Given the preimage and image, find the dilation scale factor

Mathematics

1 answer:

77julia77 [94]3 years ago

8 0

Given:

The preimage and image of a triangle in the given figure.

To find:

The dilation scale factor.

Solution:

From the given figure it is clear that the vertices of the triangle ABC are A(-2,-2), B(-1,2) and C(2,1).

The vertices of the triangle A'B'C' are A'(-4,-4), B'(-2,4) and C'(4,2).

If a figure is dilated by factor K with (0,0) as the center of dilation, then

$(x,y)\to (kx,ky)$

Let the scale factor be K, then the image of point A is:

$A(-2,-2)\to A'(k(-2),k(-2))$

$A(-2,-2)\to A'(-2k,-2k)$

From the given figure it is clear that the image of point A is A'(-4,-4).

$A'(-2k,-2k)=A'(-4,-4)$

On comparing both sides, we get

$-2k=-4$

$k=\dfrac{-4}{-2}$

$k=2$

Therefore, the dilation scale factor is 2.

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Really need help with this pls help!!

worty [1.4K]

Let's carry this math sentence over to its natural, "shapey" element. We're going to look at each term not as an ordinary number, but as the area of some shape.

x² (read as "x squared") can be seen as the area of a square with side lengths of x. 2x can similarly be seen as the area of a rectangle with a length of x and a width of 2. (Picture 1)

What's our question actually asking, though? Something about perfect squares. More specifically, we're looking for something to add on that'll make this thing a perfect square. We're trying to find a missing piece we can slot in to make a square, in other words. Problem is, our shapes don't look much like a square if we put them together right now. We need to do a little cutting and gluing first.

First, we're gonna cut the 2x rectangle lengthwise, getting two rectangles with an area of x, a length of 1, and a width of x. Next, we're going to attach them to the x² square, creating this shape that looks, strangely, like a square with a little bit missing from it (picture 2). What we're trying to do is complete this square, to find the area of that little missing chunk.

As it turns out, we have all the information we need for this. Notice that, using the lengths of the x rectangles, we can find that the square's dimensions are 1 x 1, which means that its area is 1 x 1 = 1.

If we tack this new area on to our original expression, we've "completed the square!" We now have a perfect square with side lengths of (x + 1) and an area of (x + 1)² (picture 3).

So, our final expression is x² + 2x + 1, and the missing constant - the area of the "missing square" we had to find to complete our larger one - is 1.