Question

challenge:
3)
ik =
hi =
hk =

Explanation:

Step1: Use the geometric mean theorem (altitude-on-hypotenuse theorem)

In a right - triangle, if an altitude is drawn to the hypotenuse, then the length of the altitude is the geometric mean of the lengths of the two segments of the hypotenuse. Also, each leg of the right - triangle is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to that leg.

For right - triangle \( \triangle JHI\) with altitude \( IL\perp JH\) and \( IK\perp JH\). First, for \(IK\):
In right - triangle \( \triangle IJL\) and \( \triangle IKL\) (not the main application, but we know that in right - triangle \( \triangle JHI\) with \( \angle I = 90^{\circ}\) and \( KL\perp JH\), \( IL\) is the altitude. But for \(IK\), we use the formula \(IK=\sqrt{JL\times LI}\). Since \(JL = 15\) and \(LI = 3\), then \(IK=\sqrt{15\times3}=\sqrt{45}=3\sqrt{5}\)

Step2: Use the geometric mean theorem for \(HI\)

The formula for the leg of a right - triangle (here \(HI\)) is \(HI=\sqrt{JI\times LI}\). Since \(JI=15 + 3=18\) and \(LI = 3\), then \(HI=\sqrt{18\times3}=\sqrt{54}=3\sqrt{6}\)

Step3: Use the Pythagorean theorem for \(HK\)

We know that \(HI = 3\sqrt{6}\) and \(IK=3\sqrt{5}\). In right - triangle \( \triangle HIK\), by the Pythagorean theorem \(HK=\sqrt{HI^{2}-IK^{2}}\)

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Answer:

\(IK = 3\sqrt{5}\), \(HI=3\sqrt{6}\), \(HK = 3\)