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Construct a $∆ABC$ in which $AB + AC = 5.6\ cm, BC = 4.5\ cm$ and $\angle B = 45^o$.
Given:
A $∆ABC$ in which $AB + AC = 5.6\ cm, BC = 4.5\ cm$ and $\angle B = 45^o$.
To do:
We have to construct the given triangle.
Solution:
Steps of construction:
(i) Draw a line segment $BC = 4.5\ cm$.
(ii) At $B$, draw a ray $BX$ making an angle of $45^o$ and cut off $BE = 5.6\ cm$ and join $CE$.
(iii) Draw the perpendicular bisector of $CE$ which intersects $BE$ at $A$.
(iv) Join $AC$.
Therefore,
$\triangle ABC$ is the required triangle.
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