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Ask QuestionPosted by Pawan Kumar Nanda 7 years, 9 months ago
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Aishaa Agrwll 7 years, 9 months ago
Posted by Aditya Makwana 7 years, 9 months ago
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Posted by Aruhi Sharma 7 years, 9 months ago
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Posted by Tanu Priya 7 years, 9 months ago
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Posted by Shruti Sahu 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Given radius, OP = OQ = 5 cm
Length of chord, PQ = 4 cm
OT {tex}\perp{/tex} PQ,
{tex}\therefore{/tex} PM = MQ = 4 cm [Perpendicular draw from the centre of the circle to a chord bisect the chord]
In right {tex}\triangle{/tex}OPM, OP2 = PM2 + OM2
{tex}\Rightarrow{/tex} 52 = 42 + OM2
{tex}\Rightarrow{/tex} OM2 = 25 – 16 = 9
Hence OM = 3cm
In right {tex}\triangle{/tex}PTM, PT2 = TM2 + PM2 {tex}\rightarrow{/tex} (1)
OPT = 90o [Radius is perpendicular to tangent at point of contact]
In right {tex}\triangle{/tex}OPT, OT2 = PT2 + OP2 {tex}\rightarrow{/tex} (2)
From equations (1) and (2), we get
OT2 = (TM2 + PM2) + OP2
{tex}\Rightarrow{/tex} (TM + OM)2 = (TM2 + PM2) + OP2
{tex}\Rightarrow{/tex} TM2 + OM2 + 2 {tex}\times{/tex} TM {tex}\times{/tex} OM = TM2 + PM2 + OP2
{tex}\Rightarrow{/tex} OM2 + 2 {tex}\times{/tex} TM {tex}\times{/tex} OM = PM2 + OP2
{tex}\Rightarrow{/tex} 32 + 2 {tex}\times{/tex} TM {tex}\times{/tex} 3 = 42 + 52
{tex}\Rightarrow{/tex} 9 + 6TM = 16 + 25
{tex}\Rightarrow{/tex} 6TM = 32
{tex}\Rightarrow{/tex} TM ={tex}\frac{16}3{/tex}
Equation (1) becomes,
PT2 = TM2 + PM2
= ({tex}\frac{16}3{/tex})2 + 42
= ({tex}\frac{256}9{/tex}) + 16 = {tex}\frac{256+ 144}9{/tex}
= ({tex}\frac{400}9{/tex}) = ({tex}\frac{20}3{/tex})2
Hence PT ={tex}\frac{20}3{/tex}
Thus, the length of tangent PT is ({tex}\frac{20}3{/tex}) cm.
Posted by Roshan Kumar 7 years, 9 months ago
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Ayushman Giri 7 years, 9 months ago
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Pamel Insan 7 years, 9 months ago
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Posted by Tushar Singh 7 years, 9 months ago
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Posted by Hiba Hamza 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
Natural numbers between 101 and 999 divisible by both 2 and 5, i.e., divisible by L.C.M of 2 and 5 i.e., 10, are 110, 120, 130 ,...., 990.
so, this forms an AP 110,120....,990., where first term({tex}a{/tex})=110 and common difference({tex}d{/tex})=120-110=10, and last term({tex}l{/tex})=990
Let {tex}l = {a_n} = a + (n - 1)d{/tex}
Then, {tex}990 = 110 + (n - 1) \times 10{/tex}
{tex} \Rightarrow n = 89{/tex}
Posted by Neha Kumari 7 years, 9 months ago
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Gaurav Sharma 7 years, 9 months ago
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Aishaa Agrwll 7 years, 9 months ago
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