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Mathematics

1 answer:

siniylev [52]3 years ago

8 0

To prove two triangles are similar, we only need two pairs of congruent angles.

The first pair of congruent angles could be the two right angles AED and GFC as shown by the square markers.

As for the other pair of congruent angles, we could pick on angle DAE and angle FGC, which are corresponding angles. The corresponding angles are congruent because of the vertical parallel line segments BA and FG

Once you have two pairs of congruent angles, you have effectively wrapped up the similarity proof. The specific theorem you use is the AA (angle angle) similarity theorem.

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Allison claims that the (triangle)ABQ is similar to (triangle)RPQ, given that AB and PR are parallel. Which of Allison's claims

S_A_V [24]

I draw the two triangles, see the picture attached.

As you can see, angle 1 and 2 are vertically opposite angles because they are formed by the same two crossing lines and they face each other.

Angles <span>ABQ and QPR, as well as angles BAQ and QRP, are alternate interior angles because they are formed by </span><span>two parallel lines crossed by a transversal, and they are inside the two lines on opposite sides of the transversal.</span>

Hence, Allison's correct claims are:
1 = 2 because they are vertically opposite angles.
BAQ = QRP because they are alternate interior angles.

Therefore Allison, in order to prove her claim, can use the AA similarity theorem: if two angles of a triangle are congruent to two angles of the other triangle, then the two triangles are similar.

3 0

3 years ago

What is 12 3/20 in simplest form?

Levart [38]

Answer:

12 3/20 is the simplest form

4 0

3 years ago

For the graph x = 12 find the slope of a line that is perpendicular to it and the slope of a line parallel to it. Explain your a

blagie [28]

Check the picture below.

that's the line of x = 12, just a straight vertical line, notice the green line, that's parallel to it, and the red line, that's perpendicular to it.

let's pick two points for each to get their slopes, hmm say for the green one (5,2) and (5,4)

$\bf (\stackrel{x_1}{5}~,~\stackrel{y_1}{2})\qquad 
(\stackrel{x_2}{5}~,~\stackrel{y_2}{4})
\\\\\\
% slope = m
slope = m\implies 
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{4-2}{5-5}\implies \stackrel{und efined}{\cfrac{2}{0}}$

and for the red one hmmm (3,2) and (7,2)

$\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{2})\qquad 
(\stackrel{x_2}{7}~,~\stackrel{y_2}{2})
\\\\\\
% slope = m
slope = m\implies 
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{2-2}{7-3}\implies \cfrac{0}{4}\implies 0$