an isosceles triangle ABC has its vertices on a circle.if /AB/=13CM,/BC/=13cm and /AC/=10CM,calculate the radius of the circle,to nearest whole cm
Place the circle in such a way that the midpoint of CA becomes the origin, and
B lies on the y-axis
by Pythagoras BO = 12 cm
so we have points A(5,0), B(0,12) and C(-5,0)
let the centre be (0,k) and the radius r
equation:
x^2 + (y-k)^2 = r^2
using (5,0) --> 25 + k^2 = r^2
using (0,12) -> 0 + (12-k)^2 = r^2
12-k = r
k = 12-r
in 25+k^2 = r^2
25 + (12-r)^2 = r^2
25 + 144 - 24r + r^2 = r^2
24r = 169
r = 169/24 = 7.04166
The radius is appr 7 cm
from diagram
sin=5/13=0.3846
=22.6
now angle at center is twice the angle at any point of the circumference
22.6*2=45.2
sin45.2=5/r
now find r
It is correct and good
Explain it with diagram pls
It remain the diagram
Yes
Thanks but i mean using diagram
Diagram please
To find the radius of the circle on which the isosceles triangle ABC is inscribed, we can use the fact that the perpendicular bisectors of the sides of a triangle meet at the center of the circumcircle.
Step 1: Draw a diagram depicting the isosceles triangle ABC inscribed in a circle.
Step 2: Since triangle ABC is isosceles, we know that angle ABC is congruent to angle BCA. Therefore, the perpendicular bisector of side AC will pass through the midpoint of side BC. Label this midpoint as M.
Step 3: Find the length of side BC. From the given information, we know that BC is 13 cm.
Step 4: Find the length of the perpendicular bisector of side AC, which is the same as the distance from point M to the center of the circle (radius of the circle). To find this distance, we can apply the Pythagorean theorem to the right triangle BMC (formed by the perpendicular bisector, BM, and half of side BC).
- First, find the length of BM. Since triangle BMC is a right triangle, we can use the Pythagorean theorem: BM^2 = BC^2 - MC^2.
- BM^2 = 13^2 - (1/2 * BC)^2 (MC = 1/2 * BC)
- BM^2 = 169 - 42.25
- BM^2 = 126.75
- BM ≈ √126.75 ≈ 11.26 cm
Step 5: The length of BM is the radius of the circle. So, the radius of the circle is approximately 11.26 cm, which, to the nearest whole cm, is 11 cm.
Therefore, the radius of the circle, to the nearest whole cm, is 11 cm.