An isoscelless triangle of sides of 13cm, 13cm, 10cm is inscribed in a circle is inscribed in a circle. What is the radius of the circle?
To find the radius of the circle inscribed in an isosceles triangle, we can use the formula:
radius = (side length) / (2 * sin(angle))
In this case, the triangle is isosceles with sides of 13cm, 13cm, and 10cm. Since the two equal sides are opposite to the same angle at the base, we can use the Law of Cosines to find the measure of that angle.
Using the Law of Cosines: c^2 = a^2 + b^2 - 2 * a * b * cos(C), where c is the side opposite angle C, and a and b are the other two sides of the triangle.
With c = 10cm, a = b = 13cm, and solving for cos(C), we have:
cos(C) = (a^2 + b^2 - c^2) / (2 * a * b)
= (13^2 + 13^2 - 10^2) / (2 * 13 * 13)
= (169 + 169 - 100) / (2 * 13 * 13)
= 338 / (2 * 13 * 13)
= 338 / (338)
= 1
Since cos(C) = 1, the angle C is 0 degrees. Therefore, the triangle is degenerate and cannot be inscribed in a circle. So, there is no answer for the radius of the circle in this case.
draw radii from center to the three intersections on the circumference
all length r
sine of half angle up at intersection of 13 cm legs = 5/13
so half angle = 22.6 deg
isosceles triangle now with legs r meeting base 13 at 22.6 angle
cos 22.6 = r/6.5
r = 6.5 cos 22.6
r = 6
From the diagram
Sin<=5/13=0.3846
<=22.6
22.6*2=45.2(< at centre=2 times< at circumference
sin45.2=5/r
r=7.04cm