The radius of a circle

Form a triangle similar to the one in the diagram as follows. Let the base go from the point of tangency of the leftmost circle with the 60 cm side of the main triangle to the center of the rightmost circle. Let the hypotenuse run parallel to the hypotenuse of the main triangle from the center of the rightmost circle to the 60 cm side of the main triangle. The vertical side then runs between the points where the base and hypotenuse intersect the 60 cm side.

The base then has a length of $7r,$ where $r$ is the radius of each of the circles, and so by similarity the height of our secondary triangle has a length of $\frac{60}{80}*7r = \frac{21}{4}r.$ But we can also measure this vertical side in a second way. Since the hypotenuse of our secondary triangle is a perpendicular distance of $r$ from the hypotenuse of the main triangle, by similarity the distance between the upper vertices of the two triangles will be $\frac{100}{80}*r = \frac{5}{4}r.$ Thus the length of the vertical side of our secondary triangle can also be represented by $60 - r - \frac{5}{4}r = 60 - \frac{9}{4}r,$ and so

$\dfrac{21}{4}r = 60 - \dfrac{9}{4}r \Longrightarrow \dfrac{30}{4}r = 60 \Longrightarrow \boxed{r = 8 cm}.$

Imagine a similar triangle, which height is the diameter of the last circle, then the base of the triangle must be 80-7r (r: radius) // y: height and x: base, then we know by similarity that y: 6x/8, x: 80-7r , and y: 2r, we find x in the equation and it equals to 22.0689 then y: 16.55 and therefore the radius is 8.2758

AB = 80; AC = 60; BC = 100 (this is a Pitagoric triangle!) Area of the triangle ABC = (AB x AC)/2 = (60 x 80)/2=2400 Consider the point D as the center of the circle closest to the vertex B. The aforementioned area is also the sum of the areas of the triangles ABD, BCD, and ACD. Note that: Height of the triangle ABD = r; Height of the triangle BCD = r; and Height of the triangle ACD = 7r. Therefore: (80 x r)/2 + (100 x r)/2 + (60 x 7r)/2 = 2400; 40r + 50r + 210r = 2400; 300r = 2400 and r=8.

This question should not be Level 1 for about 25 points.

Tan 2 Q = 60/ 80 = 3/ 4 = 2 t/ (1 - t^2) where t = Tan Q.

3 t^2 + 8 t - 3 = 0 => (3 t - 1)(t + 3) = 0 => t = 1/ 3 as Q is positive.

Tan Q = t = r/ x = 1/ 3 => x = 3 r.

7 r + 3 r = 80 => r = 80/ 10 = 8

Answer: 8 cm

All I did was realize that at least one more full circle would fit across the bottom if it wasn't for the hypotenuse. From there you can divide 80cm by 10, number of radii in 5 circles. If it had said "not drawn to scale" I would have been wrong.

Consider last circle touching hypotenuse

Using similarity--80-8 r/80=2 r/60

we get r=7.5 thus giving 8

It's just common sense really. You go through each radius and you will notice non of them are big enough except from 8

The drawing is clearly wrong. 8 cm means 10 halfs of the lengths of 5 circles, which are clearly shorter than 80cm.