The perimeter of triangle APM is 152 and ZPAM = 90°. A circle of radius 19 with centre O is drawn with diameter on AP so that it is tangent to AM and PM. Find AM.
Answers
Given :- The perimeter of triangle APM is 152 and ZPAM = 90°. A circle of radius 19 with centre O is drawn with diameter on AP so that it is tangent to AM and PM. Find AM.
Solution :-
from image we get,
→ AM = MN = Let x { Tangents from external points to the circle are equal .}
and,
→ PO = OA = ON = 19 { Radius of circle. }
so,
→ PN = Perimeter of ∆APM - (PA + AM + MN)
→ PN = 152 - (38 + 2x)
now, since ∆PON ~ ∆PAM
→ Perimeter of ∆PON / Perimeter of ∆PAM = ON/AM
→ (PO + ON + PN) / 152 = 19/AM
→ [19 + 19 + {152 - (38 + 2x)}] / 152 = 19/x
→ (152 - 2x) / 152 = 19/x
→ 152x - 2x² = 2888
→ 2x² - 152x + 2888 = 0
→ x² - 76x + 1444 = 0
→ x² - 38x - 38x + 1444 = 0
→ x(x - 38) - 38(x - 38) = 0
→ (x - 38)(x - 38) = 0
→ x = 38 (Ans.)
Hence, AM is equal to 38 .
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