Question

Anyone Answer plzz​

Answer 1

The figure in the attachment depicts the question.

In ∆ POX and ∆ QOX,

1. OP = OQ
2. XP = XQ
3. OX = OX

¶ ∆ POX ~= ∆ QOX by SSS congruence.

¶ Angle POX = Angle QOX [ CPCT ] .............{|}

Also

In ∆ POR and ∆ QOR

1. OP = OQ
2. {|}
3. OR = OR

¶ ∆ POR ~= ∆QOR by SSS congruence.

¶ Angle PRO = Angle QRO [ CPCT ] ..............{||}

¶ PR = QR [ CPCT ] .............{|||}

.

.

.

.

Angle PRO = 90° = Angle QRO = Angle PRX

Let OR be X.

XR = 4 - X

.

.

.

.

PR^2 = 25 - X^2 = -7 -X^2 + 8X

X = 4 cm.

PR = 3cm.

Length of common core = 3 • 2 = 6 cm.

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Answer 2

hey here is your answer

hope it helps

mark it as brainliest ☺️