**RD Sharma Class 10 Solutions Chapter 11 Exercise 11.2: **In this exercise, students can learn how to design a triangle that is comparable to a given triangle. All of the answers to the **RD Sharma Class 10 Solutions** have been written by Kopykitab experts to dispel any concerns and provide the correct techniques for solving a question. Students can use the **RD Sharma Class 10 Solutions Chapter 11** Exercise 11.2 PDF to aid them with the correct steps in solving questions.

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RD Sharma Class 10 Solutions Chapter 11 Exercise 11.2

### Access answers to RD Sharma Solutions Class 10 Maths Chapter 11 Exercise 11.2- Important Question with Answers

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 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.

(ii) With centre B and radius 4 cm and with centre C and radius 5 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 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 :

(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 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 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.

(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.

(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 triangle whose 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.

(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 that 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 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 the 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.

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