RD Sharma Solutions – Class 10 Maths Constructions Ex 11.2

# RD Sharma Solutions – Class 10 Maths Constructions Ex 11.2

## RD Sharma Solutions Chapter 11

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### 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 BC = 5 cm line segment.
(ii) Draw two arcs intersecting at A, one having a radius of 4 cm and the other with a radius of 6 cm.
(iii) Assemble with AB and AC. The triangle is then ABC.
(iv) Cut off three equal segments of a ray BX that forms an acute angle with BC, resulting in BB1 = B1B2= B2B3.
(v) Become a member of B3C.
(vi) If B’C’ is parallel to B3C and C’A’ is parallel to CA, the needed triangle is A’BC’. 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 BC = 7 cm line segment.
(ii) Cut off BA = 5 cm by drawing a ray BX at a 50° angle.
(iii) Become a member of AC. The triangle is then ABC.
(iv) Make an acute angle with BC by drawing a ray BY and cutting off 7 equal pieces to make BB, =B1B2=B2B3=B3B4=B4Bs=B5B6=B6B7
(v) Join forces with B7 and C.
(vi) Make B5C’ and C’A’ parallel to B7C and CA, respectively.
The needed triangle is then A’BC’. Question 3.
Construct a triangle similar to a given ∠ABC such that each of its sides is $$\frac { 2 }{ 3 }$$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 BC = 6 cm line segment.
(ii) Draw a ray BX at a 50° angle and a ray CY at a 60° angle with BC intersecting at A. The triangle is then ABC.
(iii) Draw another ray BZ from B, making an acute angle below BC and intersecting three equal sections to form BB1 =B1B2 = B2B2.
(iv) Become a member of B3C.
(v) Draw B2C’ parallel to B3C and C’A’ parallel to CA, starting with B2.
The needed triangle is then A’BC’. 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 $$\frac { 3 }{ 4 }$$th of the corresponding sides of ΔABC.
Solution:
Steps of construction :
(i) Draw a BC = 6 cm line segment.
(ii) Draw two arcs intersecting at A, one having a radius of 4 cm and the other with a radius of 5 cm.
(iii) Join AB and AC. The triangle is then ABC.
(iv) Cut off four equal segments of a ray BX that forms an acute angle with BC, resulting in BB1= B1B2 = B2B3 = B3B4.
(v) Join B4 and C together.
(vi) Draw C3C’ parallel to B4C from B3C, and C’A’ parallel to CA from C’.
The needed triangle is then A’BC’. Question 5.
Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are $$\frac { 7 }{ 5 }$$ of the corresponding sides of the first triangle.
Solution:
Steps of construction :
(i) Draw a BC = 5 cm line segment.
(ii) Draw arcs intersecting at A with centre B and radius 6 cm and centre C and radius 7 cm.
(iii) Join AB and AC. The triangle is then ABC.
(iv) Cut off 7 equal pieces of a ray BX that forms an acute angle with BC, resulting in BB1 = B1B2 = B2B3 = B3B4 = B4B5 = B5B6 = B6B7.
(v) Join B5 and C together.
(vi) Draw B7C’ parallel to B5C and C’A’ parallel to CA from B7. The needed triangle is then A’BC’. 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 ($$\frac { 5 }{ 4 }$$)th ot the corresponding sides of ΔABC.
Solution:
Steps of construction :
(i) Draw a line segment with the length AB = 4.5 cm.
(ii) Cut off AC = AB = 4.5 cm at A by drawing a ray AX perpendicular to AB.
(iii) Become a member of BC. The triangle is then ABC.
(iv) Create an acute angle with AB by drawing a ray AY and cutting off 5 equal sections, resulting in AA1 = A1A2 = A2A3 =A3A4 = A4A5.
(v) Join A4 and B together.
(vi) Draw 45B’ parallel to A4B and B’C’ parallel to BC starting at 45.
The needed triangle is then AB’C’. 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 $$\frac { 5 }{ 3 }$$ times the corresponding sides of the given triangle. (C.B.S.E. 2008)
Solution:
Steps of construction :
(i) Draw a BC = 5 cm line segment.
(ii) Cut off BA = 4 cm at B by drawing perpendicular BX.
(iii)join Ac, and the triangle ABC is formed.
(iv) BB1 = B1B2 = B2B3 = B3B4 = B4B5 = BB1 = B1B2 = B2B3 = B3B4 = B3B4 = B4B5 = BB1 = B1B2 = B2B3 = B3B4 = B3B4 = B3B4 = B3B4 = B3B4 = B3B
(v) Join B3 and C together.
(vi) Draw B5C’ parallel to B3C and C’A’ parallel to CA, starting at B5.
The needed triangle is then A’BC’. Question 8.
Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are $$\frac { 3 }{ 2 }$$ times the corresponding sides of the isosceles triangle.
Solution:
Steps of construction :
(i) Draw a line segment BC = 8 cm and its perpendicular bisector DX, then cut DA = 4 cm off.
(ii) Become a member of AB and AC. The triangle is then ABC.
(iii) Cut off three equal segments of a ray DY that forms an acute angle with OA, resulting in DD1 = D1D2 =D2D3 = D3D4
(iv) Become a member of D2.
(v) Make D3A’ parallel to D2A and A’B’ parallel to AB meeting BC at C’ and B’.
The needed triangle is then B’A’C’. Question 9.
Draw a ΔABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a trianglewhose sides are $$(\frac { 3 }{ 4 } )$$th of the corresponding sides of the ΔABC.
Solution:
Steps of construction :
(i) Draw a BC = 6 cm line segment.
(ii) Cut off BA = 5 cm at B by drawing a ray BX at a 60° angle with BC.
(iii) Become a member of AC. The triangle is then ABC.
(iv) Make an acute angle with BC by drawing a ray BY and cutting off four equal pieces, BB1= B1B2 B2B3=B3B4.
(v) Join B4 and C together.
(vi) Draw B3C’ parallel to B4C and C’A’ parallel to CA, starting at B3.
The needed triangle is then A’BC’. 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 with the length AB = 4.6 cm.
(ii) At A, draw a ray AX with a 60° angle.
(iii) Draw an arc intersecting AX at C with a radius of 5.1 cm and a centre B.
(iv) Become a member of BC. The triangle is then ABC.
(v) Cut off 5 equal sections of AA1 = A1A2 = A2A3 = A3A4=A4A5 by drawing a ray AX from A and establishing an acute angle with AB.
(vi) Join A4 and B together.
(vii) Draw A5B’ parallel to A4B and B’C’ parallel to BC starting at A5.
The needed triangle is then C’AB’. Question 11.
Construct a triangle similar to a given ΔXYZ with its sides equal to $$(\frac { 3 }{ 2 })$$ th of the corresponding sides of ΔXYZ. Write the steps of construction. [CBSE 1995C] Solution:
Steps of construction :
(i) Make an XYZ triangle with some appropriate data.
(ii) Cut off 5 equal pieces of a ray YL that forms an acute angle with XZ, resulting in YY1= Y1Y2 = Y2Y3 = Y3Y4.
(iii) Assemble Y4 and Z.
(iv) Draw Y3Z’ parallel to Y4Z and Z’X’ parallel to ZX starting at Y3.
The needed triangle is then X’YZ’. 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 $$\frac { 3 }{ 4 }$$ 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 BC = 5.5 cm line segment.
(ii) Draw two arcs that meet at A, one having a radius of 5 cm and the other with a radius of 6.5 cm.
(iii) Form a partnership between BA and CA.
The given triangle is ABC.
(iv) Cut 5 equal sections from BX by drawing a ray BX at an acute angle at B.
(v) Join C5 and make a 3D drawing || 5C meets BC at D.
Draw DE || CA from D to E, where it meets AB. The needed triangle is EBD. 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 QR = 7 cm line segment.
(ii) Draw a ray QX at Q with a 60° angle and a cut of PQ = 6 cm. Join the PR team.
(iii) Cut 5 equal sections from a ray QY that forms an acute angle.
(iv) Connect 5, R and 3, S, drawing 3, S parallel to 5, R and QR at S.
(v) Draw ST || RP through S, meeting PQ at T.
∴Draw ST || RP through S, meeting PQ at T. Question 15.
Draw a ΔABC in which base BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct another triangle whose sides are $$\frac { 3 }{ 4 }$$ of the corresponding sides of ΔABC. [CBSE 2017] Solution:
Steps of construction:

1. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°.
2. Draw a ray BX, which makes an acute angle ∠CBX below the line BC.
3. Locate four points B1, B2, B3and B4 on BX such that BB1 = B1B2=B2B3 = B3B4.
4. Join B4C and draw a line through B3 parallel to B4C intersecting BC to C’.
5. 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 $$\frac { 5 }{ 3 }$$ times the corresponding sides of the given triangle. [CBSE 2017] Solution:
Steps of construction:

1. Draw a right triangle ABC in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. ∠B = 90°.
2. Draw a line BX, which makes an acute angle ∠CBX below the line BC.
3. Locate 5 points B1, B2, B3, B4 and B5 on BX such that BB1 = B1B2=B2B3=B3B4=B4B5.
4. Join B3 to C and draw a line through B5 parallel to B3C, intersecting the extended line segment BC at C’.
5. 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 with the length AB = 5 cm.
(ii) Cut off AE = 3 cm at A by drawing a perpendicular.
(iii) Draw EF || AB from E.
(iv) Draw a ray from B that meets EF at C at an angle of 60.
(v) Become a member of CA. The triangle is then ABC.
(vi) Cut three equal pieces from A by drawing a ray AX that forms an acute angle with AB and creating A1= A1A2 = A2A3.
(vii) Join A2 and B together.
(viii) Draw A’B’ parallel to A2B and B’C’ parallel to BC starting from A.
The needed triangle is then C’AB’. 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 Q12 Q13 ## Related content

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