Please solve q 71 ?

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Please solve q 71 ?

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Answer 1 · 2018-05-15T00:47:05

(1) $\frac{7 r}{32}$ $(7r)/32$

Explanation:

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

![https://www.desmos.com/calculator]( useruploads.socratic.org )

I used the equations:

${(x - 2)}^{2} + {(y - 0)}^{2} = 2^{2}, y \geq 0$ $(x-2)^2+(y-0)^2=2^2, y >=0$ and ${(x - 5)}^{2} + {(y - 0)}^{2} = 2^{2}, y \geq 0$ $(x-5)^2+(y-0)^2=2^2, y >=0$

to obtain the $1 : 7$ $1:7$ for $C_{1} P_{1} : P Q$ $C_1P_1:PQ$

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

${(x - 3.5)}^{2} + {(y - 0.4375)}^{2} = {0.4375}^{2}$ $(x-3.5)^2+(y-0.4375)^2=0.4375^2$

made the small circle be tangent at the three points.

From this information, we obtain the following ratio:

$\frac{r_{small}}{r_{large}} = \frac{0.4375}{2}$ $r_"small"/r_"large"= 0.4375/2$

Multiply by $\frac{16}{16}$ $16/16$

$\frac{r_{small}}{r_{large}} = \frac{7}{32}$ $r_"small"/r_"large"= 7/32$

$r_{small} = \frac{7 r_{large}}{32}$ $r_"small" = (7r_"large")/32$

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 · 2018-05-16T05:18:41

$\frac{7 r}{32}$ $(7r)/32$

Explanation:

enter image source here
let $P Q = 7 a$ $PQ=7a$
given $C_{1} P_{1} : P Q = 1 : 7, \Rightarrow C_{1} P_{1} = 1 a$ $C_1P_1:PQ=1:7, => C_1P_1=1a$
let $r$ $r$ be the radius of the semicircle,
$\Rightarrow P Q = P C_{1} + C_{1} P_{1} + P_{1} C_{2} + C_{2} Q$ $=> PQ=PC_1+C_1P_1+P_1C_2+C_2Q$
$\Rightarrow 7 a = r + 1 a + r + r$ $=> 7a=r+1a+r+r$
$\Rightarrow 3 r = 6 a, \Rightarrow r = 2 a$ $=> 3r=6a, => color(red)(r=2a)$
$\Rightarrow C_{1} C_{2} = 7 a - 2 r = 7 a - 4 a = 3 a$ $=> C_1C_2=7a-2r=7a-4a=3a$
$\Rightarrow P_{1} P_{2} = 3 a - 2 a = a, \Rightarrow P_{1} A = P_{2} A = \frac{a}{2}$ $=> P_1P_2=3a-2a=a, => P_1A=P_2A=a/2$ , (symmetrical)
Let $O and h$ $O and h$ be the center and the radius of the small circle,
In $DeltaC_1OA, " "C_1O^2=OA^2+C_1A^2$
$=> (r-h)^2=h^2+((3a)/2)^2$
$=> (2a-h)^2=h^2+((3a)/2)^2$
$=> 4a^2-4ah+h^2=h^2+9/4a^2$
$=> h=(4a^2-9/4a^2)/(4a)$
$=> h=(7/4a)/4=(7a)/16=(7r)/32$

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Please solve q 71 ?

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