question_answer

Construct an isosceles triangle with base \[8\text{ }cm\] and altitude\[4\text{ }cm\]. Construct another triangle whose sides are \[\frac{2}{3}\] times the corresponding sides of the isosceles triangle.

Answer:

Steps of construction-

(i) Draw \[BC=8\text{ }cm\].

(ii) Construct \[XY\], the perpendicular bisector of line segment \[BC\], meeting \[BC\]at \[M\].

(iii) Cut \[MA=4\text{ }cm\]on \[XM\]. Join \[BA\] and \[CA\], \[\Delta \text{ }ABC\] is obtained.

(iv) At B, draw an acute angle in downward direction. Draw 3 arcs \[{{B}_{1}},{{B}_{2}}\] and \[{{B}_{3}}\] on it.

(v) Join \[{{B}_{3}}C\] and at \[{{B}_{2}}\] draw line parallel to \[{{B}_{3}}C\], cutting \[BC\] at \[C'\].

(vi) At \[C'\] draw \[AC\] parallel to \[AC\].

(vii) \[\Delta \text{ }A'C'B\] is required triangle.