Answer :

Let ABCD be a parallelogram, with AB = CD ; AB || CD and BC = AD ; BC || AD.

Construct AE ⊥ CD and extend CD to F such that, BF ⊥ CF.

Given: sum of squares of adjacent side = 130

⇒ CD^{2} + BC^{2} = 130 and

Length of one diagonal = 14 cm [let it be AC]

To Find: length of the other diagonal, BD

In ΔAED and ΔBCF

AE = BF [Distance between two parallel lines i.e. AB and CD]

AD = BC [opposite sides of a parallelogram are equal]

∠AED = ∠BFC [Both 90°]

ΔAED ≅ ΔBCF [By Right Angle - Hypotenuse - Side Criteria]

⇒ DE = CF [Corresponding sides of congruent triangles are equal] [1]

In ΔBFD, By Pythagoras theorem i.e.

(Hypotenuse)^{2} = (base)^{2} + (Perpendicular)^{2}

BD^{2} = DF^{2} + BF^{2}

⇒ BD^{2} = (CD + CF)^{2} + BF^{2} [2]

In ΔAEC, By Pythagoras theorem

AC^{2} = AE^{2} + CE^{2}

⇒ AC^{2} = AE^{2} + (CD - AE)^{2}

⇒ AC^{2} = BF^{2} + (CD - CF)^{2} [As, AE = BF and CF = AE] [2]

In ΔBCF, By Pythagoras theorem,

BC^{2} = BF^{2} + CF^{2}

BF^{2} = BC^{2} - CF^{2} [3]

Adding [2] and [3]

BD^{2} + AC^{2} = 2BF^{2} + (CD + CF)^{2} + (CD - CF)^{2}

⇒ BD^{2} + AC^{2}= 2BC^{2} - 2CF^{2} + CD^{2} + CF^{2} + 2CD.CF + CD^{2} + CF^{2} - 2CD.CF

⇒ BD^{2} + AC^{2} = 2BC^{2} + 2CD^{2}

⇒ BD^{2} + 14^{2} = 2(130)

⇒ BD^{2} + 196 = 260 [Using given data]

⇒ BD^{2} = 64

⇒ BD = 8 cm

Hence, length of other diagonal is 8 cm.

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