Respuesta :

Given:

C is the circumcenter of ΔPQR

PC = 3x + 7

RC = 5x - 15

QC = 51 - x

To find:

QC

Solution:

All vertices or a triangle are equidistant from its circumcenter.

[tex]P C=R C=Q C[/tex]

Let us take [tex]R C=Q C[/tex].

[tex]5 x-15=51-x[/tex]

Add 15 on both sides.

[tex]5 x-15+15=51-x+15[/tex]

[tex]5 x=66-x[/tex]

Add x on both sides.

[tex]5 x+x=66-x+x[/tex]

[tex]6 x=66[/tex]

Divide by 6 on both sides, we get

x = 11

Substitute x = 11 in QC.

QC = 51 - 11

      = 40

Therefore QC = 40.

The value of QC is 40.

Given that,

C is the circumcenter of triangle PQR is,

PC = 3x+ 7, RC = 5x -15, and QC = 51 -x.

We have to determine

The value of QC.

According to the question,

The point of intersection of the perpendicular bisectors of the sides of a triangle is called its circumcenter.

The circumcenter is equidistant to all the three vertices of a triangle.

PC = RC = QC

Substitute all the values in the equation and solve for the value of x,

[tex]\rm RC = QC \\\\5x-15 = 51-x\\\\5x +x = 51+15\\\\6x = 66\\\\x = \dfrac{66}{6}\\\\x = 11[/tex]

Therefore,

The value of QC is,

[tex]\rm QC = 51-x\\\\QC = 51-11\\\\Qc = 40[/tex]

Hence, The required value of QC is 40.

For more details refer to the link given below.

https://brainly.com/question/4362857

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