Constuct a Triangle Abc with Ab=5.5 Cm , Ac=6 Cm and ∠Bac=105°. Hence: 1) Construct the Locus of Point Equdistant from Ba and Bc. 2) Construct the Locus of Points Equidistant from B and C. - Mathematics

Sum

Constuct a triangle ABC with AB=5.5 cm , AC=6 cm and ∠BAC=105°. Hence:

1) Construct the locus of point equdistant from BA and BC.

2) Construct the Locus of points equidistant from B and C.

3) Mark the point which satisfies the above two loci As P. Measure and write the lemgth of PC

Solution

Steps of construction:

1) Draw AB=5.5 cm

2) construct ∠BAR=105°

3) With centre A and redius 6 cm, cut off arc on AR at C.

4) Join BC. ABC is the reqqiued triangle.

(1) Draw angle bisector BD of  ∠ABC, which is the locus of points equidistant from BA and BC.

(2) Draw perpendicular bisctor EF of BC, which is the loucs of point equidistant from B and C.

(3) BD and EF intersect each other at point P.Thus, P satisfies the above two loci By Mesurment,PC=4.8 cm

Concept: Circumscribing and Inscribing a Circle on a Triangle
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Selina Concise Maths Class 10 ICSE
Chapter 19 Constructions (Circles)
Exercise 19 | Q 22 | Page 293