Let $T$ be the triangle formed by the straight line $3x + 5y - 45 = 0$ and 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

We need to find the circumradius of triangle T formed by the line 3x + 5y - 45 = 0 and the coordinate axes.

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The triangle T is formed by:

The given line: 3x + 5y - 45 = 0

The x-axis (y = 0)

The y-axis (x = 0)

To find where the line intersects the axes:

Y-intercept (where line meets y-axis, x = 0):

3(0) + 5y - 45 = 0

5y = 45

y = 9

So point A = (0, 9)

X-intercept (where line meets x-axis, y = 0):

3x + 5(0) - 45 = 0

3x = 45

x = 15

So point B = (15, 0)

The third vertex is the origin O = (0, 0) where the coordinate axes meet.

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Triangle T has vertices at O(0, 0), A(0, 9), and B(15, 0).

Notice that:

Side OA lies along the y-axis

Side OB lies along the x-axis

Since the coordinate axes are perpendicular, angle AOB = 90°

This means triangle T is a right triangle with the right angle at the origin.

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For any right triangle, the circumradius equals half the hypotenuse.

In a right triangle, the hypotenuse is actually a diameter of the circumcircle. Since radius = diameter ÷ 2, we get:

Circumradius = $\frac{\text{Hypotenuse}}{2}$

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The hypotenuse is AB. Using the distance formula:

AB = $\sqrt{(15-0)^2 + (0-9)^2}$

AB = $\sqrt{15^2 + (-9)^2}$

AB = $\sqrt{225 + 81}$

AB = $\sqrt{306}$

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Circumradius L = $\frac{AB}{2} = \frac{\sqrt{306}}{2}$

Since $\sqrt{306} \approx 17.49$:

L $\approx \frac{17.49}{2} \approx 8.75$

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The closest integer to 8.75 is 9.

Therefore, the integer closest to L is 9.

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When you see a triangle formed by a line and the coordinate axes, it's always a right triangle with the right angle at the origin. For such triangles, the circumradius is simply half the distance between the two intercepts.