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Sia ? 6 years, 6 months ago
Let a be the first term and d be the common difference of the given AP. Therefore, the sum of first n terms is given by
{tex}S _ { n } = \frac { n } { 2 } \cdot \{ 2 a + ( n - 1 ) d \}{/tex}
{tex}\therefore{/tex} S10 = {tex}\frac{{10}}{2}{/tex}{tex}\cdot{/tex}(2a+9d) {tex}\Rightarrow{/tex}5(2a+9d)=210
{tex}\Rightarrow{/tex}2a+9d=42. ...(i)
Sum of last 15 terms = (S50 - S35).
{tex}\therefore{/tex} (S50 - S35) = 2565
{tex}\Rightarrow{/tex} {tex}\frac{{50}}{2}{/tex}(2a+49d)- {tex}\frac{{35}}{2}{/tex}(2a+34d)=2565
{tex}\Rightarrow{/tex} 25(2a+49d)-35(a+17d)=2565
{tex}\Rightarrow{/tex} (50a-35a)+(1225d-595d)=2565
{tex}\Rightarrow{/tex} 15a+630d = 2565 {tex}\Rightarrow{/tex} a + 42d = 171 ...... (ii)
Therefore, on solving (i) and (ii), we get a=3 and d=4.
Hence, the required AP is 3,7,11,15,19.....
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Sia ? 6 years, 6 months ago
Let ABC be the right angled triangle such that ∠B = 90° , BC = 6 cm, AB = 8 cm. Let O be the centre and r be the radius of the in circle.
AB, BC and CA are tangents to the circle at P, Nand M.
{tex}\therefore{/tex} OP = ON = OM = r (radius of the circle)
Area of the {tex}\triangle{/tex}ABC = {tex}\frac{1}{2} \times 6 \times 8{/tex}= 24 cm2
By Pythagoras theorem,
CA2 = AB2 + BC2
{tex}\Rightarrow{/tex} CA2 = 82 + 62
{tex}\Rightarrow{/tex} CA2 = 100
{tex}\Rightarrow{/tex} CA = 10 cm
Area of the {tex}\triangle{/tex}ABC = Area {tex}\triangle{/tex}OAB + Area {tex}\triangle{/tex}OBC + Area {tex}\triangle{/tex}OCA
24 = {tex}\frac{1}{2} r \times \mathrm{AB}+\frac{1}{2} r \times \mathrm{B} \mathrm{C}+\frac{1}{2} r \times \mathrm{C} \mathrm{A}{/tex}
24 = {tex}\frac{1}{2} r{/tex}(AB + BC + CA)
{tex}\Rightarrow r=\frac{2 \times 24}{(\mathrm{AB}+\mathrm{BC}+\mathrm{CA})}{/tex}
{tex}\Rightarrow r=\frac{48}{8+6+10}{/tex}
{tex}\Rightarrow r=\frac{48}{24}{/tex}
{tex}\Rightarrow{/tex} r = 2 cm
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Posted by Samar Agrahari 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 Santu Choudhary 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Sum of the volumes of three given spheres = {tex}\frac43\mathrm\pi{/tex}[(6)3 + (8)3 + (10)3]
= {tex}\frac43\mathrm\pi{/tex}[216 + 512 + 1000]
= {tex}\frac43\mathrm\pi{/tex} × 1728 = 4{tex}\pi{/tex} × 576 = 2304π cm3
Let R be the radius of single solid sphere, since volume remains the same
∴ {tex}\frac43\mathrm\pi{/tex}R3 = 2304 {tex}\pi{/tex}
⇒ R3 ={tex}\frac{2304\times3}4=\;576\times3{/tex}
⇒ R3 = 1728 = (12)3
∴ R = 12 cm
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Amit Gupta 7 years, 2 months ago
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