Step I: On BE, cut off 3 equal parts making \[{{B}_{1}},{{B}_{2}}\] and \[{{B}_{3}}\] |
Step II: Now, draw C'A' parallel to CA. Then, \[\Delta A'BC'\] is the required A whose sides are of the corresponding sides of the \[\Delta ABC\]. |
Step III: From point B draw an arc of \[2.3\text{ }cm\]and from point C draw an arc of \[2.9\text{ }cm\]cutting each other at point A. |
Step IV: Take\[BC=5\text{ }cm\]. |
Step V: Join \[{{B}_{3}}C\] and from \[{{B}_{2}}\] draw \[{{B}_{2}}C'\] parallel to \[{{B}_{3}}C,\] such that BC is 2/3 of BC. |
Step VI: On B make an acute \[\angle CBE\]downwards. |
Step VII: Join AB and AC. Then ABC is the required triangle. |
A) IV, III, VII, I, VI, V, II
B) IV, V, I, VI, III, VII, II
C) IV, III, VII, VI, I, V, II
D) IV, VII. Ill, VI, V, I, II
Correct Answer: C
Solution :
Correct sequence of steps is IV, III, VII, VI, I, V, II.You need to login to perform this action.
You will be redirected in
3 sec