If z is the circumcenter of triangle qrs find each measure

Applying the circumcenter theorem and the Pythagorean Theorem, the following measures are calculated as:

a. QR = 50

b. RZ = 41

c. XS = 40

d. ZS = 41

e. WZ = 32.5

Based on the circumcenter theorem of a triangle, the following holds true given that Z is the circumcenter of triangle QRS:

Given:

a. Find QR:

QR = 2 × WQ = 2 × 25

QR = 50

b. Find RZ in right ΔXRZ:

XZ = 9 (given)

XR = 1/2(RS) = 1/2(80) = 40

RZ = √(XZ² + XR²)

RZ = √(9² + 40²)

RZ = 41

c. Find XS:

XS = 1/2(RS) = 1/2(80)

XS = 40

d. Find RZ:

RZ = ZS = QZ = 41 (equidistant)

ZS = 41

e. Find WZ in right ΔWRZ:

WR = WQ = 25

WR = 25

RZ = 41

WZ = √(RZ² - WR²)

WZ = √(41² - 25²)

WZ = 32.5

Learn more about the circumcenter theorem on:

link

Step-by-step explanation:

Circumcenter of A triangle is the center of a circle passing through the vertices of the given triangle.

So, the distance of Z from the vertices Q, R and S will be the radii of the circle.

Therefore, ZQ = ZR = ZS

Since, circumcenter of a triangle is defined by the point of intersection of the perpendicular bisectors of the sides of the given triangle.

Therefore, points W, X and Y are the midpoints of sides QR, RS and QS.

Using these facts,

a). QR = 2(WQ)

          = 2(25)

          = 50

b). Apply Pythagoras theorem in ΔRXZ,

    (RZ)² = (RX)² + (ZX)²

    (RZ)² = (40)² + (9)²

     RZ = √1681 = 41

c). XS =

          =

          = 40

d). Since, ZQ = ZR = ZS

    ZS = 41

e). By applying Pythagoras theorem in ΔRWZ,

   (RZ)² = (WZ)² + (WR)²

   (41)² = (WZ)² + (25)²

   (WZ)² = (41)² - (25)²

    WZ = √1056

   WZ =

          ≈ 32.5

the transverse axis of a hyperbola is the line that contains the two vertices and the two focuses.

in this example (hyperbola of equation x^2/2 - y^2/4 = 1), the transverse axis is the  x-axis  . 

step-by-step explanation:

answer: plz put a pic in so i can answer it

step-by-step explanation: