Construct an isosceles triangle whose base is $ 8 \mathrm{~cm} $ and altitude $ 4 \mathrm{~cm} $ and then another triangle whose sides are $ 3 / 2 $ times the corresponding sides of the isosceles triangle.
Given:
An isosceles triangle whose base is \( 8 \mathrm{~cm} \) and altitude \( 4 \mathrm{~cm} \).
To do:
We have to construct an isosceles triangle whose base is \( 8 \mathrm{~cm} \) and altitude \( 4 \mathrm{~cm} \) and then another triangle whose sides are \( 3 / 2 \) times the corresponding sides of the isosceles triangle.
Solution:
Steps of construction:
(i) Draw a line segment $BC = 8\ cm$ and draw its perpendicular bisector $DX$ and cut off $DA = 4\ cm$.
(ii) Join $AB$ and $AC$.
$ABC$ is the required triangle.
(iii) Draw a ray $DY$ making an acute angle with $OA$ and cut off three equal parts making $DD_1 = D_1D_2 =D_2D_3 = D_3D_4$
(iv) Join $D_2A$
(v) Draw $D_3A’$ parallel to $D_2A$ and $A’B’$ parallel to $AB$ meeting $BC$ at $C’$ and $B’$ respectively.
$B’A’C’$ is the required triangle.
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