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faust18 [17]
1 year ago
9

I need help with #1 of this problem. It has writings on it because I just looked up the answer because I’m confused but I want t

o know the answer and how to do it with work provided please

Mathematics
1 answer:
belka [17]1 year ago
4 0

In the figure below

1) Using the theorem of similar triangles (ΔBXY and ΔBAC),

\frac{BX}{BA}=\frac{BY}{BC}=\frac{XY}{AC}

Where

\begin{gathered} BX=4 \\ BA=5 \\ BY=6 \\ BC\text{ = x} \end{gathered}

Thus,

\begin{gathered} \frac{4}{5}=\frac{6}{x} \\ \text{cross}-\text{multiply} \\ 4\times x=6\times5 \\ 4x=30 \\ \text{divide both sides by the coefficient of x, which is 4} \\ \text{thus,} \\ \frac{4x}{4}=\frac{30}{4} \\ x=7.5 \end{gathered}

thus, BC = 7.5

2) BX = 9, BA = 15, BY = 15, YC = y

In the above diagram,

\begin{gathered} BC=BY+YC \\ \Rightarrow BC=15\text{ + y} \end{gathered}

Thus, from the theorem of similar triangles,

\begin{gathered} \frac{BX}{BA}=\frac{BY}{BC}=\frac{XY}{AC} \\ \frac{9}{15}=\frac{15}{15+y} \end{gathered}

solving for y, we have

\begin{gathered} \frac{9}{15}=\frac{15}{15+y} \\ \text{cross}-\text{multiply} \\ 9(15+y)=15(15) \\ \text{open brackets} \\ 135+9y=225 \\ \text{collect like terms} \\ 9y\text{ = 225}-135 \\ 9y=90 \\ \text{divide both sides by the coefficient of y, which is 9} \\ \text{thus,} \\ \frac{9y}{9}=\frac{90}{9} \\ \Rightarrow y=10 \end{gathered}

thus, YC = 10.

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