homework
5.01: swbat identify and apply the properties of parallelograms and rectangles to find missing angles and sides
show all work and annotations for full credit. use your quadrilateral property reference sheet!
**#1.) in parallelogram abcd, ( mangle b = (9x + 3)^circ ), and ( mangle c = (3x + 33)^circ ). what is the ( mangle d )? explain how you know.
image of parallelogram abcd with vertices labeled a, b, c, d
#2.) in rectangle quad, ( mangle adu = (x + 42)^circ ) and ( mangle qdu = (2x - 12)^circ ). fill in all angles in the rectangle.
image of rectangle quad with vertices labeled d, q, u, a and a diagonal from d to u
*#3.) (4.03 review) graph the segment with endpoints c (1, -5) and d (4, -2). what are the coordinates of c and d under the transformation ( r_{-90^circ} )?
image of a coordinate grid table with columns: pre - image coordinates, pre - image quadrant, image quadrant, image coordinates

Question

Accepted Answer

### Problem #1 # Explanation: ## Step1: Recall parallelogram property In a parallelogram, consecutive angles are supplementary (sum to $180^\circ$). So, $m\angle B + m\angle C = 180^\circ$. Substitute the given expressions: $(9x + 3) + (3x + 33) = 180$. ## Step2: Solve for $x$ Combine like terms: $12x + 36 = 180$. Subtract 36 from both sides: $12x = 144$. Divide by 12: $x = 12$. ## Step3: Find $m\angle B$ Substitute $x = 12$ into $m\angle B = (9x + 3)^\circ$: $9(12) + 3 = 108 + 3 = 111^\circ$. ## Step4: Find $m\angle D$ In a parallelogram, opposite angles are equal. $\angle B$ and $\angle D$ are opposite? Wait, no—$\angle B$ and $\angle D$? Wait, no, consecutive angles: $\angle B$ and $\angle C$ are supplementary, $\angle C$ and $\angle D$ are supplementary? Wait, no, opposite angles: $\angle B = \angle D$? Wait, no, in parallelogram $ABCD$, $AB \parallel CD$ and $AD \parallel BC$. So $\angle B$ and $\angle D$ are opposite? Wait, no, vertices are $A, B, C, D$ in order. So $\angle A$ and $\angle C$ are opposite, $\angle B$ and $\angle D$ are opposite? Wait, no, consecutive angles: $\angle B$ and $\angle C$ are consecutive (adjacent), so they are supplementary. Then $\angle D$ is opposite to $\angle B$? Wait, no, $\angle A$ and $\angle C$ are opposite, $\angle B$ and $\angle D$ are opposite. Wait, no, let's correct: In parallelogram $ABCD$, $AB \parallel CD$, $AD \parallel BC$. So $\angle B$ and $\angle C$ are same - side interior angles (consecutive), so supplementary. $\angle D$ and $\angle C$ are also consecutive? Wait, no, $\angle D$ is adjacent to $\angle C$ and $\angle A$. Wait, maybe I mixed up. Wait, in parallelogram, opposite angles are equal, consecutive angles are supplementary. So $\angle B$ and $\angle D$ are opposite? Wait, no, $\angle A = \angle C$, $\angle B = \angle D$. Wait, no, let's label the parallelogram: $A - B - C - D - A$. So $AB \parallel CD$, $AD \parallel BC$. So $\angle A$ and $\angle B$ are consecutive (supplementary), $\angle B$ and $\angle C$ are consecutive (supplementary), $\angle C$ and $\angle D$ are consecutive (supplementary), $\angle D$ and $\angle A$ are consecutive (supplementary). And opposite angles: $\angle A = \angle C$, $\angle B = \angle D$. Wait, so if $\angle B = 111^\circ$, then $\angle D = \angle B = 111^\circ$? Wait, no, wait, we found $m\angle B = 111^\circ$, then $\angle D$ is opposite to $\angle B$? Wait, no, $\angle A$ and $\angle C$ are opposite, $\angle B$ and $\angle D$ are opposite. So yes, $\angle B = \angle D$. Wait, but let's check with $\angle C$. $m\angle C = 3x + 33 = 3(12)+33 = 36 + 33 = 69^\circ$. Then $\angle B + \angle C = 111 + 69 = 180$, which is correct (consecutive angles supplementary). Then $\angle D$ is opposite to $\angle B$? Wait, no, $\angle D$ is adjacent to $\angle C$, so $\angle D + \angle C = 180$? Wait, no, $\angle D$ and $\angle C$ are consecutive, so they should be supplementary. Wait, $\angle D + \angle C = 180$, so $\angle D = 180 - 69 = 111^\circ$, which matches $\angle B$. So yes, $\angle D = 111^\circ$. # Answer: $m\angle D = 111^\circ$ ### Problem #2 # Explanation: ## Step1: Recall rectangle property In a rectangle, all angles are $90^\circ$, and the diagonal bisects the angles? Wait, no, in rectangle $QUAD$, $\angle ADU + \angle QDU = \angle ADQ = 90^\circ$ (since $\angle ADQ$ is a right angle in the rectangle). So $(x + 42) + (2x - 12) = 90$. ## Step2: Solve for $x$ Combine like terms: $3x + 30 = 90$. Subtract 30: $3x = 60$. Divide by 3: $x = 20$. ## Step3: Find $m\angle ADU$ Substitute $x = 20$ into $m\angle ADU = (x + 42)^\circ$: $20 + 42 = 62^\circ$. ## Step4: Find $m\angle QDU$ Substitute $x = 20$ into $m\angle QDU = (2x - 12)^\circ$: $2(20)-12 = 40 - 12 = 28^\circ$. ## Step5: Angles in rectangle All four angles of a rectangle are $90^\circ$ ( $\angle A$, $\angle D$, $\angle U$, $\angle Q$ are $90^\circ$ each). The angles formed by the diagonal: $\angle ADU = 62^\circ$, $\angle QDU = 28^\circ$, and their counterparts (since the diagonal is common, $\angle DAU$ and $\angle DQU$ will have angles equal to these due to triangle congruence, but the rectangle's internal angles are all $90^\circ$). # Answer: $\angle A = \angle D = \angle U = \angle Q = 90^\circ$, $m\angle ADU = 62^\circ$, $m\angle QDU = 28^\circ$, $m\angle DAU = 28^\circ$ (since $\triangle ADU$ is right - angled, $90 - 62 = 28$), $m\angle DQU = 62^\circ$ (since $\triangle QDU$ is right - angled, $90 - 28 = 62$). ### Problem #3 # Explanation: ## Step1: Recall rotation rule The rule for a $90^\circ$ clockwise (or $- 90^\circ$ counter - clockwise? Wait, $R_{-90^\circ}$ is a $90^\circ$ clockwise rotation? Wait, the rotation rule: For a point $(x,y)$, a $90^\circ$ clockwise (or $-90^\circ$ counter - clockwise) rotation about the origin transforms it to $(y, - x)$. Wait, no: - $90^\circ$ counter - clockwise ($R_{90^\circ}$): $(x,y)\to(-y,x)$ - $90^\circ$ clockwise ($R_{-90^\circ}$): $(x,y)\to(y, - x)$ ## Step2: Rotate point $C(1,-5)$ Using $R_{-90^\circ}$ rule $(x,y)\to(y, - x)$: $x = 1$, $y=-5$, so $C' = (-5, - 1)$? Wait, no, wait: Wait, maybe I mixed up. Let's confirm: For $90^\circ$ clockwise (rotation matrix $\begin{pmatrix}0&1\-1&0\end{pmatrix}$): $\begin{pmatrix}0&1\-1&0\end{pmatrix}\begin{pmatrix}x\y\end{pmatrix}=\begin{pmatrix}y\-x\end{pmatrix}$ For $90^\circ$ counter - clockwise (rotation matrix $\begin{pmatrix}0&-1\1&0\end{pmatrix}$): $\begin{pmatrix}0&-1\1&0\end{pmatrix}\begin{pmatrix}x\y\end{pmatrix}=\begin{pmatrix}-y\x\end{pmatrix}$ Wait, the problem says $R_{-90^\circ}$, which is equivalent to $90^\circ$ clockwise. So for point $C(1,-5)$: $x = 1$, $y=-5$, so $C'=(y, - x)=(-5, - 1)$? Wait, no, $y=-5$, $-x=-1$, so $(-5, - 1)$? Wait, no, wait: $(x,y)=(1, - 5)$, applying $R_{-90^\circ}$ (clockwise $90^\circ$): new $x = y=-5$, new $y=-x=-1$, so $C'=(-5, - 1)$? Wait, no, that can't be. Wait, let's take a simple point, say $(1,0)$. Rotating $90^\circ$ clockwise, it should go to $(0, - 1)$. Using the rule $(x,y)\to(y, - x)$: $(1,0)\to(0, - 1)$, which is correct. Another point: $(0,1)$ rotated $90^\circ$ clockwise goes to $(1,0)$, using the rule: $(0,1)\to(1,0)$, correct. So for $C(1,-5)$: $x = 1$, $y = - 5$, so $C'=(y, - x)=(-5, - 1)$ ## Step3: Rotate point $D(4,-2)$ Using the same rule $(x,y)\to(y, - x)$: $x = 4$, $y=-2$, so $D'=(-2, - 4)$ ## Step4: Determine quadrants - Pre - image $C(1,-5)$: $x = 1>0$, $y=-5<0$, so Quadrant IV. - Pre - image $D(4,-2)$: $x = 4>0$, $y=-2<0$, so Quadrant IV. - Image $C'(-5, - 1)$: $x=-5<0$, $y=-1<0$, so Quadrant III. - Image $D'(-2, - 4)$: $x=-2<0$, $y=-4<0$, so Quadrant III. # Answer: | Pre - Image Coordinates | Pre - Image Quadrant | Image Quadrant | Image Coordinates | |-------------------------|----------------------|----------------|-------------------| | $C(1,-5)$ | IV | III | $C'(-5, - 1)$ | | $D(4,-2)$ | IV | III | $D'(-2, - 4)$ |

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QUESTION IMAGE

Question

Explanation:

Problem #1

Step1: Recall parallelogram property

Step2: Solve for \(x\)

Step3: Find \(m\angle B\)

Step4: Find \(m\angle D\)

Step1: Recall rectangle property

Step2: Solve for \(x\)

Step3: Find \(m\angle ADU\)

Step4: Find \(m\angle QDU\)

Step5: Angles in rectangle

Step1: Recall rotation rule

Step2: Rotate point \(C(1,-5)\)

Step3: Rotate point \(D(4,-2)\)

Step4: Determine quadrants

Answer:

Problem #2