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
