RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.2

RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.2

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.2

Other Exercises

• RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.1
• RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.2
• RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.3

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 BB₁ = B₁B₂= B₂B₃.
(v) Join B₃C.
(vi) Draw B’C’ parallel to B₃C 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, =B₁B₂=B₂B₃=B₃B₄=B₄B_s=B₅B₆=B₆B₇
(v) Join B₇ and C
(vi) Draw B₅C’ parallel to B₇C 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 \(\frac { 2 }{ 3 }\)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 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 BB₁ =B₁B₂ = B₂B₂
(iv) Join B₃C.
(v) From B₂, draw B₂C’ parallel to B₃C 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 \(\frac { 3 }{ 4 }\)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 BB₁=  B₁B₂ = B₂B₃ = B₃B₄.
(v) Join B₄ and C.
(vi) From B₃C draw C₃C’ parallel to B₄C 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 \(\frac { 7 }{ 5 }\) 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 BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B₇.
(v) Join B₅ and C.
(vi) From B₇, draw B₇C’ parallel to B₅C 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 (\(\frac { 5 }{ 4 }\))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 AA₁ = A₁A₂ = A₂A₃ =A₃A₄ = A₄A₅
(v) Join A₄ and B.
(vi) From 45, draw 45B’ parallel to  A₄B 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 \(\frac { 5 }{ 3 }\) 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 BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅
(v) Join B₃ and C.
(vi) From B₅, draw B₅C’ parallel to B₃C 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 \(\frac { 3 }{ 2 }\) 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 DD₁ = D₁D₂ =D₂D₃ = D₃D₄
(iv) Join D₂
(v) Draw D₃A’ parallel to D₂A 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 \((\frac { 3 }{ 4 } )\)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 BB₁= B₁B₂  B₂B₃=B₃B₄.
(v) Join B₄ and C.
(vi) From B₃, draw B₃C’ parallel to B₄C 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 AA₁ = A₁A₂ = A₂A₃ = A₃A₄=A₄A₅.
(vi) Join A₄ and B.
(vii) From A₅, drawA₅B’ parallel to A₄B 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 \((\frac { 3 }{ 2 })\) 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 YY₁= Y₁Y₂ = Y₂Y₃ = Y₃Y₄.
(iii) Join Y₄ and Z.
(iv) From Y₃, draw Y₃Z’ parallel to Y₄Z 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 \(\frac { 3 }{ 4 }\) 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 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 \(\frac { 3 }{ 4 }\) 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 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 A₁= A₁A₂ = A₂A₃.
(vii) Join A₂ and B.
(viii) From A , draw A^B’ parallel to A₂B and B’C’ parallel toBC.
Then ΔC’AB’ is the required triangle.

Hope given RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.2 are helpful to complete your math homework.

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