# RD Sharma Class 10 Solutions Chapter 11 Constructions Exercise 11.1 and Exercise 11.2 and Exercise 11.3

### RD Sharma Class 10 Solutions Constructions Exercise 11.1

Question 1.

Determine a point which divides a line segment of length 12 cm internally in the ratio 2 : 3. Also justify your construction.

Solution:

Steps of construction :

(i) Draw a line segment AB = 12 cm.

(ii) Draw a ray AX at A making an acute angle with AB.

(iii) From B, draw another ray BY parallel to AX.

(iv) Cut off 2 equal parts from AX and 3 equal parts from BY.

(v) Join 2 and 3 which intersects AB at P.

P is the required point which divides AB in the ratio of 2 : 3 internally.

Question 2.

Divide a line segment of length 9 cm internally in the ratio 4 : 3. Also, give justification of the construction.

Solution:

Steps of construction :

(i) Draw a line segment AB = 9 cm.

(ii) Draw a ray AX making an acute angle with AB.

(iii) From B, draw another ray BY parallel to AX.

(iv) Cut off 4 equal parts from AX and 3 parts from BY.

(v) Join 4 and 3 which intersects AB at P.

P is the required point which divides AB in the ratio of 4 : 3 internally.

Question 3.

Divide a line segment of length 14 cm internally in the ratio 2 : 5. Also justify your construction.

Solution:

Steps of construction :

(i) Draw a line segment AB = 14 cm.

(ii) Draw a ray AX making an acute angle with AB.

(iii) From B, draw another ray BY parallel to AX.

(iv) From AX, cut off 2 equal parts and from B, cut off 5 equal parts.

(v) Join 2 and 5 which intersects AB at P.

P is the required point which divides AB in the ratio of 2 : 5 internally.

Question 4.

Draw a line segment of length 8 cm and divide it internally in the ratio 4 : 5.

Solution:

Steps of construction :

(i) Draw a line segment AB = 8 cm.

(ii) Draw a ray AX making an acute angle with ∠BAX = 60° withAB.

(iii) Draw a ray BY parallel to AX by making an acute angle ∠ABY = ∠BAX.

(iv) Mark four points A_{1}, A_{2}, A_{3}, A_{4} on AX and five points B_{1}, B_{2}, B_{3}, B_{4}, B_{s} on BY in such a way that AA_{1} = A_{1}A_{2} = A_{2}A_{3} = A_{3}A_{4} .

(v) Join A_{4}B_{5}.

(vi) Let this line intersect AB at a point P.

Thus, P is the point dividing the line segment AB internally in the ratio of 4 : 5.

RD Sharma Class 10 Solutions Chapter 11 Constructions Exercise 11.1

### Q1.

RD Sharma Class 10 Solutions Chapter 11 Constructions Exercise 11.1 Q2

RD Sharma Class 10 Solutions Chapter 11 Constructions Exercise 11.1 Q3

## RD Sharma Class 10 Solutions Chapter 11 Constructions Exercise 11.2

### RD Sharma Class 10 Solutions Constructions Exercise 11.2

Question 1.

Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are (2/3) of the corresponding sides of it.

Solution:

Steps of construction :

(i) Draw a line segment BC = 5 cm.

(ii) With centre B and radius 4 cm and with centre C and radius 6 cm, draw arcs intersecting each other at A.

(iii) Join AB and AC. Then ABC is the triangle.

(iv) Draw a ray BX making an acute angle with BC and cut off 3 equal parts making BB_{1} = B_{1}B_{2}= B_{2}B_{3}.

(v) Join B_{3}C.

(vi) Draw B’C’ parallel to B_{3}C and C’A’ parallel to CA then ΔA’BC’ is the required triangle.

Question 2.

Construct a triangle similar to a given ΔABC such that each of its sides is (5/7)^{th }of the corresponding sides of ΔABC. It is given that AB = 5 cm, BC = 7 cm and ∠ABC = 50°.

Solution:

Steps of construction :

(i) Draw a line segment BC = 7 cm.

(ii) Draw a ray BX making an angle of 50° and cut off BA = 5 cm.

(iii) Join AC. Then ABC is the triangle.

(iv) Draw a ray BY making an acute angle with BC and cut off 7 equal parts making BB, =B_{1}B_{2}=B_{2}B_{3}=B_{3}B_{4}=B_{4}B_{s}=B_{5}B_{6}=B_{6}B_{7 }(v) Join B_{7} and C

(vi) Draw B_{5}C’ parallel to B_{7}C and C’A’ parallel to CA.

Then ΔA’BC’ is the required triangle.

Question 3.

Construct a triangle similar to a given ∠ABC such that each of its sides is rd of the corresponding sides of ΔABC. It is given that BC = 6 cm, ∠B = 50° and ∠C = 60°.

Solution:

Steps of construction :

(i) Draw a line segment BC = 6 cm.

(ii) Draw a ray BX making an angle of 50° and CY making 60° with BC which intersect each other at A. Then ABC is the triangle.

(iii) From B, draw another ray BZ making an acute angle below BC and intersect 3 equal parts making BB_{1} =B_{1}B_{2} = B_{2}B_{2}

(iv) Join B_{3}C.

(v) From B_{2}, draw B_{2}C’ parallel to B_{3}C and C’A’ parallel to CA.

Then ΔA’BC’ is the required triangle.

Question 4.

Draw a ΔABC in which BC = 6 cm, AB = 4 cm and AC = 5 cm. Draw a triangle similar to ΔABC with its sides equal to th of the corresponding sides of ΔABC.

Solution:

Steps of construction :

(i) Draw a line segment BC = 6 cm.

(ii) With centre B and radius 4 cm and with centre C and radius 5 cm, draw arcs’intersecting eachother at A.

(iii) Join AB and AC. Then ABC is the triangle,

(iv) Draw a ray BX making an acute angle with BC and cut off 4 equal parts making BB_{1}= B_{1}B_{2 }= B_{2}B_{3} = B_{3}B_{4}.

(v) Join B_{4} and C.

(vi) From B_{3}C draw C_{3}C’ parallel to B_{4}C and from C’, draw C’A’ parallel to CA.

Then ΔA’BC’ is the required triangle.

Question 5.

Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are of the corresponding sides of the first triangle.

Solution:

Steps of construction :

(i) Draw a line segment BC = 5 cm.

(ii) With centre B and radius 6 cm and with centre C and radius 7 cm, draw arcs intersecting eachother at A.

(iii) Join AB and AC. Then ABC is the triangle.

(iv) Draw a ray BX making an acute angle with BC and cut off 7 equal parts making BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4} = B_{4}B_{5} = B_{5}B_{6} = B_{6}B_{7}.

(v) Join B_{5} and C.

(vi) From B_{7}, draw B_{7}C’ parallel to B_{5}C and C’A’ parallel CA. Then ΔA’BC’ is the required triangle.

Question 6.

Draw a right triangle ABC in which AC = AB = 4.5 cm and ∠A = 90°. Draw a triangle similar to ΔABC with its sides equal to ()th ot the corresponding sides of ΔABC.

Solution:

Steps of construction :

(i) Draw a line segment AB = 4.5 cm.

(ii) At A, draw a ray AX perpendicular to AB and cut off AC = AB = 4.5 cm.

(iii) Join BC. Then ABC is the triangle.

(iv) Draw a ray AY making an acute angle with AB and cut off 5 equal parts making AA_{1} = A_{1}A_{2} = A_{2}A_{3} =A_{3}A_{4} = A_{4}A_{5}_{ }(v) Join A_{4} and B.

(vi) From 45, draw 45B’ parallel to A_{4}B and B’C’ parallel to BC.

Then ΔAB’C’ is the required triangle.

Question 7.

Draw a right triangle in which the sides (other than hypotenuse) are of lengths 5 cm and 4 cm. Then construct another triangle whose sides are times the corresponding sides of the given triangle. (C.B.S.E. 2008)

Solution:

Steps of construction :

(i) Draw a line segment BC = 5 cm.

(ii) At B, draw perpendicular BX and cut off BA = 4 cm.

(iii )join Ac , then ABC is the triangle

(iv) Draw a ray BY making an acute angle with BC, and cut off 5 equal parts making BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4} = B_{4}B_{5}

(v) Join B_{3} and C.

(vi) From B_{5}, draw B_{5}C’ parallel to B_{3}C and C’A’ parallel to CA.

Then ΔA’BC’ is the required triangle.

Question 8.

Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are times the corresponding sides of the isosceles triangle.

Solution:

Steps of construction :

(i) Draw a line segment BC = 8 cm and draw its perpendicular bisector DX and cut off DA = 4 cm.

(ii) Join AB and AC. Then ABC is the triangle.

(iii) Draw a ray DY making an acute angle with OA and cut off 3 equal parts making DD_{1} = D_{1}D_{2} =D_{2}D_{3} = D_{3}D_{4 }(iv) Join D_{2 }(v) Draw D_{3}A’ parallel to D_{2}A and A’B’ parallel to AB meeting BC at C’ and B’ respectively.

Then ΔB’A’C’ is the required triangle.

Question 9.

Draw a ΔABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a trianglewhose sides are th of the corresponding sides of the ΔABC.

Solution:

Steps of construction :

(i) Draw a line segment BC = 6 cm.

(ii) At B, draw a ray BX making an angle of 60° with BC and cut off BA = 5 cm.

(iii) Join AC. Then ABC is the triangle.

(iv) Draw a ray BY making an acute angle with BC and cut off 4 equal parts making BB_{1}= B_{1}B_{2} B_{2}B_{3}=B_{3}B_{4}.

(v) Join B_{4} and C.

(vi) From B_{3}, draw B_{3}C’ parallel to B_{4}C and C’A’ parallel to CA.

Then ΔA’BC’ is the required triangle.

Question 10.

Construct a triangle similar to ΔABC in which AB = 4.6 cm, BC = 5.1 cm,∠A = 60° with scale factor 4 : 5.

Solution:

Steps of construction :

(i) Draw a line segment AB = 4.6 cm.

(ii) At A, draw a ray AX making an angle of 60°.

(iii) With centre B and radius 5.1 cm draw an arc intersecting AX at C.

(iv) Join BC. Then ABC is the triangle.

(v) From A, draw a ray AX making an acute angle with AB and cut off 5 equal parts making AA_{1} = A_{1}A_{2} = A_{2}A_{3} = A_{3}A_{4}=A_{4}A_{5}.

(vi) Join A_{4} and B.

(vii) From A_{5}, drawA_{5}B’ parallel to A_{4}B and B’C’ parallel to BC.

Then ΔC’AB’ is the required triangle.

Question 11.

Construct a triangle similar to a given ΔXYZ with its sides equal to th of the corresponding sides of ΔXYZ. Write the steps of construction. [CBSE 1995C]

Solution:

Steps of construction :

(i) Draw a triangle XYZ with some suitable data.

(ii) Draw a ray YL making an acute angle with XZ and cut off 5 equal parts making YY_{1}= Y_{1}Y_{2} = Y_{2}Y_{3} = Y_{3}Y_{4}.

(iii) Join Y_{4} and Z.

(iv) From Y_{3}, draw Y_{3}Z’ parallel to Y_{4}Z and Z’X’ parallel to ZX.

Then ΔX’YZ’ is the required triangle.

Question 12.

Draw a right triangle in which sides (other than the hypotenuse) are of lengths 8 cm and 6 cm. Then construct another triangle whose sides are times the corresponding sides of the first triangle.

Solution:

(i) Draw right ΔABC right angle at B and BC = 8 cm and BA = 6 cm.

(ii) Draw a line BY making an a cut angle with BC and cut off 4 equal parts.

(iii) Join 4C and draw 3C’ || 4C and C’A’ parallel to CA.

The BC’A’ is the required triangle.

Question 13.

Construct a triangle with sides 5 cm, 5.5 cm and 6.5 cm. Now construct another triangle, whose sides are 3/5 times the corresponding sides of the given triangle. [CBSE 2014]

Solution:

Steps of construction:

(i) Draw a line segment BC = 5.5 cm.

(ii) With centre B and radius 5 cm and with centre C and radius 6.5 cm, draw arcs which intersect each other at A

(iii) Join BA and CA.

ΔABC is the given triangle.

(iv) At B, draw a ray BX making an acute angle and cut off 5 equal parts from BX.

(v) Join C5 and draw 3D || 5C which meets BC at D.

From D, draw DE || CA which meets AB at E.

∴ ΔEBD is the required triangle.

Question 14.

Construct a triangle PQR with side QR = 7 cm, PQ = 6 cm and ∠PQR = 60°. Then construct another triangle whose sides are 3/5 of the corresponding sides of ΔPQR. [CBSE 2014]

Solution:

Steps of construction:

(i) Draw a line segment QR = 7 cm.

(ii) At Q draw a ray QX making an angle of 60° and cut of PQ = 6 cm. Join PR.

(iii) Draw a ray QY making an acute angle and cut off 5 equal parts.

(iv) Join 5, R and through 3, draw 3, S parallel to 5, R which meet QR at S.

(v) Through S, draw ST || RP meeting PQ at T.

∴ ΔQST is the required triangle.

Question 15.

Draw a ΔABC in which base BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct another triangle whose sides are of the corresponding sides of ΔABC. [CBSE 2017]

Solution:

Steps of construction:

- Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°.
- Draw a ray BX, which makes an acute angle ∠CBX below the line BC.
- Locate four points B
_{1}, B_{2}, B_{3}and B_{4 }on BX such that BB_{1}= B_{1}B_{2}=B_{2}B_{3}= B_{3}B_{4}. - Join B
_{4}C and draw a line through B_{3}parallel to B_{4}C intersecting BC to C’. - Draw a line through C’ parallel to the line CA to intersect BA at A’.

Question 16.

Draw a right triangle in which the sides (other than the hypotenuse) arc of lengths 4 cm and 3 cm. Now, construct another triangle whose sides are times the corresponding sides of the given triangle. [CBSE 2017]

Solution:

Steps of construction:

- Draw a right triangle ABC in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. ∠B = 90°.
- Draw a line BX, which makes an acute angle ∠CBX below the line BC.
- Locate 5 points B
_{1}, B_{2}, B_{3}, B_{4}and B_{5}on BX such that BB_{1}= B_{1}B_{2}=B_{2}B_{3}=B_{3}B_{4}=B_{4}B_{5}. - Join B
_{3}to C and draw a line through B_{5}parallel to B_{3}C, intersecting the extended line segment BC at C’. - Draw a line through C’ parallel to CA intersecting the extended line segment BA at A’.

Question 17.

Construct a ΔABC in which AB = 5 cm, ∠B = 60°, altitude CD = 3 cm. Construct a ΔAQR similar to ΔABC such that side of ΔAQR is 1.5 times that of the corresponding sides of ΔACB.

Solution:

Steps of construction :

(i) Draw a line segment AB = 5 cm.

(ii) At A, draw a perpendicular and cut off AE = 3 cm.

(iii) From E, draw EF || AB.

(iv) From B, draw a ray making an angle of 60 meeting EF at C.

(v) Join CA. Then ABC is the triangle.

(vi) From A, draw a ray AX making an acute angle with AB and cut off 3 equal parts making A A_{1}= A_{1}A_{2} = A_{2}A_{3}.

(vii) Join A_{2} and B.

(viii) From A , draw A^B’ parallel to A_{2}B and B’C’ parallel toBC.

Then ΔC’AB’ is the required triangle.

RD Sharma Class 10 Solutions Chapter 11 Constructions Exercise 11.2 Q1

RD Sharma Class 10 Solutions Chapter 11 Constructions Exercise 11.2 Q2

RD Sharma Class 10 Solutions Chapter 11 Constructions Exercise 11.2 Q3

RD Sharma Class 10 Solutions Chapter 11 Constructions Exercise 11.2 Q4

Chapter 11 Constructions Exercise 11.2 Q5

RD Sharma Class 10 Solutions Chapter 11 Constructions Exercise 11.2 Q6

Constructions Exercise 11.2 Q7

Chapter 11 Constructions Exercise 11.2 Q8

Constructions Exercise 11.2 Q9

Constructions Exercise 11.2 Q10

Q11

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Q13

## RD Sharma Class 10 Solutions Chapter 11 Constructions Exercise 11.3

### RD Sharma Class 10 Solutions Constructions Exercise 11.3

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.

Solution:

Steps of construction :

(i) Draw a circle with O centre and 6 cm radius.

(ii) Take a point P, 10 cm away from the centre O.

(iii) Join PO and bisect it at M.

(iv) With centre M and diameter PO, draw a circle intersecting the given circle at T and S.

(v) Join PT and PS.

Then PT and PS are the required tangents.

Question 2.

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.

Solution:

Steps of construction :

(i) Draw a circle with centre O and radius 3 cm.

(ii) Draw a diameter and produce it to both sides.

(iii) Take two points P and Q on this diameter with a distance of 7 cm each from the centre O.

(iv) Bisect PO at M and QO at N

(v) With centres M and N, draw circle on PO and QO as diameter which intersect the given circle at S, T and S’, T’ respectively.

(vi) Join PS, PT, QS’ and QT’.

Then PS, PT, QS’ and QT’ are the required tangents to the given circle.

Question 3.

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. [CBSE 2013]

Solution:

Steps of construction :

(i) Draw a line segment AB = 8 cm.

(ii) With centre A and radius 4 cm and with centre B and radius 3 cm, circles are drawn.

(iii) Bisect AB at M.

(iv) With centre M and diameter AB, draw a circle which intersects the two circles at S’, T’ and S, T respectively.

(v) Join AS, AT, BS’and BT’.

Then AS, AT, BS’ and BT’ are the required tangent.

Question 4.

Draw two tangents to a circle of raidus 3.5 cm from a point P at a distance of 6.2 cm from its centre.

Solution:

Steps of construction :

(i) Draw a circle with centre O and radius 3.5 cm

(ii) Take a point P which is 6.2 cm from O.

(iii) Bisect PO at M and draw a circle with centre M and diameter OP which intersects the given circle at T and S respectively.

(iv) Join PT and PS.

PT and PS are the required tangents to circle.

Question 5.

Draw a pair of tangents to a circle of radius 4.5 cm, which are inclined to each other at an angle of 45°. [CBSE 2013]

Solution:

Steps of construction :

Angle at the centre 180° – 45° = 135°

(i) Draw a circle with centre O and radius 4.5 cm.

(ii) At O, draw an angle ∠TOS = 135°

(iii) At T and S draw perpendicular which meet each other at P.

PT and PS are the tangents which inclined each other 45°.

Question 6.

Draw a right triangle ABC in which AB = 6 cm, BC = 8 cm and ∠B = 90°. Draw BD perpendicular from B on AC and draw a circle passing through the points B, C and D. Construct tangents from A to this circle.

Solution:

Steps of Construction :

Draw a line segment BC = 8 cm

From B draw an angle of 90°

Draw an arc = 6cm cutting the angle at A.

Join AC.

ΔABC is the required A.

Draw ⊥ bisector of BC cutting BC at M.

Take M as centre and BM as radius, draw a circle.

Take A as centre and AB as radius draw an arc cutting the circle at E. Join AE.

AB and AE are the required tangents.

Justification :

∠ABC = 90° (Given)

Since, OB is a radius of the circle.

∴ AB is a tangent to the circle.

Also AE is a tangent to the circle.

Question 7.

Draw two concentric circles of radii 3 cm and 5 cm. Construct a tangent to the smaller circle from a point on the larger circle. Also, measure its length. [CBSE 2016]

Solution:

Given, two concentric circles of radii 3 cm and 5 cm with centre O. We have to draw pair of tangents from point P on outer circle to the other.

Steps of construction :

(i) Draw two concentric circles with centre O and radii 3 cm and 5 cm.

(ii) Taking any point P on outer circle. Join OP.

(iii) Bisect OP, let M’ be the mid-point of OP.

Taking M’ as centre and OM’ as radius draw a circle dotted which cuts the inner circle as M and P’.

(iv) Join PM and PP’. Thus, PM and PP’ are the required tangents.

(v) On measuring PM and PP’, we find that PM = PP’ = 4 cm.

Actual calculation:

In right angle ΔOMP, ∠PMO = 90°

∴ PM^{2} = OP^{2} – OM^{2 }[by Pythagoras theorem i.e. (hypotenuse)^{2} = (base)^{2} + (perpendicular)^{2}]

⇒ PM^{2} = (5)^{2} – (3)^{2} = 25 – 9 = 16

⇒ PM = 4 cm

Hence, the length of both tangents is 4 cm.

### RD Sharma Class 10th Solutions Chapter 11 Constructions Exercise 11.3 Q1

RD Sharma Class 10 Solutions Chapter 11 Constructions Exercise 11.3 Q2

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