Question

1. use the diagram to complete each equation.\\(\frac{b}{a} = \frac{y}{h}\\) \\(\frac{b}{a} = \frac{h}{x}\\)\\(\frac{c}{a} = \frac{b}{h}\\) \\(\frac{c}{a} = \frac{a}{x}\\) \\(\frac{c}{b} = \frac{b}{y}\\)

Explanation:

Step1: Analyze Similar Triangles

The diagram shows a right triangle with an altitude \( h \) to the hypotenuse \( c \), creating two smaller similar right triangles and the original triangle. By the geometric mean theorem (or altitude-on-hypotenuse theorem), corresponding sides of similar triangles are proportional.

Step2: Solve \(\frac{b}{a}=\frac{y}{h}\)

In similar triangles, the ratio of a leg of the large triangle to the other leg equals the ratio of the corresponding segments of the hypotenuse to the altitude. Here, \( \frac{b}{a}=\frac{y}{h} \) (matching the first blank with \( y \)).

Step3: Solve \(\frac{b}{a}=\frac{h}{x}\)

This is already given as correct (proportionality from similar triangles: leg/leg = altitude/segment of hypotenuse).

Step4: Solve \(\frac{c}{a}=\frac{b}{h}\)

Using similar triangles, hypotenuse/leg = leg/altitude? Wait, no—correctly, hypotenuse/leg = leg/segment? Wait, re - evaluating: From similar triangles, \( \frac{c}{a}=\frac{a}{x} \)? Wait, no, the original problem has blanks. Wait, the first blank in \(\frac{b}{a}=\frac{\underline{}}{h}\) is \( y \) (since in similar triangles, the ratio of the longer leg \( b \) to the shorter leg \( a \) is equal to the ratio of the segment of the hypotenuse adjacent to \( b \) (which is \( y \)) to the altitude \( h \)).

For \(\frac{c}{a}=\frac{b}{\underline{}}\), using the proportion from similar triangles (the large triangle and the triangle with base \( b \) and height \( h \)), we have \(\frac{c}{a}=\frac{b}{h}\)? Wait, no, actually, from the geometric mean, \( \frac{c}{a}=\frac{a}{x} \) and \( \frac{c}{b}=\frac{b}{y} \). Wait, let's re - establish the proportions:

Let the original right triangle have legs \( a \), \( b \), hypotenuse \( c \), altitude \( h \) to hypotenuse, dividing hypotenuse into \( x \) (adjacent to \( a \)) and \( y \) (adjacent to \( b \)).

By similar triangles:

• Triangle with legs \( a \), \( h \) and hypotenuse \( x \) is similar to triangle with legs \( h \), \( y \) and hypotenuse \( b \), and to the original triangle with legs \( a \), \( b \) and hypotenuse \( c \).

So, \( \frac{b}{a}=\frac{y}{h} \) (corresponding sides: \( b \) corresponds to \( y \), \( a \) corresponds to \( h \) in similar triangles)
\( \frac{b}{a}=\frac{h}{x} \) (corresponding sides: \( b \) corresponds to \( h \), \( a \) corresponds to \( x \))
\( \frac{c}{a}=\frac{a}{x} \), but the problem has \( \frac{c}{a}=\frac{b}{\underline{}} \). Wait, maybe a typo, but following the given structure, for \( \frac{c}{a}=\frac{b}{\underline{}} \), if we use the proportion \( \frac{c}{b}=\frac{b}{y} \), and cross - multiplying, but for \( \frac{c}{a}=\frac{b}{h} \), let's check:

From \( \frac{b}{a}=\frac{y}{h} \) and \( \frac{c}{b}=\frac{b}{y} \), we can substitute \( y=\frac{b\times h}{a} \) into \( \frac{c}{b}=\frac{b}{y} \), we get \( \frac{c}{b}=\frac{b}{\frac{b\times h}{a}}=\frac{a}{h} \), so \( \frac{c}{a}=\frac{b}{h} \) (by cross - multiplying \( c\times h=a\times b \), which is the area formula, and also a geometric mean result: \( a\times b = c\times h \)).

For \( \frac{c}{a}=\frac{a}{\underline{}} \), from similar triangles, \( \frac{c}{a}=\frac{a}{x} \), so the blank is \( x \).

For \( \frac{c}{b}=\frac{\underline{}}{y} \), from similar triangles, \( \frac{c}{b}=\frac{b}{y} \), so the blank is \( b \).

Answer:

For \(\frac{b}{a}=\frac{\underline{}}{h}\), the answer is \( y \); for \(\frac{c}{a}=\frac{b}{\underline{}}\), the answer is \( h \); for \(\frac{c}{a}=\frac{a}{\underline{}}\), the answer is \( x \); for \(\frac{c}{b}=\frac{\underline{}}{y}\), the answer is \( b \) (filling in the blanks in the given equations as per the geometric mean theorem for right triangles with an altitude to the hypotenuse).