Asked by Joey
An isosceles triangle is inscribed in a circle. The shortest side is the base which is 16 cm long. If the radius of the circle is 10 cm, what is the length of side "a"?
Answers
Answered by
Reiny
I will assume "a" is one of the equal sides of the triangle
make a sketch, by drawing in the altitude to the base of the triangle.
draw in the radius to the base vertex.
You will have a right-angled triangle with sides 8 and x and hypotenuse 10
x^2 + 8^2 = 10^2
x = 6 ( you might have recognized the 3-4-5 right-angled triangle multiplied by a factor of 2 )
So the altitude is 6+10 = 16
Now you have a large right-angled triangle with sides 8 and 16 with hypotenuse "a"
a^2 = 16^2+8^2 = 320
a = √320 = 8√5
make a sketch, by drawing in the altitude to the base of the triangle.
draw in the radius to the base vertex.
You will have a right-angled triangle with sides 8 and x and hypotenuse 10
x^2 + 8^2 = 10^2
x = 6 ( you might have recognized the 3-4-5 right-angled triangle multiplied by a factor of 2 )
So the altitude is 6+10 = 16
Now you have a large right-angled triangle with sides 8 and 16 with hypotenuse "a"
a^2 = 16^2+8^2 = 320
a = √320 = 8√5
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