Question

Two circle od radii 26 cm and 10 cm are concentric.Find the length of a chord of the outer circle which touches the inner​

Answer 1

Step-by-step explanation:

Answer

Given−

OisthecentreoftwoconcentriccirclesandAB,which

isachordoftheoutercircle,touchestheinnercircleatP.

Theradiusoftheinnercircleis10cmand

theradiusoftheoutercircleis26cm.

Tofindout−

thelengthofAB=?

Solution−

WejoinOP.

ThenOPisaradiusthroughPwhichisthepointofcontact

ofABtotheinnercircle.i.ePO=10cm.

AlsowejoinOA.ThenOAistheradiusofthe

outercircle=26cm.

NowOP⊥AB⟹∠OPA=90

o

sincetheradiusofacirclethroughthe

pointofcontactofatangenttothecircleisperpedicular

tothetangent.

∴ΔOPAisarightonewithOAashypotenuse.

So,applyingPythagorastheorem,

AP=

OA

2

−OP

2

=

26

2

−10

2

cm=24cm

ButAB=2APsincetheperpendicular,droppedfromthecenter

ofacircletoanyofitschord,bisectsthelatter.

i.ePisthemidpointofAB.

∴AB=2×24cm=48cm.

Answer 2

Answer:

let r be 26 cm

and R be 10 cn

A center of both circles

AB perpendicular to DC

as radius from Centre is perpendicular

to the tangent at the point of

Thus ∆ ABC

R^2+BC ^2= r^2

BC ^2 = 26^2 - 10^2

BC ^2 = 676-100

BC^3 = 576

BC = √576

BC = 24cm

similarly, DB = BC

Thus DC = 2BC

DC =24×2

DC = 48cm = chord length