3 years ago

C is the circumcenter of triangle PQR. PC=3x+7, RC=5x-15, and QC=51-x What does QC=

Mathematics

1 answer:

Readme [11.4K]3 years ago

8 0

Given:

C is the circumcenter of ΔPQR

PC = 3x + 7

RC = 5x - 15

QC = 51 - x

To find:

Solution:

All vertices or a triangle are equidistant from its circumcenter.

$P C=R C=Q C$

Let us take $R C=Q C$ .

$5 x-15=51-x$

Add 15 on both sides.

$5 x-15+15=51-x+15$

$5 x=66-x$

Add x on both sides.

$5 x+x=66-x+x$

$6 x=66$

Divide by 6 on both sides, we get

x = 11

Substitute x = 11 in QC.

QC = 51 - 11

= 40

Therefore QC = 40.

You might be interested in

Find all solutions to 0= 2x^4+5x^3+38x^2+125x-300

Mariana [72]

$2x^4+5x^3+38x^2+125x-300=0\\\\2x^4+5x^3-12x^2+50x^2+125x-300=0\\\\x^2(2x^2+5x-12)+25(2x^2+5x-12)+0\\\\(2x^2+5x-12)(x^2+25)=0\\\Updownarrow\\1^o\ 2x^2+5x-12=0\ \vee\ 2^o\ x^2+25=0\\\\\\1^o\ \Delta=5^2-4\cdot2\cdot(-12)=25+96=121\\\\\sqrt\Delta=\sqrt{121}=11\\\\x_1=\frac{-5-11}{2\cdot2}=\frac{-16}{4}=-4;\ x_2=\frac{-5+11}{2\cdot2}=\frac{6}{4}=\frac{3}{2}\\\\\\2^o\ x^2=-25-false\\\\\\Answer:x=-4\ or\ x=\frac{3}{2}$