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Sia ? 6 years, 6 months ago
We know that ∠ADO' = 90° ( since O'D is perpendicular to AC)
∠ACO= 90° ( OC(radius)perpendicular to AC(tangent))
In triangles ADO'and ACO ,
∠ADO' = ∠ACO ( each 90°)
∠DAO = ∠CAO (common)
by AA criterion, triangles ADO' and ACO are similar to each other.
{tex}{AO'\over AO }= {DO'\over CO}{/tex} ( corresponding sides of similar triangles)
AO = AO' + O'X + OX
= 3AO' (since AO'=O'X=OX because radii of the two circles are equal )
{tex}{AO' \over AO }= {AO' \over3AO} ={1\over3}{/tex}
{tex} \\ {DO'\over CO}={AO'\over AO} = {1\over 3 }{/tex}
{tex}{DO'\over CO}={1\over 3 }.{/tex}
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Sia ? 6 years, 6 months ago
{tex}a, 7, b, 23{/tex} and {tex}c{/tex} are in A.P.
Let the common difference be d.
{tex}a + d = 7{/tex} ....(i)
{tex}a + 3d = 23{/tex} ....(ii)
From (i) and (ii), we get
{tex}2d = 16{/tex}
{tex}d = 8{/tex}
Put d = 8 in (1) we get
{tex}a + 8 = 7{/tex}
{tex}a = -1{/tex}
{tex}b = a + 2d{/tex}
{tex}b = - 1{/tex} + 2 {tex}\times{/tex} 8
or,{tex} b = - 1 + 16{/tex}
or, {tex}b = 15{/tex}
{tex}c = a + 4d{/tex}
= - 1 + 4 {tex}\times{/tex} 8
= - 1 + 32
{tex}c = 31{/tex}
{tex}\therefore{/tex} {tex}a = - 1,b = 15, c =31{/tex}
Posted by Keerthika Aishu 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
let the three consecutive terms of AP be {tex}a_1,a_2,a_3{/tex}
where {tex}a_1=(2p+1),a_2=13,a_3=(5p-3){/tex}
we know that difference between any two consecutive terms of an AP is equal
{tex}\therefore{a_2} - {a_1} = {a_3} - {a_2}{/tex}
{tex} \Rightarrow 13 - (2p + 1) = (5p - 3) - 13{/tex}
{tex} \Rightarrow 12 - 2p = 5p - 16{/tex}
{tex} \Rightarrow 7p = 28{/tex}
{tex} \Rightarrow p = 4{/tex}
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Sia ? 6 years, 6 months ago
Since P is mid-point of line segment AB, then
{tex}\frac { - 6 + ( - 2 ) } { 2 } = \frac { a } { 2 }{/tex}
{tex}\Rightarrow \quad \frac { - 8 } { 2 } = \frac { a } { 2 }{/tex}
{tex}\Rightarrow{/tex} a = - 8
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Sia ? 6 years, 6 months ago
Check previous year papers : https://mycbseguide.com/cbse-question-papers.html
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Sia ? 6 years, 4 months ago
Given, {tex}\alpha {/tex} and {tex}\beta{/tex} are the zeroes of polynomial {tex}{x^2} - p(x + 1) + c{/tex}
which can be written as {tex}{x^2} - px + c - p{/tex}
So, sum of zeroes, {tex}\alpha + \beta = p{/tex} {tex}[\because {/tex} sum of coefficients = {tex}\frac{{ - (coefficient(x))}}{{\operatorname{co} efficient({x^2})}}{/tex}]
and product of zeroes {tex}\alpha \beta = c - p{/tex} {tex}[\because {/tex} product of cofficients= {tex}\frac{{constant\_term}}{{coefficient({x^2})}}{/tex},]
Also, {tex}(\alpha + 1)(\beta + 1) = 0{/tex}
{tex}\alpha\beta + \alpha + \beta + 1 = 0{/tex}
{tex} \Rightarrow c - p + p + 1 = 0{/tex}
{tex}\Rightarrow c = - 1{/tex}
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