The claim made by Kyle is not entirely true and unjustifiable because for the figures or solids to be similar, not only the corresponding sides should have the finite ratio but also the corresponding angles should be equal. Although, the claim depends on the type of quadrilateral.
Answer:

Step-by-step explanation:
Given
--- proper representation of the question
Required
Evaluate
Convert numbers to improper fractions

Take LCM


Convert to improper fraction

Answer:
(9, 27 )
Step-by-step explanation:
Since the dilatation is centred at the origin, multiply each of the coordinates of the original point by the scale factor of 3
(3, 9 ) → (3(3), 3(9)) → (9, 27 )
Answer:
x = -3
y = 0
Step-by-step explanation:
<u>Given</u><u> </u><u>equations</u><u> </u><u>:</u><u>-</u><u> </u>
<u>-x</u><u> </u><u>+</u><u> </u><u>2</u><u>y</u><u> </u><u>=</u><u> </u><u>3</u><u> </u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u> </u><u>(</u><u> </u><u>i</u><u> </u><u>)</u>
<u>2</u><u>x</u><u> </u><u>-</u><u> </u><u>3</u><u>y</u><u> </u><u>=</u><u> </u><u>-</u><u>6</u><u> </u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u> </u><u>(</u><u> </u><u>ii</u><u> </u><u>)</u>
<u>From</u><u> </u><u>(</u><u> </u><u>i</u><u> </u><u>)</u><u> </u><u> </u>
<u>-x</u><u> </u><u>+</u><u> </u><u>2</u><u>y</u><u> </u><u>=</u><u> </u><u>3</u><u> </u>
<u>-x</u><u> </u><u>=</u><u> </u><u>3</u><u> </u><u>-</u><u> </u><u>2</u><u>y</u><u> </u>
<u>x</u><u> </u><u>=</u><u> </u><u>2</u><u>y</u><u> </u><u>-</u><u> </u><u>3</u><u> </u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u> </u><u>(</u><u> </u><u>iii</u><u> </u><u>)</u>
<u>From</u><u> </u><u>(</u><u> </u><u>ii</u><u> </u><u>)</u><u> </u>
<u>2</u><u>x</u><u> </u><u>-</u><u> </u><u>3</u><u>y</u><u> </u><u>=</u><u> </u><u>-</u><u>6</u><u> </u>
<u>2</u><u>x</u><u> </u><u>=</u><u> </u><u>-</u><u>6</u><u> </u><u>+</u><u> </u><u>3</u><u>y</u><u> </u>
<u>
</u>
<u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u> </u><u>(</u><u> </u><u>iv</u><u> </u><u>)</u>
<u>Equating</u><u> </u><u>(</u><u> </u><u>iii</u><u> </u><u>)</u><u> </u><u>and</u><u> </u><u>(</u><u> </u><u>iv</u><u> </u><u>)</u>
<u>x</u><u> </u><u>=</u><u> </u><u>x</u><u> </u>
<u>
</u>
4y - 6 = -6 + 3y
4y - 3y = -6 + 6
y = 0
Putting value of y in ( iii )
x = 2y - 3
x = 2 ( 0 ) - 3
x = -3
No because a square pyramid must hav a square for its base and a square is not a triangle
answer is not possible