Construct a $\triangle ABC$ in which $BC = 3.4\ cm, AB - AC = 1.5\ cm$ and $\angle B = 45^o$.
Given:
A $∆ABC$ in which $BC = 3.4\ cm, AB - AC = 1.5\ cm$ and $\angle B = 45^o$.
To do:
We have to construct the given triangle.
Solution:
Steps of construction:
(i) Draw a line segment $BC = 3.4\ cm$.
(ii) At $B$, draw a ray $BX$ making an angle of $45^o$ and cut off $BE = 1.5\ cm$.
(iii) Join $EC$.
(iv) Draw the perpendicular bisector of $CE$ which intersects $BE$ produced at $A$.
(v) Join $AC$.
Therefore,
$\triangle ABC$ is the required triangle.
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