Construct a triangle ABC in which \[BC=6\text{ }cm,\text{ }AB=5\text{ }cm\] and \[\angle ABC=60{}^\circ \]. |
Then construct another triangle whose sides are \[\frac{3}{4}\] times the corresponding sides of \[\Delta \text{ }ABC\] |
Answer:
Steps of Construction - (i) Draw a line segment \[BC=6\text{ }cm\]. (ii) Construct \[\angle XBC=60{}^\circ \] (iii) With B as centre and radius equal to 5 cm draw an arc which intersect XB at A. (iv) Join AC. Thus, \[\Delta \text{ }ABC\] is obtained. (v) Draw D on BC such that \[BD=\frac{3}{4}BC=\left( \frac{3}{4}\times 6 \right)cm=\frac{9}{2}cm=4.5\,cm\] (vi) Draw \[DE\parallel CA\], cutting BA at E. Then, \[\Delta \text{ }BDE\]is the required triangle similar to \[\Delta \text{ }ABC\]such that each side of \[\Delta \text{ }BDE\] is \[\frac{3}{4}\] times the corresponding side of\[\Delta \text{ }ABC\].
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