Construct a ∆ in which BC = 6cm , B= 75° and sum of other two sides is 9 cm
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(1) Construct BC = 6 cm
(2) Construct an angle of 60° at B on BC.
(3) Taking B as centre, with 9 cm radius, draw an arc , which cuts the ray of 60° angle, at X
(4) Join X & C
(5) Then draw perpendicular bisector of XC
(6) Wherever this perpendicular bisector intersects BX, name that point as A.
(7) Join A & C
So, tri ABC will be the required triangle.
Justification:
Since AC = AX ( as point ‘A’on perpendicular bisector remains equidistant from X & C )
But AX + AB = 9cm ( we constructed)
=> AC + AB = 9 cm
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