Question

question 1 of 10
what is the value of x in the figure below? in this diagram, △abd ~ △cad
a. 20
b. √10
c. 20/15
d. 5
e √300
f 45/4

Explanation:

Step1: Use the property of similar triangles

Since \(\triangle ABD\sim\triangle CAD\), we have the proportion \(\frac{CD}{AD}=\frac{AD}{BD}\). Also, by the geometric mean theorem in a right - angled triangle \(ABC\) (right - angled at \(A\) with altitude \(AD\)), \(AC^{2}=CD\times CB\).

We know that \(AC = 15\) and \(CB=20\). Let \(CD=x\), then \(BD = 20 - x\). Using the formula \(AC^{2}=CD\times CB\) (derived from similar triangles \(\triangle ABD\sim\triangle CAD\) and \(\triangle ABC\) properties).

Step2: Substitute the values into the formula

Substitute \(AC = 15\) and \(CB = 20\) into \(AC^{2}=CD\times CB\). We get \(15^{2}=x\times20\).

$$
LATEXBLOCK0
$$

Another way:
Since \(\triangle ABD\sim\triangle CAD\), we know that \(\frac{CD}{AC}=\frac{AC}{CB}\) (corresponding sides of similar triangles).
Substitute \(AC = 15\) and \(CB = 20\) into \(\frac{CD}{AC}=\frac{AC}{CB}\), i.e., \(\frac{x}{15}=\frac{15}{20}\)

$$
LATEXBLOCK1
$$

Answer:

F. \(\frac{45}{4}\)