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Construct a Triangle Abc, Such that Ab= 6 Cm, Bc= 7.3 Cm and Ca= 5.2 Cm. Locate a Point Which is Equidistant from A, B and C. - Mathematics

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प्रश्न

Construct a triangle ABC, such that AB= 6 cm, BC= 7.3 cm and CA= 5.2 cm. Locate a point which is equidistant from A, B and C.

आकृति
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उत्तर

Steps of construction: 

(i) Draw a line segment BC= 7.3 cm. 

(ii) With Bas centre and radius 6 cm draw an arc. 

(iii) With C as centre and radius 5.2 cm draw another arc which intersects the first arc at A. 

(iv) Join AB and AC. 

(v) Draw perpendicuIar bisector of BC , AB and AC. 
In triangIe ABC, P is the point of intersection of AB , AC and BC. 

Therefore, PA = PB, PB = PC, PC = PA. 

Thus, circum-centre of a triangle is the point which is equidistant from all its vertices. 

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अध्याय 16: Loci - Exercise 16.1

APPEARS IN

फ्रैंक Mathematics - Part 2 [English] Class 10 ICSE
अध्याय 16 Loci
Exercise 16.1 | Q 26

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संबंधित प्रश्न

Ruler and compasses may be used in this question. All construction lines and arcs must be clearly shown and be of sufficient length and clarity to permit assessment.

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  4. Mark the point Q, in your construction, which would make ΔQBC equal in area to ΔABC, and isosceles.
  5. Measure and record the length of CQ.

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