Q. 24.0( 3 Votes )

Draw a parallelog

Answer :

Thinking process:


I. Firstly we draw a line segment, then either of one end of the line segment with length 5 cm and making an angle 60° with this end. We know that is parallelogram both opposite sides are equal and parallel, then again draw a line with 5 cm making an angle with 60° from other end of line segment. Now, join both parallel line by a line segment whose measurement is 3 cm, we get a parallelogram. After that we draw a diagonal and get a triangle BOC.


II. Now, we construct the triangle BD’C’ similar to ∆BDC with scale factor 4/3.


III. Now, draw the line segment D’A’ parallel to DA.


IV. Finally, we get the required parallelogram A ‘BC’D.


Steps of construction:


1. Draw a line segment AB=3 cm.



2. Now, draw a ray BY making an acute ABY=60°.


To draw the angle 60° at B:


a. With B as centre and with any radius, draw another arc cutting the line AB at E.



b. With E as centre and with the same radius, draw an arc cutting the first arc (drawn in step a) at point F.



c. Now join the ray BY which forms an angle of 60° with the line AB.



3. With B as centre and radius equal to 5 cm draw an arc cut the point C on BY.



4. Again draw a ray AZ making an acute ZAX1=60°.(BY││AZ, YBX1=ZAX1=60°)


Following the similar steps a, b and c as in point 2, we will draw a ray AZ making an angle 60° with AX �1.



5. With A as centre and radius equal to 5 cm draw an arc cut the point D on AZ.



6. Now, join CD and finally make a parallelogram ABCD.



7. Join BD, which is a diagonal of parallelogram ABCD.



8. From B draw any ray BX downwards making an acute CBX.



9. Locate 4 points B1,B2,B3,B4 on BX, such that BB1=B1B2=B2B3=B3B4.



10. Join B3C and from B4 draw a line B4C’||B3C intersecting the extended line segment BC at C’.



11. From point C’ draw C’D’││CD intersecting the extended line segment BD at D’. Then, ∆D’BC’ is the required triangle whose sides are 4/3 of the corresponding sides of ∆DBC.



12. Now draw a line segment D’A’ parallel to DA, where A’ lies on the extended side BA i.e., a ray BX1.



13. Finally, we observe that A’BC’D’ is a parallelogram in which A’D’=6.5 cm A’B= 4 cm and A’BD’=60° divide it into triangles BC’D’ and A’BD’ by the diagonal BD’.



Justification that rBCD is similar to rBC’D’:


Let


BB1=B1B2=B2B3= B3B4= x


The triangles ∆BCB3 and ∆BC’B4 are similar by


[by AA congruency criteria, CBB3 = C’BB4 as it is common in both triangles and BB3C=BB4C’ and are corresponding angles of the same transverse as CB3 || C’B4]


---- (i)


Similarly, ∆BCD and ∆BC’D’ are similar.


[by AA congruency criteria, B is common in both triangles and BCD=BC’D’ and are corresponding angles of the same transverse as CD || C’D’]


Hence


---- (ii)


From (i) and (ii) we can say that the constructed triangle ∆BC’D’ is of scale times of the ∆BCD.


Justification that A’BC’D’ is a parallelogram:


As per the construction,


CD || C’D’ and BC’ || A’D’ ---- (i)


Also,


As per the given data, ABC = 60°.


Consider the parallelogram ABCD, the sum of complementary angles is 180°.


So ABC + BCD = 180°


BCD = 180° - ABC = 180° - 60° = 120°.


Therefore BCD = 120°.


Now, as CD || C’D’,


BCD = BC’D’ (as corresponding angles)


Hence BC’D’ = 120° ---- (ii)


As BC||AD and AD||A’D’, hence BC||A’D’,


ABC = D’A’X1 (corresponding angles)


So D’A’X1 = ABC = 60°


Now consider D’A’X1 + D’A’B = 180° (linear angle)


60° + D’A’B = 180°


D’A’B = 180° - 60° = 120° ---- (iii)


From (i), (ii) and (iii), we can say that A’BC’D’ is a parallelogram.


(As opposite sides and opposite angles are equal).

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