Question

applying the geometric mean (altitude) theorem
what is the value of k?
k =

Explanation:

Step1: Apply the geometric mean (altitude) theorem

The geometric mean (altitude) theorem states that in a right - triangle, the altitude \(h\) to the hypotenuse is the geometric mean between the segments of the hypotenuse. Here, for the right - triangle \(LMN\) with altitude \(NO = 4\) to the hypotenuse \(ML\) (where \(ML=k + 8\)), and one segment of the hypotenuse is \(k\). The formula is \(NO^{2}=MO\times LO\). But another way (using similar triangles): In right - triangle \(LMN\) with right - angle at \(N\) and altitude \(NO\) to hypotenuse \(ML\), we have \(\triangle LNO\sim\triangle NMO\). Then \(\frac{NO}{MO}=\frac{LO}{NO}\), which gives \(NO^{2}=MO\times LO\). Also, using the geometric mean theorem for the leg: \(MN^{2}=MO\times ML\). Let \(MO = k\) and \(ML=k + 8\), \(MN = 7\). So \(7^{2}=k(k + 8)\).

Step2: Expand and solve the quadratic equation

Expand \(49=k^{2}+8k\). Rearrange to \(k^{2}+8k - 49=0\). Using the quadratic formula \(k=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c = 0\). Here \(a = 1\), \(b = 8\), \(c=-49\). Then \(k=\frac{-8\pm\sqrt{8^{2}-4\times1\times(-49)}}{2\times1}=\frac{-8\pm\sqrt{64 + 196}}{2}=\frac{-8\pm\sqrt{260}}{2}=\frac{-8\pm2\sqrt{65}}{2}=-4\pm\sqrt{65}\). Since \(k>0\) (length), \(k=-4+\sqrt{65}\approx - 4+8.062 = 4.062\approx4\).

Answer:

\(4\)