AB = 10 cm.
∴ AF = FB = 5 cm.
CD = 24 cm.
∴ CE = DE = 12 cm.
माना OE = x cm
∴ OF = (17 – x) cm
Δ O D E , O D = O E 2 + D E 2 \Delta \mathrm{ODE}, \mathrm{OD}=\sqrt{\mathrm{OE}^{2}+\mathrm{DE}^{2}} Δ ODE , OD = OE 2 + DE 2 से
= x 2 + 1 2 2 =\sqrt{x^{2}+12^{2}} = x 2 + 1 2 2
Δ \Delta Δ OAF से,
O A = O F 2 + A F 2 \mathrm{OA}=\sqrt{\mathrm{OF}^{2}+\mathrm{AF}^{2}} OA = OF 2 + AF 2
= ( 17 − x ) 2 + 5 2 =\sqrt{(17-x)^{2}+5^{2}} = ( 17 − x ) 2 + 5 2
OA = OD
∴ x 2 + 1 2 2 = ( 17 − x ) 2 + 5 2 \therefore \quad \sqrt{x^{2}+12^{2}}=\sqrt{(17-x)^{2}+5^{2}} ∴ x 2 + 1 2 2 = ( 17 − x ) 2 + 5 2
⇒ x 2 + 144 = 289 − 34 x + x 2 + 25 \Rightarrow x^{2}+144=289-34 x+x^{2}+25 ⇒ x 2 + 144 = 289 − 34 x + x 2 + 25
⇒ 34 x = 289 + 25 − 144 = 170 \Rightarrow 34 x=289+25-144=170 ⇒ 34 x = 289 + 25 − 144 = 170
⇒ x = 170 34 = 5 \Rightarrow x=\frac{170}{34}=5 ⇒ x = 34 170 = 5
∴ \therefore ∴ समीकरण (i) से,
O D = x 2 + 1 2 2 = 5 2 + 144 \mathrm{OD}=\sqrt{x^{2}+12^{2}}=\sqrt{5^{2}+144} OD = x 2 + 1 2 2 = 5 2 + 144
= 169 = 13 c m . =\sqrt{169}=13 \mathrm{~cm} . = 169 = 13 cm .