Construct an equilateral triangle whose perimeter is 15 cm?

steps of construction:

1. let us draw a straight line PQ=15 cm.

2. at P draw an angle RPQ=60 deg, at Q draw an angle SQP=60 deg.

3. now we will draw the angle bisectors of ∠RPQ and ∠SQP. let it intersect at point A.

4. draw the perpendicular bisector of PA and QA. let it intersect PQ at B and C respectively.

5. join AB and AC.

thus ABC is the required triangle.

  • -1

(i) Draw XY = 15 cm.
(ii) Draw MXY = 60° and NYX = 60°
(iii) Draw angle bisectors of MXY and NYX, meeting at a point, say A.
(iv) Draw perpendicular bisector of XA and YA, meeting XY at B and C respectively.
(v) Join A to B and A to C.
ABC is the required triangle.

  • 1
What are you looking for?