Question

If XY=32,XZ=30, JQ=12, the radius circumscribed circle of XYZ is 26 find QK

Answer 1

(QK) is also equal to 22.

To find (QK), use the property that in a right triangle, the diameter of the circumscribed circle is twice the length of the hypotenuse.

Given (XY = 32) and (XZ = 30), are the legs of the right triangle (XYZ).

The hypotenuse, (YZ), can be found using the Pythagorean Theorem:

YZ = √(XY² + XZ²)

YZ = √(32² + 30²)

YZ = √1024 + 900

YZ = √1924

YZ ≈ 44

The radius of the circumscribed circle is half the hypotenuse:

R = YZ / 2

R = 44 / 2

R = 22

So, the radius of the circumscribed circle is 22.

To find (QK), use the fact that (QK) is a radius of the circumscribed circle.

i.e QK = R

If R = 22, QK = 22

Therefore, (QK) is also equal to 22.