  Question

Construct a triangle $$ABC$$ in which $$BC=7\ cm, \angle B=75^{o}$$ and $$AB+AC=13\ cm$$.

Solution

R.E.F. Image.Given,to construct a $$\Delta ABC$$ where $$BC = 7 cm,\,\angle B = 75^{\circ}$$ & $$AB + AC = 13 cm.$$construction steps :-(i) Draw a line segment BC = 7 cm as a base.(ii) Now mark $$\angle B = 75^{\circ}$$ and extend it till x.(iii) Now, by taking radius 13 cm draw an arc from 'B' on the line of angle just projected.(iv) Now mark it as 'D', & join C to D.(v) Draw $$\perp ^{lar}$$ bisectors for DC and mark them as P & Q.(vi) Now extend a line from 'B' to the $$\perp ^{lar}$$ bisector that cuts on C & D line.(vii) Now mark that point as 'A'.(viii) Thus, required $$\Delta ABC$$ has been formed.(ix) Now, try taking exactly 1/2 of the angle and mark an arc on B.Conclusion :-$$\therefore$$ The required $$\Delta ABC$$ will BC = 7 cm,$$\angle B = 45^{\circ}$$ & AB + AC = 13 cm is constructed. Mathematics

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