A circle of radius 6 is externally tangent to a circle with radius 24. The area of the triangular region bounded by the three common tangent lines of these two circles is R, the value of largest digit in R is _______

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answer is 9.

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Detailed Solution

$C_{1} N = 24, C_{2} M = 6 C_{1} C_{2} = C_{1} S + C_{2} S = 24 + 6 = 30 \frac{C_{2} M}{C_{2} P} = \frac{C_{1} N}{C_{1} P} \Rightarrow \frac{6}{l} = \frac{24}{30 + l} \Rightarrow 180 + 6 l = 24 l \Rightarrow l = 10 S P = S C_{2} + C_{2} P = 6 + l = 16$

$M P = \sqrt{{(C_{2} P)}^{2} - {(M P)}^{2}} = \sqrt{10^{2} - 6^{2}} = 8 \frac{C_{2} M}{M P} = \frac{S Q}{P S} \Rightarrow \frac{6}{8} = \frac{S Q}{16} \Rightarrow S Q = 12$

Area of $Δ l e Q R P = \frac{1}{2} \times 24 \times 16 = 192$

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