tell this please...............
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From figure:
Since OT is perpendicular bisector of PQ.
So, PM = QM = (1/2) PQ = 8 cm.
(i) In ΔOMP:
(OP)² = (PM)² + (OM)²
(10)² = (8)² + OM²
100 - 64 = OM²
36 = OM²
OM = 6 cm.
∴ Let TP = x.
(ii) In ΔPMT:
(TP)² = (PM)² + (MT)²
⇒ x² = 8² + MT²
⇒ x² = 64 + MT²
(iii) In ΔOPT:
(OT)² = (OP)² + (TP)²
⇒ (OT)² = 10² + x²
⇒ (OM + MT)² = 100 + x²
⇒ (6 + MT)² = 100 + x²
⇒ 36 + MT² + 2(6)(MT) = 100 + x²
⇒ 36 + MT² + 12MT = 100 + (64 + MT²)
⇒ 36 + MT² + 12MT - 100 - 64 - MT² = 0
⇒ -128 + 12MT = 0
⇒ MT = 128/12
Substitute in (ii), we get
⇒ x² = 64 + (128/12)²
⇒ x² = 64 + (16384/144)
⇒ x² = (9216 + 16384)/144
⇒ x² = (25600/144)
⇒ x = 160/12
⇒ x = 13.33 cm.
Therefore, TP = x = 13.33 cm.
Hope it helps!
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siddhartharao77:
Great thanks!
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