Question

two circles have an external tangent with length 36 cm.The shortest distance between these circle is 14cm.If the radius of the longer circle is 4 times the radius of the smaller circle then the radius of the larger circle in cms is..

Answer 1

Answer:

Step-by-step explanation:4.4(I guess)

Answer 2

Answer:

20 cm

Step-by-step explanation:

Given:

Let there are two circles whose centers are $C_{1}$ and $C_{2}$

Let radius of $C_{2}$ = $r_{2}$

radius of $C_{1}$ = $r_{1}$ or 4$r_{2}$ { ∵ given, Radius of larger circle is 4 times the radius of smaller circle }

A/Q, C1C2 = 36cm

To find: radius of larger circle

Solution:

According to below picture,

draw a line parallel to tangent such that its one end lies on $C_{2}$ and other are  intersect perpendicularlly at radius of circle $C_{1}$ at point O

Now, we have ΔO$C_{1}$$C_{2}$

In which $OC_{1}$ = $r_{1} - r_{2}$

$C_{2} O$ = 36 CM

$C_{1} C_{2}$ = $r_{1}+14+r_{2}$

By Pythagoras theorem,

$(C_{1}C_{2} )^{2}$ = $(OC_{1} )^{2}$ + $(C_{2}O )^{2}$

=> $(r_{1}+14+r_{2})^{2}$ = $(r_{1}-r_{2} )^{2}$ + $(36)^{2}$

=> $(4r_{2}+14+r_{2} )^{2}$ = $(4r_{2} - r_{2} )^{2}$ + $(36)^{2}$    [ ∵ $r_{1} = 4r_{2}$ ]

=> $(5r_{2} + 14)^{2}$ = $(3r_{2} )^{2}$+ $(36)^{2}$

=> $25(r_{2} )^{2}$ + $196$ + $140r_{2}$ = $9(r_{2} )^{2}$ + $1296$

=> $16(r_{2} )^{2}$ + $140r_{2}$ - $1100 = 0$

=> $4(r_{2} )^{2}$$+ 35r_{2} - 275 = 0$

Hence, $r_{2} = 5 cm$

∴ $r_{1} = 4r_{2}$ = 20cm

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