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Let’s Construct a triangle ABC, such that $AB=6cm$, $BC=7.3cm$ and $CA=5.2cm$ locate a point which is at equal distance from A, B and C.
Let’s Construct a triangle ABC, such that $AB=6cm$, $BC=7.3cm$ and $CA=5.2cm$ locate a point which is at equal distance from A, B and C.
Class 10 | Frank | Guides | ICSE | Loci | Maths
Let’s Construct a triangle ABC, such that $AB=6cm$, $BC=7.3cm$ and $CA=5.2cm$ locate a point which is at equal distance from A, B and C.

Steps of construction:

1. Let us Draw a line segment $BC=7.3cm$.

2. With B as a center and radius $6cm$, draw an arc with the help of compass.

3. With C as a center and radius $5.2cm$, draw another arc with the help of compass, cutting the previous arc at A.

4. Now Join AB and AC.

Then, $\vartriangle ABC$ is the required triangle.

5. With B as center and radius measuring more than half of BC, draw arcs with the help of compass  on both sides of BC.

6. With C as center and the same radius as before, draw arcs with the help of compass on both sides of BC, cutting the previous arcs as shown. Now Join them.

7. Again with B as center and radius measuring more than half of BC, draw arcs with the help of compass on both sides of BC.

8. With A as center and the same radius as before, draw arcs with the help of compass on both sides of BC, cutting the previous arcs as shown. Now Join them.

9. With C as center and radius measuring more than half of BC, draw arcs with the help of compass  on both sides of BC.

10. With A as center and the same radius as before, draw arcs with the help of compass on both sides of BC, cutting the previous arcs as shown. Now Join them

So, in triangle ABC, P is the point of intersection of AB, AC and BC.

Therefore, We will get $PA=PB$, $PB=PC$, $PC=PA$

Hence, P is the point which is equal distance from A, B and C.

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