RD Chapter 11 Constructions Ex 11.2 Solutions
Question - 11 : - Construct a triangle similar to a given ΔXYZ with its sides equal to (3/2) th of the corresponding sides of ΔXYZ. Write the steps of construction.
Answer - 11 : - Steps of construction :
(i) Draw a triangle XYZwith some suitable data.
(ii) Draw a ray YL makingan acute angle with XZ and cut off 5 equal parts making YY1= Y1Y2 =Y2Y3 = Y3Y4.
(iii) Join Y4 andZ.
(iv) From Y3,draw Y3Z’ parallel to Y4Z and Z’X’ parallel to ZX.
Then ΔX’YZ’ is the required triangle.
Answer - 12 : -
(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]
Answer - 13 : -
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]
Answer - 14 : -
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 3/4 of the corresponding sides of ΔABC. [CBSE 2017]
Answer - 15 : -
Steps of construction:
1. Draw a triangle ABCwith side BC = 6 cm, AB = 5 cm and ∠ABC = 60°.
2. Draw a ray BX,which makes an acute angle ∠CBXbelow the line BC.
3. Locate four points B1,B2, B3and B4 on BX such that BB1 =B1B2=B2B3 = B3B4.
4. Join B4Cand draw a line through B3 parallel to B4C intersectingBC 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 5/3 times the corresponding sides of the given triangle. [CBSE 2017]
Answer - 16 : -
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.
Answer - 17 : -
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.