Question

PLEASE help me! Im stuck :(

Answer 1

a) QR = 50

b) RZ = 41

c) XS = 40

d) ZS = 41

e) WZ = 4√(66) = 32.5 (nearest tenth)

Explanation:

A circumcenter of a triangle is:

• The center of a circle that passes through each vertex of a triangle.
• The point at which the perpendicular bisectors of the sides of the triangle intersect.

A perpendicular bisector is a line that divides another line segment into two equal parts at a right angle.

Given Z is the circumcenter of ΔQRS.

• ZW is the perpendicular bisector of QR, so QW = WR.
• ZX is the perpendicular bisector of RS, so RX = XS.
• ZY is the perpendicular bisector of QS, so QY = YS.

Part (a)

As ZW is the perpendicular bisector of QR and QW = 25, then:

• QR = 2 × 25 = 50

Part (b)

As ZX is the perpendicular bisector of RS, and RS = 80, then RX = 40.

Therefore, triangle RXZ is a right triangle with legs RX = 40 and ZX = 9.

To calculate RZ, use Pythagoras Theorem:

⇒ RX² + ZX² = RZ²

⇒ 40² + 9² = RZ²

⇒ 1681 = RZ²

⇒ RZ = √(1681)

⇒ RZ = 41

Part (c)

As ZX is the perpendicular bisector of RS, XS is half the length of RS.

Therefore:

• XS = 80 ÷ 2 = 40

Part (d)

As ZX is the perpendicular bisector of RS, ΔRZS is an isoceles triangle with base RS and where RZ = ZS.

Therefore:

• ZS = RZ = 41

Part (e)

As Z is the circumcenter ΔQRS, Z is the center of a circle that passes through the vertices Q, R and S. Therefore, ZQ, ZR and ZS are the radii of the circle, and so ZQ = ZR = ZS.

As ZW is the perpendicular bisector of QR, triangle QZW is a right triangle with leg QW = 25 and hypotenuse ZQ = 41.

To calculate WZ, use Pythagoras Theorem:

⇒ WZ² + QW² = ZQ²

⇒ WZ² + 25² = 41²

⇒ WZ² = 41² - 25²

⇒ WZ² = 1056

⇒ WZ = √(1056)

⇒ WZ = 4√(66)

⇒ WZ = 32.5 (nearest tenth)