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Sum of the squares of adjacent sides of a parallelogram is 130 sq.cm and length of one of its diagonals is 14 cm. Find the length of the other diagonal.

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#### Solution

It is given that,

AB^{2} + AD^{2} = 130 sq. cm

BD = 14 cm

Diagonals of a parallelogram bisect each other.

i.e. O is the midpoint of AC and BD.

In ∆ABD, point O is the midpoint of side BD.

\[{AB}^2 + {AD}^2 = 2 {AO}^2 + 2 {BO}^2 \left( \text{by Apollonius theorem} \right)\]

\[ \Rightarrow 130 = 2 {AO}^2 + 2 \left( 7 \right)^2 \]

\[ \Rightarrow 130 = 2 {AO}^2 + 2 \times 49\]

\[ \Rightarrow 130 = 2 {AO}^2 + 98\]

\[ \Rightarrow 2 {AO}^2 = 130 - 98\]

\[ \Rightarrow 2 {AO}^2 = 32\]

\[ \Rightarrow {AO}^2 = 16\]

\[ \Rightarrow AO = 4 cm\]

Since point O is the midpoint of side AC.

Hence, the length of the other diagonal is 8 cm.

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