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## RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.2

**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.

(i)

**(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 23rd 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 34th of the corresponding sides of ΔABC.**

**Solution:**

Steps of construction :

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

Steps of construction :

(i)

**(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 75 of the corresponding sides of the first triangle.**

**Solution:**

Steps of construction :

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 (54)th ot the corresponding sides of ΔABC.**

**Solution:**

**Steps of construction :**

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

(i)

**(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 53 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.

Steps of construction :

(i)

**(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 32 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 (34)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.

(i)

**(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 (32) 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 34 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.

(i)

**(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 34 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 53 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.

(i)

**(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.

**All Chapter RD Sharma Solutions For Class10 Maths**

—————————————————————————–**All Subject NCERT Exemplar Problems Solutions For Class10**

**All Subject NCERT Solutions For Class 10**

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