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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.

1197982_1284015_ans_87d9bb8de7184136a935e8e6bd3d037d.jpg

Mathematics

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