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3 years ago

Prove that the circumcenter of a triangle is equidistant from its vertices.

Mathematics

1 answer:

ruslelena [56]3 years ago

5 0

Answer:

Step-by-step explanation:

The circumcenter is equidistant from the three vertices of the triangle. From the figure shown, we will prove DA = DB = DC. 2) DA = DB, DC = DB(If a point is on the perp. bisector of a segment, it is equidistant from each endpoint of the segment.)

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In the following figure, the square ABCD is inscribed in the circle centered at point P. What is the area of the circle?

Ainat [17]

Answer:

$A_{Circle} \approx 39.24\: units^2$

Step-by-step explanation:

Square ABCD is inscribed in a circle with center P such that BC = 5 units.

BD will be diagonal of the square as well as diameter of the circle.

$BD = BC\times\sqrt 2$

$\implies BD = 5\sqrt 2$

-> Diameter of the circle $(d) = 5\sqrt 2$

-> Radius of the circle $(r) = 2.5\sqrt 2=3.535\: units$

$A_{Circle} =\pi(r)^2$

$\implies A_{Circle} =3.14(3.535)^2$

$\implies A_{Circle} =39.2381465$

$\implies A_{Circle} \approx 39.24\: units^2$

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