A B C D is a quadrilateral . Prove that AB + BC + CD + DA > AC + BD.
Answers
Answer:
Given that,
A point P is 25 cm away from the centre of a circle. Centre of the circle is O. And, the length drawn from point P to the circle is 24 cm. TP is 24 cm.
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\underline{\bf{\dag} \:\mathfrak{Using\;Circle\; Theorem\: :}}
†UsingCircleTheorem:
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Tangent drawn from an external point (P) is perpendicular to the radius at the point of Contact, O.
Therefore, OT \perp⊥ PT.
\rule{250px}{.3ex}
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\underline{\pink{\bigstar\:\sf{By\: Using\; Pythagoras\: Theorem\;In\;\triangle\;OTP\; :}}}
★ByUsingPythagorasTheoremIn△OTP:
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\begin{gathered}:\implies\sf (OP)^2 = (OT)^2 + (PT)^2 \\\\\\:\implies\sf (OT)^2 = (OP)^2 - (PT)^2 \\\\\\:\implies\sf (OT)^2 = (25)^2 - (24)^2 \\\\\\:\implies\sf (OT)^2 = 625 - 576 \\\\\\:\implies\sf (OT)^2 = 49 \\\\\\:\implies\sf OT = \sqrt{49} \\\\\\:\implies\underline{\boxed{\frak{\pink{OT = 7\;cm}}}}\;\bigstar\end{gathered}
:⟹(OP)
2
=(OT)
2
+(PT)
2
:⟹(OT)
2
=(OP)
2
−(PT)
2
:⟹(OT)
2
=(25)
2
−(24)
2
:⟹(OT)
2
=625−576
:⟹(OT)
2
=49
:⟹OT=
49
:⟹
OT=7cm
★
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\therefore{\underline{\textsf{Hence, \; required\; radius\;of\;the\;circle\;is\;\textbf{7 cm}.}}}∴
Hence, requiredradiusofthecircleis7 cm.
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