Draw a line XY=10.4 i.e. the perimeter.
Construct an angle equal to ∠B=45° and another angle equal to ∠C=120°
Bisect these angles and name the intersecting point as A.
Construct perpendicular bisectors of AX and AY and name the PQ and RS respectively.
Name the intersecting point of PQ and XY as B and RS the intersecting point of RS and XY as C.
Join AB and AC.
ABC is the required triangle.
As B is on line PQ which is the perpendicular bisector of AX,
AB+BC+CA = XB+BC+CY=XY
Then,∠BAX=∠AXB (as in triangle AXB, AB is equal to XB)
As ∠ABC is the external angle of triangle AXB
Then,∠ABC=∠BAX + ∠AXB(exterior angle sum property)
∠ABC=2∠AXB=45° or ∠B.
∠ACB=2∠CAY=120° or ∠C.
Thus the construction is justified.
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