. Taking A and B as centre, draw arcs of 6 cm and 5 cm radius respectively. Let these
arcs intersect each other at point C.△ABC is the required triangle having length of sides as 5 cm , 6 cm , 7 cm respectively.
Step 2 Draw a ray AX making acute angle with line AB on the opposite side of vertex C.
Step 3
Locate 7 points, A 1, A 2, A 3, A 4A 5, A 5, A 7( as 7 is greater between 5 and 7), on line AX
such that AA 1 = A 1A 2 = A 2A 3 = A 3A 4 = A 4A 5 = A 5A 6 = A 6A 7
Step 4 Join BA 5 and draw a line through A 7 parallel to BA 5 to intersect extended line segment AB at point B ′.
Step 5 Draw a line through B ′ parallel to BC intersecting the extended line segment AC at C ′.△AB ′C ′ is the required triangle.

The construction can be justi ed by proving that
7 7 7
AB ′ = 5 AB, BC ′ = 5 BC, AC ′ = 5 AC

In △ABC and ΔAB ′C ′
∠ABC = ∠AB ′C ′( Corresponding angles )
∠BAC = ∠B ′AC ′( Common )
∴ △ABC − △AB ′C ′
AB BC AC
⇒ = = …
AB ′ B ′C ′ AC ′
In △AA 5B and △AA 7B ′

∠A 5AB = ∠A 7AB ′( Common )
∠AA sB = ∠AA 7B ′( Corresponding angles )
∴ △AA 5B − △AA 7B ′ (AA similarity criterion)
AB AA 5
⇒ = AA
AB ′ 7

AB 5
⇒ = 7
AB ′
On comparing equations (1) and (2), we obtain
AB BC AC 5
= = = 7
AB ′ BC ′ AC ′
7 7 7
⇒ AB ′ = 5 AB, B ′C = 5 BC, AC = 5 AC

Page : 220 , Block Name : Exercise 11.1

Q4
In the question, give the justification of the construction also , construct an isosceles triangle whose base is 8cm and altitude 4cm and then another
1
triangle whose sides are 1 2 times the corresponding sides of the isosceles triangle.

Answer. Let us assume that ∠ABC is an isosceles triangle having CA and CB of equal lengths, base AB of 8 crn, and AD is the altitude
3
of 4 cm. A△AB ′C ′whose sides are 2 times of can be drawn as follows.
Step 1 Draw a line segment Aa of 8 cm. Draw arcs of same radius on both sides of the line segment while taking point A and as its
centre. Let these arcs intersect each other at O and O'. Join OO' Let OO' intersect AB at D.
Step 2 Taking D as centre, draw an arc of 4 cm radius which cuts the extended line segment OO' at point C. An isosceles AA 3C is
formed, having CD (altitude) as 4 cm and AB (base) as 8 cm.
Step 3 Draw a ray AX making an acute angle with line segment AB on the opposite side of vertex C.
Step 4 Locate 3 points (as 3 is greater between 3 and 2) A 1, A 2, and A 3 on AX such that AA 1 = A 1A 2 = A 2A 3.
Step 5 Join BA 2 and draw a line through A 3 parallel to BA 2 to intersect extended line segment AB at point B ′.
Step 6 Draw a line through B ′ parallel to BC intersecting the extended the segment AC at C ′. ΔAB ′C ′ Is the required triangle.

The construction can be justi ed by proving that
3 3 3
AB ′ = 2 AB, BC ′ = 2 BC, AC ′ = 2 AC
In △ABC and ΔAB ′C ′
∠ABC = ∠AB ′C ′ (Corresponding angles)
∠BAC = ∠B ′AC ′( Common )
∴ △ABC − ΔAB ′C ′ (AA similarity criterion)
AB BC AC
⇒ = =
AB ′ BC ′ AC ′

In △AA 2B and ΔAA 3B ′
∠A 2AB = ∠A 3AB ′( Corresponding angles )

∠AA 2B = ∠AA 3B ′ (Corresponding angles)
∴ △AA 2B − △AA 3B ′ (AA similarity criterion)
AB AA 2
⇒ AB = AA
3

AB 2
⇒ = 3
AB ′
On comparing equations (1) and (2), we obtain
AB BC AC 2
AB
= = = 3
BC ′ AC ′
3 3 3
⇒ AB ′ = 2 AB, BC = 2 BC, AC = 2 AC

Page : 220 , Block Name : Exercise 11.1

Q5 In the question, give the justi cation of the construction also , draw a triangle ABC with side BC = 6 cm, AB=5 cm and ∠ABC=60 ∘ .
3
Then construct a triangle whose sides are 4 of the corresponding sides of the triangle ABC.

3
Answer. A △A ′BC ′ whose sides 4 of the corresponding sides of can be drawn as follows.
Step 1 Draw a △ABC with side BC = 6 crn, AB = 5 cm and ∠ABC = 60 ∘ .
Step 2 Draw a ray BX making an acute angle with BC on the opposite side of vertex A. Step 3 Locate 4 points (as 4 is greater in 3 and 4),
B 1, B 2, B 3, B 4, on line segment BX.
Step 4 Join B 4Cand draw a line through B 3 , parallel to B 4C intersecting BC at C ′
Step 5 Draw a line through C' parallel to AC intersecting AB at A'.△A ′BC ′ is the required triangle.

The construction can be justi ed by proving that,
3 3 3
AB = 4 AB, BC ′ = 4 BC, AC ′ = 4 AC
In △A ′BC ′ and △ABC ,
∠A ′C ′B = ∠ACB (Corresponding angles)
∠A ′BC ′ = ∠ABC( Common)
∴ △A ′BC ′ ∼ △ABC( AA similarity criterion )
AB BC AC
⇒ AB = BC = AC
In △BB 3C ′ and △BB 4C

∠B 3BC ′ = ∠B 4BC(Common)
∠BB 3C ′ = ∠BB 4C( Corresponding angles)

∴ △BB 3C ′ ∼ ΔBB 4C( AA similarity criterion)
BC ′ BB 3
⇒ BC = BB
4

BC ′ 3
⇒ BC = 4

From equations (1) and (2), we obtain
AB BC A ′C ′ 3
AB = BC = AC = 4
3 3 3
⇒ AB = 4 AB, BC ′ = 4 BC, AC ′ = 4 AC

Page : 220 , Block Name : Exercise 11.1

Q6
In the question, give the justification of the construction also , draw a triangle ABC with side BC = 7cm, ∠B = 45 ∘ , ∠A = 105 ∘ . Then, construct a
4
triangle whose sides are 3 times the corresponding sides of Δ ABC.

Answer.
∠B = 45 ∘ , ∠A = 105 ∘
Sum of all interior angles in a triangle is 180 ∘ .
∠A + ∠B + ∠C = 180 ∘
105 ∘ + 45 ∘ + ∠C = 180 ∘
∠C = 180 ∘ − 150 ∘
∠C = 30 ∘
The required triangle can be drawn as follows.
Step 1
Draw a △ABC with side BC = 7cm, ∠B = 45 ∘ , ∠C = 30 ∘
Step 2
Draw a ray BX making an acute angle with BC on the opposite side of vertex A .
Step 3
JoinB 3C. Draw a line through B 4 parallel to B 3C intersecting extended BC at C ′.
Step 4
Through C ′, draw a line parallel to AC intersecting extended line segment at C ′.
△A ′BC ′ is the required triangle.

The construction can be justi ed by proving that
4 4 4
A ′B = 3 AB, BC ′ = 3 BC, AC ′ = 3 AC

In △ABC and △A ′BC ′
∠ABC = ∠A ′BC ′( Correspon )
∠ACB = ∠A ′C ′B( Corresponding angles )
∴ △ABC − △A ′BC ′ (AA similarity criterion)

AB BC AC
⇒ = =
A ′B BC ′ A ′C ′

In ΔBB 3C and ΔBB 4C ′

∠B 3BC = ∠B 4BC ′( Common )

∠BB 3C = ∠BB 4C ′( Corresponding angles)

∴ △BB 3C − ΔBB 4C ′( AA similarity criterion)
BC BB 3
⇒ = BB
BC ′ 4
BC 3
⇒ = 4
BC ′
On comparing equations (1) and (2), we obtain
AB BC AC 3
= = = 4
A ′B BC ′ A ′C ′
4 4 4
⇒ AB = 3 AB, BC ∗ = 3 BC, AC ′ = 3 AC

Page : 220 , Block Name : Exercise 11.1

Q7 In the question, give the justi cation of the construction also , draw a right triangle in which the sides (other than hypotenuse) are
5
of lengths 4 cm and 3 cm . Then construct another triangle whose sides are 3 times the corresponding sides of the given triangle.

Answer. It is given that sides other than hypotenuse are of lengths 4 cm and 3 cm . Clearly,these will be perpendicular to each other.
The required triangle can be drawn as follows.
Step 1
Draw a line segment AB=4 cm . Draw a ray SA making 90 ∘ with it.
Step 2
Draw an arc of 3 cm radius while taking A as its centre to intersect SA at C . Join BC .△ABC is the required triangle.
Step 3
Draw a ray AX making an acute angle with AB , opposite to vertex C.
Step 4
Locate 5 points as 5 is greater in 5 and 3, A 1, A 2, A 3, As, on line segment AX such that AA 1 = A 1A 2 = A 2A 3 = A 3A 4 = A 4A 5
Step 5
Join A 3 B . Draw a line through A 5 parallel to A 3 B intersecting extended line segment AB at B ′
Step 6
AB AA 3
⇒ = AA
AB ′ 5

AB 3
⇒ = 5
AB ′

On comparing equations (1) and (2), we obtain
AB BC AC 3
AB
= = = 5
BC ′ AC ′
5 5 5
⇒ AB ′ = 3 AB, BC ′ = 3 BC, AC ′ = 3 AC

Through B', draw a line parallel to BC intersecting extended line segment AC at C .
△AB ′C ′ is the required triangle.

5 5 5
The construction can be justi ed by proving that AB ′ = 3 AB, BC ′ = 3 BC, AC ′ = 3 AC In ΔABC and ΔAB ′C ′
∠ABC = ∠AB ′C ′( Corresponding angles )
∠BAC = ∠B ′AC ′( Common )
∴ △ABC − ΔAB ′C ′ (AA similarity criterion)
AB BC AC
⇒ = =
AB ′ B ′C ′ AC ′
In △AA 3B and △AA 5B ′

∠A 3AB = ∠A 5AB ′( Common )
∠AA 3B = ∠AA 5B ′( Corresponding angles )

∴ △AA 3B − △AA 5B ′ (AA similarity criterion)

Page : 220 , Block Name : Exercise 11.1

Exercise 11.2

Q1 In question , give also the justi cation of the construction : 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.

Answer. A pair of tangents to the given circle can be constructed as follows.
Step 1 Taking ny point O of the given plane as centre, draw a circle of 6 cm radius Locate a point P, 10 Jn 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 the required tangents.

The lengths of tangents PQ and PR are 8 cm each.

The construction can be justi ed 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.

∠PQO is an angle in the semi-circle. We know that angle in a semicircle is a right
angle.
∴ ∠PQO = 90 ∘
⇒ ∠PQ ⊥ 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

Page : 221 , Block Name : Exercise 11.2

Q2 In question , give also the justi cation of the construction : 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.

Answer. 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 5 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-pont 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.
Step 5 Join PQ and PR. PQ and PR are the required tangents.

It can be observed that PQ and PR are of length 4.47cm each.
In ΔPQO ,
since PQ is a tangent,
∠PQO = 90 ∘
PO = 6cm
QO = 4cm
Applying Pythagoras theorem in ΔPQO, we obtain
PQ 2 + QO 2 = PQ 2
PQ 2 + (4) 2 = (6) 2
PQ 2 + 16 = 36
PQ 2 + 16 = 36
PQ 2 = 36 − 16
PQ = 2√5
PQ = 4.47cm

The construction can be justi ed by proving that PQ and PR are the tangents to the circle (whose centre is O and radius is 4 crn). For
this, let us join OQ and OR.

∠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.

Page : 221 , Block Name : Exercise 11.2

Q3 In question , give also the justi cation of the construction : Draw a circle of radius 3 crn, 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.

Answer. 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 crn 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 m d-points of OR and OS respectively.
Step 4 Taking T and U as its centre and Kith 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.

The construction can be justi ed 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.

Page : 221 , Block Name : Exercise 11.2

Q4 In question , give also the justi cation of the construction: Draw a pair of tangents to a circle of radius 5 cm which are inclined to
each other at an angle of 60^{\circ} .

Answer. The tangents can be constructed in the following manner:
Step 1 Draw a circle of radius 5 cm and with centre as O.
Step 2 Take a point A on the circumference of the circle end join OA. Draw a perpendicular to OA at point A.

(
Step 3 Draw a radius OB, making an angle of 120 ∘ 180 ∘ − 60 ∘ with OA. )
Step 4 Draw a perpendicular to OB at point B. Let both the perpendiculars intersect at point P, PA and PB are the required tangents at
angle of 60 ∘

The construction can be justi ed by proving that ∠APB = 60 ∘
By our construction
∠OAP = 90 ∘
∠OBP = 90 ∘
And ∠AOB = 120 ∘
We know that the sum of all interior angles of a quadrilateral = 360 ∘
∠OAP + ∠AOB + ∠OBP + ∠APB = 360 ∘
90 ∘ + 120 ∘ + 90 ∘ + ∠APB = 360 ∘
∠APB = 60 ∘

Page : 221 , Block Name : Exercise 11.2

Q5 In question , give also the justi cation of the construction: Draw a line segment AB of length 8 cm . Taking A as centre, draw a
circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm . Construct tangents to each circle from the centre of
the other circle.

Answer. The tangents can be constructed on the given circles as follows.
Step 1 Draw a line segment AB of 8 cm. Taking A and B as centre, draw two circles of 4 cm and 3 cm radius.
Step 2 Bisect the line AS. Let the mid-point of AB be C. Taking C as centre, draw a circle of AC radius which will intersect the circles at
points P, Q, R, and S. Join AP, AQ, AS, and AR, These are the required tangents.

The construction can be justi ed by proving that AS and AR are the tangents of the circle (whose centre is A and radius is 3 cm ) and
BP and BQ are the tangents of the circle (whose centre is A and radius is 4 cm ). For this , join AP, AQ , BS and BR.

∠ASB is an angle in the serni-circle. We know that an angle in a semi-circle is a right angle.
∴ ∠ASB = 90 ∘
⇒ BS ⊥ AS
Since BS is the radius of the circle, AS has to be a tangent of the circle. Similarly, AR, BP, and BQ are the tangents.

Page : 221 , Block Name : Exercise 11.2

Q6
In question , give also the justification of the construction: Let ABC be a right triangle in which AB = 6cm, BC = 8cm and ∠B = 90 ∘ . BD is the
perpendicular from B on AC . The circle through B, C, D is drawn. Construct the tangents
from A to this circle.

Answer. Consider the following situation. If a circle is drawn through B, D, and C, BC will be its diameter as ∠BDC is of measure 90 ∘ .

The centre E of this circle will be the mid- point of BC.

The required tangents can be constructed on the given circle as follows.
Step 1 Join AE and bisect it. Let F be the mid-point of AE.
Step 2 Taking F as centre and FE as its radius, draw a circle which will intersect the circle at point B and G. join AG. AB and AG are the
required tangents.

The construction can be justi ed by proving that AG and Aa are the tangents to the circle. For this, join EC.

∠AGE is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle.
∴ ∠AGE = 90 ∘
⇒ EG ⊥ AG
Since EG is the radius of the circle, AG has to be a tangent of the circle.
Already, ∠B = 90 ∘
⇒ AB ⊥ BE
Since BE is the radius of the circle, AB has to be a tangent of the circle.

Page : 222 , Block Name : Exercise 11.2

Q7 In question , give also the justi cation of the construction: Draw a circle with the help of a bangle. Take a point outside the circle.
Construct the pair of tangents from this point to the circles.

Answer. The required tangents can be constructed on the given circle as follows. Step 1 Draw a circle with the help of a bangle.
Step 2 Take a point P outside this circle and take two chords QR and ST.
Step 3 Draw perpendicular bisectors of these chords. Let them intersect each other at point O.
Step 4 Join PO and bisect it. Let U be the mid-point of PO. Taking U as centre, draw a circle of radius OU, which will intersect the circle
at V and W. Join PV and PW.
PV and PW are the required tangents.

The construction can be justi ed by proving that PV and PW are the tangents to the circle. For this, rst of all, it has to be proved that
O is the centre of the circle. Let us join OV and OW.

We know that perpendicular bisector of a chord passes through the centre. Therefore, the perpendicular bisector of chords QR and ST
pass through the centre. It is clear that the intersection point of these perpendicular bisectors is the centre of the circle ∠PVO is an
angle in the serni-circle. We know that an angle in a serni-circle is a right angle.
∴ ∠PVO = 90 ∘
⇒ OV ⊥ PV
Since OV is the radius of the circle, PV has to be e tangent of the circle. Similarly, PW is a tangent of the circle.

Page : 222 , Block Name : Exercise 11.2