Question

Please solve q 71 ?

Answer 1

(1) `(7r)/32`

Explanation:

I used a free online graphing calculator to duplicate the drawing:

I used the equations:

`(x-2)^2+(y-0)^2=2^2, y >=0` and `(x-5)^2+(y-0)^2=2^2, y >=0`

to obtain the `1:7` for `C_1P_1:PQ`

By putting the radius of the small circle on an adjustable slider and making its center dependant on the radius, I discovered the equation:

`(x-3.5)^2+(y-0.4375)^2=0.4375^2`

made the small circle be tangent at the three points.

From this information, we obtain the following ratio:

`r_"small"/r_"large"= 0.4375/2`

Multiply by `16/16`

`r_"small"/r_"large"= 7/32`

`r_"small" = (7r_"large")/32`

Please observe that the above equation describes the selection (1).

I know that this is not the geometric solution that you desire but you, now, know the correct answer and can work out a solution.

Answer 2

`(7r)/32`

Explanation:

let `PQ=7a`
given `C_1P_1:PQ=1:7, => C_1P_1=1a`
let `r` be the radius of the semicircle,
`=> PQ=PC_1+C_1P_1+P_1C_2+C_2Q`
`=> 7a=r+1a+r+r`
`=> 3r=6a, => color(red)(r=2a)`
`=> C_1C_2=7a-2r=7a-4a=3a`
`=> P_1P_2=3a-2a=a, => P_1A=P_2A=a/2`, (symmetrical)
Let `O and h` be the center and the radius of the small circle,
In `DeltaC_1OA, " "C_1O^2=OA^2+C_1A^2`
`=> (r-h)^2=h^2+((3a)/2)^2`
`=> (2a-h)^2=h^2+((3a)/2)^2`
`=> 4a^2-4ah+h^2=h^2+9/4a^2`
`=> h=(4a^2-9/4a^2)/(4a)`
`=> h=(7/4a)/4=(7a)/16=(7r)/32`