An isosceles triangle ABC has its vertices on a circle. If AB=13cm,BC =13cm and AC=10cm, calculate:the height BM of the triangle. Calculate the radius of the circle, to the nearest whole cm
The height can be calculated using the Pythagorean Theorem : )
c^2 = a^2 + b^2
BM divides the base into two 5cm sections, giving you two 5-12-13 right triangles. So, BM = 12.
Let the center of the circle be O. BM = OM + r = 12.
Now you have an isosceles triangle MOC whose sides are 12-r, r-5, and 5.
So, now just solve
5^2 + (12-r)^2 = r^2
What are u guys doing I don't understand all
a. X = AC/2 = 10/2 = 5 cm.
Y = BM = height.
x^2 + y^2 = 13^2,
5^2 + y^2 = 169,
Y = 12 cm.
To calculate the height (BM) of the isosceles triangle, we can use the Pythagorean theorem.
In triangle ABC, let's denote the center of the circle as O, and let M be the midpoint of side AC. Since triangle ABC is isosceles, we know that AM = CM.
We can draw a perpendicular line from O to side AC, intersecting at point D. This line will bisect side AC.
By the Pythagorean theorem, we can find the height BM using the formula:
BM^2 = BD^2 + DM^2
First, let's find BD. Since BD is the perpendicular distance from the center of the circle to side AC, it is also the radius of the circle. Let's calculate BD using the theorem of right triangles in triangle OBD:
OD^2 = OB^2 - BD^2
Substituting the given lengths, we have:
OD^2 = (AB/2)^2 - BD^2
OD^2 = (13/2)^2 - BD^2
OD^2 = 169/4 - BD^2
Since OD is equal to the radius of the circle, we can substitute R for OD. Taking the square root of both sides, we have:
R = sqrt(169/4 - BD^2)
Next, let's calculate DM. DM is half the length of side AC, so DM = AC/2 = 10/2 = 5cm.
Now we can calculate BM using the formula:
BM^2 = BD^2 + DM^2
BM = sqrt(BD^2 + DM^2)
Substituting the values we have:
BM = sqrt(R^2 + 5^2)
To find the height BM and the radius R to the nearest whole centimeter, we need to substitute the values of BD and R that satisfy the given conditions.
Since AB = BC, the triangle ABC is an isosceles right triangle. The ratio of the sides in a right isosceles triangle is 1:1:√2. Therefore, AB = BC = 13cm, and by Pythagorean theorem AC = 13√2 cm.
Using these values:
BM = sqrt(R^2 + 5^2) = sqrt(169/4 - BD^2) = sqrt(13/2)^2 + 5^2) = sqrt(169/4 - 25) = sqrt(169 - 100)/2 = sqrt(69)/2
To find the height BM, we calculate:
BM = sqrt(69)/2 ≈ 4.16 cm
To find the radius R, we plug in the height BM and solve for R:
4.16 ≈ sqrt(R^2 + 5^2)
(4.16)^2 ≈ R^2 + 5^2
17.3056 ≈ R^2 + 25
R^2 ≈ 17.3056 - 25
R^2 ≈ -7.6944
Since we can't have a negative radius, it means that the given side lengths (AB, BC, AC) do not form a valid isosceles triangle within a circle. Please double-check your measurements or provide correct values for the triangle sides.
ooooops! The triangle does not hit the outside edge of the circle, thus the height must be more than 5 : ) So sorry... I haven't had my first cup of tea yet.
Oobleck has an outstanding solution : )