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Marianna [84]
2 years ago
13

The figure shows two similar triangles on a coordinate grid: A coordinate grid is shown from positive 6 to negative 6 on the x-a

xis and from positive 6 to negative 6 on the y-axis. A triangle ABC is shown with vertex A on ordered pair negative 3, 3, vertex B on ordered pair 0, 0 and vertex C on ordered pair negative 6, negative 3. A triangle A prime B prime C prime is also shown with vertex A prime on ordered pair 1, 1, vertex B prime on ordered pair 0, 0, and vertex C prime on ordered pair 2, negative 1. Which set of transformations have been performed on triangle ABC to form triangle A'B'C'? Dilation by a scale factor of 1 over 3 followed by reflection about the x-axis Dilation by a scale factor of 3 followed by reflection about the x-axis Dilation by a scale factor of 1 over 3 followed by reflection about the y-axis Dilation by a scale factor of 3 followed by reflection about the y-axis

Mathematics
1 answer:
SIZIF [17.4K]2 years ago
5 0

From what I know, figure ABC has gotten smaller so you wouldn't use either options that suggests a dilation of 3, because that would mean it got bigger. (But please correct me if I'm wrong.) Also, if you look at the little unit square measurements on the graph, looking at figure ABC, C is 6 units from B, while in figure A'B'C', C' is 2 units from B', which 6÷2=3 meaning that A'B'C' would most likely be a third of its size.

Next the figure is being mirrored over the y-axis which is the vertical line going up and down.

So the answer would be the option that says

"Dilation by a scale factor of 1 over 3 followed by a reflection about the y-axis"

I hope that it's right, because I haven't practiced this in a while, and I hope this made some sense.

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Given: KLMN is a trapezoid m∠N = m∠KML ME ⊥ KN , ME = 3√5 , KE = 8, LM:KN = 3:5 Find: KM, LM, KN, Area of KLMN
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As ME is perpendicular to KN, ∠KEM is a right angle
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for this we can use Pythogoras' theorem where the square of hypotenuse is equal to the sum of the squares of the adjacent sides.
KM² = KE² + ME²
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Q2)Find LM
It is said that ratio of LM:KN is 3:5
Therefore if we take the length of one unit as x
length of LM is 3x
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KN is greater than LM by 2 units 
If we take the figure ∠K and ∠N are equal. 
Since the angles on opposite sides of the bases are equal then this is called an isosceles trapezoid. So if we draw a line from L that cuts KN perpendicularly at D, ΔKEM and ΔLDN are congruent therefore KE = DN
since KN is greater than LM due to KE and DN , the extra 2 units of KN correspond to 16 units 
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x = 8
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Q3)Find KN 
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same height ME = LD perpendicular distance between the 2 parallel sides 
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when 2 angles and one side of one triangle is equal to two angles and one side on another triangle then the 2 triangles are congruent according to AAS theorem (AAS). Therefore KE = DN 
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Q4)Find area KLMN
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substituting values into the general equation
Area = 1/2 * ME * (KN+ LM) 
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