Discussion

Explanation:

In ASCQ,
AS = QC and also AS || QC
⇒ ASCQ is a parallelogram.

Since the figure is symmetric we get
AJ = DM = CL = BK, and
PJ = QM = RL = SK, and
JM = ML = LK = KJ

Now, in ∆DCL,
DQ = QC and QM || CL
⇒ DM = ML (By Basic Proportionality Theorem)
and also, MQ = ½ LC

Let LC = 2x ⇒ QM = x

∴ DR = DM + ML + LR = 2x + 2x + x = 5x

In right triangle DCR,
DR2 = DC2 + CR2

⇒ (5x)2 = 102 + 52

⇒ 25x2 = 125

⇒ x2 = 5

⇒ x = √5

∴ JK = 2x = 2√5

Hence, option (b).

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