Question

question 10 of 10
what is the value of ( h ) in the figure below? in this diagram, ( \triangle b a d sim \triangle c b d ).

a. 80
b. ( sqrt{80} )
c. ( \frac{16}{5} )
d. 8
e. 16
f. 5

Explanation:

Step1: Identify corresponding sides of similar triangles

Since \(\triangle BAD \sim \triangle CBD\), the ratios of corresponding sides are equal. So, \(\frac{BD}{CD}=\frac{AD}{BD}\).
We know \(CD = 4\) and \(AC = 20\), so \(AD=AC - CD=20 - 4 = 16\). Let \(BD = h\).

Step2: Set up the proportion and solve for \(h\)

Substitute the values into the proportion: \(\frac{h}{4}=\frac{16}{h}\).
Cross - multiply to get \(h^{2}=4\times16\).
Calculate \(4\times16 = 64\)? Wait, no, wait. Wait, \(AD = 20 - 4=16\)? Wait, no, the length of \(AD\): since \(AC = 20\) and \(CD = 4\), then \(AD=20 - 4 = 16\)? Wait, no, maybe I made a mistake. Wait, the diagram: \(A\) to \(D\) to \(C\), with \(AC = 20\) and \(CD = 4\), so \(AD=20 - 4 = 16\). Then the proportion from similar triangles \(\triangle BAD\sim\triangle CBD\) is \(\frac{BD}{CD}=\frac{AD}{BD}\), so \(h^{2}=AD\times CD\).
So \(h^{2}=16\times4=64\)? Wait, no, that gives \(h = 8\). Wait, let's re - check.
Wait, the right triangle altitude theorem: in a right triangle, the altitude to the hypotenuse is the geometric mean of the segments into which it divides the hypotenuse. So \(h^{2}=AD\times CD\).
\(AD=AC - CD=20 - 4 = 16\), \(CD = 4\). So \(h^{2}=16\times4 = 64\)? No, \(16\times4 = 64\), then \(h=\sqrt{64}=8\)? Wait, but \(16\times4 = 64\), so \(h = 8\).

Wait, maybe I misread the length of \(AD\). Wait, the total length \(AC = 20\), \(CD = 4\), so \(AD=20 - 4=16\). Then by the geometric mean theorem (altitude - on - hypotenuse theorem) for right triangles, \(h^{2}=AD\times CD\). So \(h^{2}=16\times4 = 64\), so \(h = 8\).

Answer:

D. 8