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Steps of Construction :
(1) From any ray DF, cut the part DE equal to the
perimeterp. Make angles ‘EDL equal to ‘x and
‘DEM equal to ‘y on the same side of the line
segment DEatD and E.
(2) Draw the bisectors BG and EHof the two angles.
(3) Let these bisectors DG and EHintersect at a
point A. At the point A, draw ∠DAB equal to ∠ADE
and∠EAC equal to ∠AED.
(4) Let AB intersect DE at B and AC intersect DE at
C.
Then,'ABC is the required triangle.

Proof : In 'ADB,‘ADB ‘DAB [by construction], ?AB DB.

Again, in 'ACE,‘AEC ‘EAC;? CA CE.
Therefore, in 'ABC,ABBCCA DBBCCE DE p.

‘ABC ‘ADB‘DAB ‘x ‘x ‘x

2

1

2

1

and.
2

1

2

1

‘ACB ‘AEC‘EAC ‘y ‘y ‘y Therefore, 'ABCis the required triangle.

Activity:

1. Two acute base adjacent angles and the perimeter of a triangle are given. Construct

the triangle in an alternative way.

Example 1. Construct a triangle ABC, in which ‘B = 60°, ‘C= 45° and the
perimeterAB + BC + CA = 11 cm.

Steps of Construction: Follow the steps below :
(1) Draw a line segment PQ = 11 cm.

(2) At P, construct an angle of ‘QPL = 60° and at Q, an angle of ‘PQM = 45° on
the same side of PQ.
(3) Draw the bisectors PG and QH of the two angles. Let the bisectors PG and QH
of these angles intersect at A.
(4) Draw perpendicular bisector of the segments PA of QA to intersect PQ at B and C.

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