[tex]\qquad \qquad\huge \underline{\boxed{\sf Answer}}[/tex]
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]
