Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are $1\frac{1}{2}$ times the corresponding sides of the isosceles triangle.

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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 $1\frac{1}{2}$ 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.

Updated on 10-Oct-2022 13:23:52