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please answer and give steps of construction too.
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(I) draw a line segment BC =7cm as a base.
(ii) now mark angel B = 75° and extend it till X.
(iii) now, by taking radius 13cm draw an arc from 'B' on the line of angel just projected .
(iv)now mark it as 'D' and join C toD.
(v) Draw perpendicular bisectors for DC and mark them as P and Q.
(vi) Now extend a line from B to the perpendicular bisector that cuts on C and D line .
(vii)Now mark that point as 'A'.
(viii) Thus required ∆ ABC has been formed.
(ix) Now try taking exactly 1/2 of the angel and mark an arc on 'B'.
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A triangle is the closed figure with 3 sides equal which makes then called equilateral triangle.
Refer to the attachement. . .
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