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Let T be the triangle formed by the straight line $3x+5y-45=0$ and T the coordinate axes. Let the circumcircle of T have radius of length L, measured in the same unit as the coordinate axes. Then, the integer closest to L is ______

Given that,  the straight line $3x + 5y – 45 = 0 \quad \longrightarrow (1)$

We know that,

• In $x – \text{axis}, y = 0$
• In $y – \text{axis}, x = 0$

Now, put $x=0$ in the equation $(1),$ we get

$3(0) + 5y – 45 = 0$

$\Rightarrow 5y – 45 = 0$

$\Rightarrow \boxed{y=9}$

We get,  the coordinate $\boxed{(x,y) = (0,9)}$

And, put $y=0$ in the equation $(1),$ we get

$3x + 5(0) – 45 = 0$

$\Rightarrow 3x – 45 = 0$

$\Rightarrow \boxed{x=15}$

We get,  the coordinate $\boxed{(x,y) = (15,0)}$

Now, we can draw the diagram as follow :

We have$, \triangle \text{AOB}$ is right angle triangle, then we can apply the Pythagoras’ theorem :

$\boxed{ \text{Hypotenuse}^{2} = \text{Perpendicular}^{2} + \text{Base}^{2}}$

Now, $\text{(AB)}^{2} = \text{(OA)}^{2} + \text{(OB)}^{2}$

$\Rightarrow (2 \text{L})^{2} = 15^{2} + 9^{2}$

$\Rightarrow 4 \text{L}^{2} = 225 +81$

$\Rightarrow 4 \text{L}^{2} = 306$

$\Rightarrow \text{L}^{2} = \frac{306}{4} = 76 \cdot 5$

$\Rightarrow \text{L} = \sqrt{76 \cdot5} = 8 \cdot746 \approx 9.$

$\therefore$ The radius of circumcircle $\text{L} = 9$ units.

$\textbf{PS:}$  In a right-angle $\triangle \text{AOB},$ hypotenuse is the diameter of a circumcircle.

Correct Answer $:9$

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