Asked by Alabi
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
Answers
Answered by
Ms Pi 3.14159265358979323
The height can be calculated using the Pythagorean Theorem : )
c^2 = a^2 + b^2
c^2 = a^2 + b^2
Answered by
Ms Pi 3.14159265358979323
Then the radius is half of the height : )
Answered by
oobleck
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
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
Answered by
Ms Pi 3.14159265358979323
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 : )
Oobleck has an outstanding solution : )
Answered by
henry2,
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.
Y = BM = height.
x^2 + y^2 = 13^2,
5^2 + y^2 = 169,
Y = 12 cm.
Answered by
Me endo
What are u guys doing I don't understand all
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