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Komok [63]
3 years ago
5

Triangle PQR is transformed to triangle P′Q′R′. Triangle PQR has vertices P(8, 0), Q(6, 2), and R(−2, −4). Triangle P′Q′R′ has v

ertices P′(4, 0), Q′(3, 1), and R′(−1, −2).
Plot triangles PQR and P′Q′R′ on your own coordinate grid.


Part A: What is the scale factor of the dilation that transforms triangle PQR to triangle P′Q′R′? Explain your answer. (4 points)


Part B: Write the coordinates of triangle P′′Q′′R′′ obtained after P′Q′R′ is reflected about the y-axis. (4 points)


Part C: Are the two triangles PQR and P′'Q′'R′' congruent? Explain your answer. (2 points)

can someone plz not be greedy and take points for free without asNSWERING DA QUESTION

THANK U
Mathematics
1 answer:
pav-90 [236]3 years ago
3 0

lo siento de antemano! Necesito los puntos para hacer una publicación sobre mi amigo desaparecido. Ella desapareció y no puedo encontrarla. Ella se ha ido desde Halloween gracias por su comprensión. 8 9 028 los números son para covienes que estoy cosiendo el quistoion. Lo siento, nuevamente.

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Read 2 more answers
There is a triangle with a perimeter of 63 cm, one side of which is 21 cm. Also, one of the medians is perpendicular to one of t
NemiM [27]

Answer:

21cm; 28cm; 14cm

Step-by-step explanation:

There is no info in the problem/s  text which one of the triangle's  side is 21 cm. That is why we have to try all possible variants.

Let the triangle is ABC . Let the AK is the angle A bisector and BM is median.

Let O is AK and BM cross point.

Have a look to triangle ABM.  AO is the bisector and AOB=AOM=90 degrees (means that AO is as bisector as altitude)

=> triangle ABM is isosceles => AB=AM  (1)

1. Let AC=21   So AM=21/2=10.5 cm

So AB=10.5 cm as well.  So BC= P-AB-AC=63-21-10.5=31.5 cm

Such triangle doesn' t exist ( is impossible), because the triangle's inequality doesn't fulfill.  AB+AC>BC ( We have 21+10.5=31.5 => AB+AC=BC)

2. Let AB=21 So AM=21 and AC=42 .So  BC= P-AB-AC=63-21-42=0 cm- such triangle doesn't exist.

3. Finally let BC=21 cm. So AB+AC= 63-21=42 cm

We know (1) that AB=AM so AC=2*AB.  So AB+AC=AB+2*AB=3*AB

=>3*AB=42=> AB=14 cm => AC=2*14=28 cm.

Let check if this triangle exists ( if the triangle's inequality fulfills).

BC+AB>AC    21+14>28 - correct=> the triangle with the sides' length 21cm,14 cm, 28cm exists.

This variant is the only possible solution of the given problem.

6 0
3 years ago
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