Question-217802

Question Number 217802 by Ismoiljon_008 last updated on 21/Mar/25

Commented by Ismoiljon_008 last updated on 21/Mar/25

a + b = c T h e n R = ?

H e l p m e, p l e a s e

Commented by MathematicalUser2357 last updated on 26/Mar/25

I just tried to get the intersection points where the center of the circle is (p, q) but got

x = \frac{- 4 (4 p + 3 q - 18) \pm 4 \sqrt{{(4 p + 3 q - 18)}^{2} - 25 ({(q - 6)}^{2} - r^{2} + p^{2})}}{25}

We have to solve for c like this :

c = \int_{\frac{- 4 (4 p + 3 q - 18) - 4 \sqrt{{(4 p + 3 q - 18)}^{2} - 25 ({(q - 6)}^{2} - r^{2} + p^{2})}}{25}}^{\frac{- 4 (4 p + 3 q - 18) + 4 \sqrt{{(4 p + 3 q - 18)}^{2} - 25 ({(q - 6)}^{2} - r^{2} + p^{2})}}{25}} (\sqrt{r^{2} - {(x - p)}^{2} +} \frac{3}{4} x - 6) d x

But I have no idea to solve this

Answered by mr W last updated on 21/Mar/25

Commented by mr W last updated on 21/Mar/25

a + b + d = t r i a n g l e = \frac{6 \times 8}{2}

c + d = \frac{3}{4} o f c i r c l e + s q u a r e R \times R = \frac{3 π R^{2}}{4} + R^{2}

s i n c e a + b = c,

\frac{3 π R^{2}}{4} + R^{2} = \frac{6 \times 8}{2}

\Rightarrow R = \sqrt{\frac{96}{3 π + 4}} ✓

Commented by Ismoiljon_008 last updated on 21/Mar/25

T h a n k y o u v e r y m u c h