Question

find the corresponding sides and corresponding angles of the pre-image.
ma =
∠m =
at =
∠a =
th =
∠t =
hm =
∠h =

Explanation:

Step1: Identify Coordinates of Pre - Image Points

First, we find the coordinates of the vertices of the pre - image (the left - hand figure with vertices M, A, T, H). Let's assume the grid has each square with side length 1.
- Coordinates of M: Let's assume from the grid, M is at (-8, 8)
- Coordinates of A: A is at (-2, 8)
- Coordinates of T: T is at (-2, 2)
- Coordinates of H: H is at (-4, 2)

Step2: Calculate Length of MA

The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). For points M(-8,8) and A(-2,8), since \(y_1 = y_2 = 8\), the distance \(MA=\vert x_2 - x_1\vert=\vert-2-(-8)\vert=\vert6\vert = 6\).

Step3: Calculate Length of AT

For points A(-2,8) and T(-2,2), since \(x_1=x_2=-2\), the distance \(AT=\vert y_2 - y_1\vert=\vert2 - 8\vert=\vert-6\vert = 6\).

Step4: Calculate Length of TH

For points T(-2,2) and H(-4,2), since \(y_1 = y_2 = 2\), the distance \(TH=\vert x_2 - x_1\vert=\vert-4-(-2)\vert=\vert-2\vert = 2\).

Step5: Calculate Length of HM

For points H(-4,2) and M(-8,8), using the distance formula: \(HM=\sqrt{(-8 + 4)^2+(8 - 2)^2}=\sqrt{(-4)^2+6^2}=\sqrt{16 + 36}=\sqrt{52}=2\sqrt{13}\approx7.21\) (but we can also use the corresponding side from the image. Let's assume the image (right - hand figure) has vertices E, R, L, U. Let's find the corresponding sides by looking at the transformation (probably translation or rotation/reflection, but since we need corresponding sides, we match the sides with the same relative position.
Looking at the image (right - hand figure) with vertices E, R, L, U:
- Let's assume E is at (2,0), R is at (2,2), L is at (8,0), U is at (8,7) (approximate from the grid). Wait, maybe a better approach is to find the corresponding sides by the shape. The pre - image is a quadrilateral, and the image is also a quadrilateral. The side MA in the pre - image should correspond to a side in the image. Let's check the horizontal and vertical sides.
Alternatively, since MA is horizontal (y - coordinate same), in the image, the side EL (assuming E(2,0), L(8,0)): distance between E(2,0) and L(8,0) is \(8 - 2=6\), so MA corresponds to EL (length 6).
AT is vertical (x - coordinate same), in the image, LU (L(8,0), U(8,7)): distance is \(7 - 0 = 7\)? Wait, maybe my coordinate assumption is wrong. Let's re - examine the grid.
Looking at the pre - image:
- MA: from x=-8 to x=-2 (y = 8), so length 6 (since -2 - (-8)=6)
- AT: from y = 8 to y = 2 (x=-2), length 6 (8 - 2 = 6)
- TH: from x=-2 to x=-4 (y = 2), length 2 (-2-(-4)=2? Wait, no, -4-(-2)=-2, absolute value 2)
- HM: from x=-4 to x=-8 (x - change -4) and y from 2 to 8 (y - change 6), so length \(\sqrt{(-4)^2+6^2}=\sqrt{16 + 36}=\sqrt{52}\)

For the angles:
- \(\angle M\): The angle at M. The vectors \(\overrightarrow{MM_A}\) (from M to A) is (6,0) and \(\overrightarrow{MM_H}\) (from M to H) is (4, - 6) (wait, M(-8,8), H(-4,2): \(\overrightarrow{MH}=(-4 + 8,2 - 8)=(4,-6)\), \(\overrightarrow{MA}=(-2 + 8,8 - 8)=(6,0)\). The angle between (6,0) and (4,-6) can be found using the dot product formula \(\cos\theta=\frac{\vec{a}\cdot\vec{b}}{\vert\vec{a}\vert\vert\vec{b}\vert}\). \(\vec{a}=(6,0)\), \(\vec{b}=(4,-6)\), \(\vec{a}\cdot\vec{b}=6\times4+0\times(-6)=24\), \(\vert\vec{a}\vert = 6\), \(\vert\vec{b}\vert=\sqrt{4^2+(-6)^2}=\sqrt{16 + 36}=\sqrt{52}\). \(\cos\theta=\frac{24}{6\sqrt{52}}=\frac{4}{\sqrt{52}}\approx\frac{4}{7.21}\approx0.555\), so \(\theta\approx56.3^{\circ}\). But maybe the figure is a quadrilateral where \(\angle M\) corresponds to an angle in the image. Alternatively, since MA and AT are both length 6, and TH is length 2, maybe the figure is a quadrilateral with MA || EL (in image), AT || LU, TH || ER, HM || RU.

Assuming the image is a translation or similar figure, the corresponding sides:
- MA corresponds to EL (length 6)
- AT corresponds to LU (length 6)
- TH corresponds to ER (length 2)
- HM corresponds to RU (length \(\sqrt{52}\))

For the angles, corresponding angles are equal. So \(\angle M\) corresponds to \(\angle U\) (if the correspondence is M - U, A - L, T - E, H - R), \(\angle A\) corresponds to \(\angle L\), \(\angle T\) corresponds to \(\angle E\), \(\angle H\) corresponds to \(\angle R\).

But let's re - do the side lengths with correct coordinates:

Pre - image vertices:
- M: (-8,8)
- A: (-2,8)
- T: (-2,2)
- H: (-4,2)

Image vertices (right - hand figure):
- E: (2,0)
- R: (2,2)
- L: (8,0)
- U: (8,7) Wait, no, maybe E(2,1), R(2,3), L(8,1), U(8,7). Let's check the vertical and horizontal distances.

MA: from x=-8 to x=-2 (y = 8), length 6.

In the image, the side from E to L: E(2,1), L(8,1), length 6. So MA corresponds to EL (length 6).

AT: from y = 8 to y = 2 (x=-2), length 6. In the image, from L(8,1) to U(8,7), length 6. So AT corresponds to LU (length 6).

TH: from x=-2 to x=-4 (y = 2), length 2. In the image, from E(2,1) to R(2,3), length 2. So TH corresponds to ER (length 2).

HM: from H(-4,2) to M(-8,8). The vector is (-4,6). In the image, from R(2,3) to U(8,7), vector is (6,4). Wait, no, maybe the transformation is a rotation. Let's calculate the length of HM: \(\sqrt{(-8 + 4)^2+(8 - 2)^2}=\sqrt{(-4)^2+6^2}=\sqrt{16 + 36}=\sqrt{52}\approx7.21\). The length of RU (from R(2,3) to U(8,7)): \(\sqrt{(8 - 2)^2+(7 - 3)^2}=\sqrt{36 + 16}=\sqrt{52}\approx7.21\). So HM corresponds to RU (length \(\sqrt{52}\)).

For the angles:
- \(\angle M\): At point M, the angle between MA (vector (6,0)) and HM (vector (4,-6)). The dot product formula: \(\cos\theta=\frac{(6,0)\cdot(4,-6)}{\vert(6,0)\vert\vert(4,-6)\vert}=\frac{24}{6\times\sqrt{52}}=\frac{4}{\sqrt{52}}\approx0.555\), so \(\theta\approx56.3^{\circ}\). The angle at U (corresponding to \(\angle M\)): vector \(\overrightarrow{UR}\) (from U to R) is (-6,-4) and \(\overrightarrow{UL}\) (from U to L) is (0,-6). Wait, maybe it's easier to say that corresponding angles are equal, so \(\angle M=\angle U\), \(\angle A=\angle L\), \(\angle T=\angle E\), \(\angle H=\angle R\).

But let's focus on the side lengths first:

MA: length 6 (since it's a horizontal line from x=-8 to x=-2 at y = 8, so the length is \(\vert-2-(-8)\vert = 6\))

AT: length 6 (vertical line from y = 8 to y = 2 at x=-2, length is \(\vert2 - 8\vert = 6\))

TH: length 2 (horizontal line from x=-2 to x=-4 at y = 2, length is \(\vert-4-(-2)\vert = 2\))

HM: length \(\sqrt{(-8+4)^2+(8 - 2)^2}=\sqrt{16 + 36}=\sqrt{52}=2\sqrt{13}\approx7.21\)

For the angles, since the figures are (probably) congruent (same shape and size), corresponding angles are equal. So:

\(\angle M\) corresponds to the angle at U, \(\angle A\) corresponds to the angle at L, \(\angle T\) corresponds to the angle at E, \(\angle H\) corresponds to the angle at R.

But let's re - check the grid. Maybe the pre - image and image are congruent, so:

MA = EL (length 6)

AT = LU (length 6)

TH = ER (length 2)

HM = RU (length \(\sqrt{52}\))

And \(\angle M=\angle U\), \(\angle A=\angle L\), \(\angle T=\angle E\), \(\angle H=\angle R\)

Answer:

MA = 6, \(\angle M=\angle U\) (or the corresponding angle in the image)

AT = 6, \(\angle A=\angle L\)

TH = 2, \(\angle T=\angle E\)

HM = \(2\sqrt{13}\) (or the length of the corresponding side in the image), \(\angle H=\angle R\)

(If we assume the image has sides EL = 6, LU = 6, ER = 2, RU=\(2\sqrt{13}\) and angles \(\angle U=\angle M\), \(\angle L=\angle A\), \(\angle E=\angle T\), \(\angle R=\angle H\))