मराठी

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.

आकृती
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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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पाठ 16: Loci - Exercise 16.1

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संबंधित प्रश्‍न

Construct a triangle ABC with AB = 5.5 cm, AC = 6 cm and ∠BAC = 105°

Hence:

1) Construct the locus of points equidistant 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 length of PC.


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II. Construct the locus of points inside the triangle which are equidistant from BA and BC.
III. Construct the locus of points inside the triangle which are equidistant from B and C.
IV. Mark the point P which is equidistant from AB, BC and also equidistant from B and C. Measure and records the length of PB.


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  3. Construct the locus of the vertices of the triangles with BC as base and which are equal in area to triangle ABC.
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  5. Measure and record the length of CQ.

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