Help with this please!!!

Mathematics

1 answer:

omeli [17]3 years ago

8 0

Answer:

Sometimes true.

Step-by-step explanation:

If I is not the midpoint of HJ, then it may lie any where on the line HJ and may not lie on HJ.

So, sometime i.e when I lies on HJ, H , I and J are collinear

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The area of triangle ABC is _____. A 60 B65 C130 D150

Romashka [77]

Is there a picture that goes with the question ?

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3 years ago

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

belka [17]

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.