Step I: Draw BC = 6 cm. |
Step II: At B construct \[\angle CBX={{60}^{o}}\] and at C construct |
Suppose BX and CY intersect at A. \[\Delta \,ABC\]so obtained is the given triangle. |
Step III: Construct an obtuse angle \[\angle CBZ\] at B on opposite side of vertex A of \[\Delta \,ABC.\] |
Step IV: Mark-off three (greater 3 of 2 in 3/2) points \[{{B}_{1}},{{B}_{2}},{{B}_{3}},\] on BZ such that\[B{{B}_{1}}={{B}_{1}}{{B}_{2}}={{B}_{2}}{{B}_{3}}.\]. |
Step V: Join \[{{B}_{2}}\] (the second point) to C and draw a line through \[{{B}_{3}}\] parallel to \[{{B}_{2}}C,\] intersecting the extended line segment BC at C'. |
Step VI: Draw a line through C' parallel to CA intersecting the extended line segment BA at A'. |
Triangle A'B'C so obtained is the required triangle such that \[\frac{A'B}{AB}=\frac{BC'}{BC}=\frac{A'C'}{AC}=\frac{3}{2}\] |
A) Step III
B) Step IV
C) Step V
D) Step II
Correct Answer: A
Solution :
Step III is incorrect since we construct an acute angle \[\angle CBZ\]at B on opposite side of vertex A of \[\Delta ABC\].You need to login to perform this action.
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