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For the first figure ~

The diagonals of a kite intersect each other at 90°

So, we can apply Pythagoras theorem here :

[tex]\qquad \sf  \dashrightarrow \: CD² = OC² + OD²[/tex]

[tex]\qquad \sf  \dashrightarrow \: CD² = {7}^{2} + {9}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \: CD² = 49 + 81[/tex]

[tex]\qquad \sf  \dashrightarrow \: CD² = 130[/tex]

[tex]\qquad \sf  \dashrightarrow \: CD=x = \sqrt{ 130}[/tex]

For the second figure ;

we have same concept of kite, and use of Pythagoras theorem !

Also, the diagonal QS bisects diagonal PR

Hence,

[tex]\qquad \sf  \dashrightarrow \: PR = 2 \times OR [/tex]

[tex]\qquad \sf  \dashrightarrow \: 10 = 2 \times OR [/tex]

[tex]\qquad \sf  \dashrightarrow \: OR = 10 \div 2[/tex]

[tex]\qquad \sf  \dashrightarrow \: OR = 5 \: mm[/tex]

now, apply pythagoras theorem ~

[tex]\qquad \sf  \dashrightarrow \: QR² = OR² + OQ²[/tex]

[tex]\qquad \sf  \dashrightarrow \: QR² = {5}^{2} + {6}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \: QR² = 25 + 36[/tex]

[tex]\qquad \sf  \dashrightarrow \: QR² = 61[/tex]

[tex]\qquad \sf  \dashrightarrow \: QR=x = \sqrt{61} \: mm[/tex]

here, 2 OR = 2 OP = PR

so, similarly OP = 5 mm

Applying pythagoras theorem again ;

[tex]\qquad \sf  \dashrightarrow \: SP² = OS² + OP²[/tex]

[tex]\qquad \sf  \dashrightarrow \: SP² = {10}^{2} + {5}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \: SP² = {100}^{} + 25[/tex]

[tex]\qquad \sf  \dashrightarrow \: SP² = 125[/tex]

[tex]\qquad \sf  \dashrightarrow \: SP = \sqrt{125}[/tex]

[tex]\qquad \sf  \dashrightarrow \: SP = y = 5\sqrt{5} \: mm[/tex]

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