Is there a method that could prove that these two triangles are similar?

Mathematics

1 answer:

dalvyx [7]3 years ago

4 0

Answer:

A Yes, a sequence of dilations and rigid motions carries one triangle onto the other

Step-by-step explanation:

Since two angles are marked with the same measures in each triangle, the triangles are similar by the AA postulate with no further action.

By some accounts it seems to be necessary to lay one figure on top of the other to demonstrate similarity. In order to do that, the smaller triangle would need to be dilated and translated (a rigid motion) to make it occupy the same space as the larger triangle.

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Order the following integers from least to greatest 6, 11, 16, -8, -5

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What are the coordinates of the intersection of the diagonals of parallelogram LMNO, with vertices L(0,-3), M(-2,1), N(1,5), O(3

ArbitrLikvidat [17]

ANSWER

$\boxed {( \frac{1}{2} , 1)}$

EXPLANATION

The given parallelogram has vertices,

L(0,-3), M(-2,1), N(1,5), O(3,1).

The diagonals of the parallelogram bisect each other.

From the diagram, we can see that, the diagonals have coordinates L(0,-3),N(1,5)

and

M(-2,1),O(3,1).

The midpoint of any of the diagonals will give us the coordinates of intersection of the diagonals.

Recall the midpoint formula,

$(\frac{x_1+x_2}{2}, \frac{y_2+y_1}{2})$

Using L(0,-3),N(1,5) gives,

$(\frac{0+1}{2}, \frac{ - 3+5}{2})$

$(\frac{1}{2}, \frac{ 2}{2})$

$(\frac{1}{2}, 1)$

Or we could have also used,M(-2,1),O(3,1) to get,

$(\frac{-2+3}{2}, \frac{ 1+1}{2})$

$(\frac{ 1}{2}, \frac{ 2}{2})$

$(\frac{ 1}{2}, 1)$

The correct answer is C