Find the area of the octagon formed from this 6 * 6 square. The vertices of the hexagon are on points dividing the side in thirds.
[emphasis mine]
I emphasized two words above to highlight an error in this question -- first Pappas asks for the area of an octagon, but then she describes the vertices of a hexagon. Well, the square has four sides and it takes two points to trisect each side, and so there are eight vertices. Thus the intended polygon is clearly an octagon. And I can see the diagram -- and yes, it's definitely an octagon. Let's forgive Pappas for the typo and fix it ourselves:
Find the area of the octagon formed from this 6 * 6 square. The vertices of the octagon are on points dividing the side in thirds.
Well, the area of the square is 36, and one-third of the side length is two. Thus we have a composite figure formed by cutting off an isosceles right triangle of leg 2 from each vertex of the square. The area of each of the four cut-off triangles must be half of two times two, which is 2. So we have:
36 - 4(2) = 28
Therefore the desired area of the octagon is 28 square units -- and of course, today's date is the 28th.
As it turns out, there is a Geometry problem on the Pappas calendar everyday this week. So expect me to describe the Pappas problem in all of this week's blog entries.
Meanwhile, this is our final week of SBAC Practice Test questions. Surely, by now all high schools in both of my districts have completed the SBAC. My intention is for all SBAC review questions after the state test should become finals review questions instead.
Indeed, these review questions plus this week's Pappas problems together make up a nice review week for the final exam. Today's Pappas problem, for example, reviews Chapter 8 on area, especially Lessons 8-3 and 8-5.
This is what I wrote last year about today's lesson:
Question 27 of the SBAC Practice Exam is on quadrilaterals:
In the given figure, quadrilateral ABCD is a rectangle, and quadrilateral ACED is a parallelogram.
Ted claims that the two shaded triangles [ABC and DCE] are congruent. Is Ted's claim correct? Include all work and/or reasoning necessary to either prove the two triangles congruent or to disprove Ted's claim.
This is a Geometry question. Let's try to come up with a two-column proof of Ted's claim:
Given: Quadrilateral ABCD is a rectangle, and quadrilateral ACED is a parallelogram.
Prove: Triangle ABC = DCE.
Proof:
Statements Reasons
1. bla, bla, bla 1. Given
2. ABCD is a pgram. 2. A rectangle is a parallelogram.
3. AB = DC, AC = DE 3. Opposite sides of a pgram are congruent.
4.AB | | DC, AC | | DE 4. Opposite sides of a pgram are parallel.
5. Angle BAC = ACD, 5. Alternate Interior Angles (parallel line consequences)
Angle ACD = CDE
6. Angle BAC = CDE 6. Transitive Property of Congruence
7. Triangle ABC = DCE 7. SAS Congruence Theorem [steps 3,6,3]
This proof is sound, but it's tricky to enter a two-column proof into the SBAC. Most likely, the SBAC expects students to enter a paragraph proof into the box. We might try the following:
Paragraph Proof:
Since ABCD is a rectangle, it's also a parallelogram. Since opposite sides of a parallelogram are congruent, we have AB = DC and AC = DE. Since opposite sides of a parallelogram are also parallel by definition,AB | | DC and AC | | DE. This implies that alternate interior angles BAC and ACD are congruent, as are ACD and CDE. By the Transitive Property of Congruence, BAC = CDE. Therefore,
Triangles ABC and DCE are congruent by SAS. QED
It's interesting to think of this problem in terms of transformations. Notice that if a diagonal divides a parallelogram into two triangles, then a 180-degree rotation about the midpoint of the diagonal maps one triangle to the other. Thus a 180-degree rotation about the midpoint ofAC maps Triangle ABC to CDA, and likewise a half-turn about the midpoint of CD maps CDA to DCE. So the composite of two isometries (itself an isometry) maps ABC to DCE, therefore these triangles are congruent. (The composite of two half-turns is always a translation.) Even though the Common Core is all about isometries, using this to answer the SBAC question is definitely not recommended.
Unfortunately, neither the girl nor the guy from the Pre-Calc class supplies the correct proof. The girl marks the congruence relations AD = BC and AB = DC on her paper, then proclaims "Yes correct." The guy doesn't even attempt this problem.
Both of these are Pre-Calc juniors. Neither has sat in a Geometry classroom in two years, since the end of their freshman year. So it's likely that both have forgotten how to write two-column proofs. These students need a refresher on Geometry proofs in order to be successful.
The marks on the girl's paper, by the way, suggests another valid proof based on SSS. If she had drawn in the congruence relations on the sides of parallelogram ACED, she would have had enough information to prove the triangles congruent:
Proof:
Statements Reasons
1. bla, bla, bla 1. Given
2. ABCD is a pgram. 2. A rectangle is a parallelogram.
3. AB = DC, AC = DE, 3. Opposite sides of a pgram are congruent.
BC = AD, AD = CE
4. BC = CE 4. Transitive Property of Congruence
5. Triangle ABC = DCE 5. SSS Congruence Property [steps 3,3,4]
This might be a better proof to show Algebra II or Pre-Calc juniors before the SBAC, since we don't have to remind them what Alternate Interior Angles are. On the other hand, my original proof is excellent for Geometry students before the final, since we want to remind them what Alternate Interior Angles are.
Question 28 of the SBAC Practice Exam is on nonlinear functions:
Choose the domain for which each function is defined.
[The four functions are f (x) = (x + 4)/x, f (x) = x/(x + 4), f (x) = x(x + 4), 4/(x^2 + 8x + 16). The possible domains are all real numbers, x != 0, x != 4, x != -4. Here the "!=" means "does not equal" in ASCII and several computer languages.]
These functions are not linear. It will be hard-pressed to consider this an Algebra I question -- even though many texts (such as the Glencoe Algebra I text) contains a lesson on rational functions (Lesson 11-2), many teachers (including those in our district) skip Chapter 11 altogether. Thus even though students don't have to graph anything, this is mostly likely considered an Algebra II problem.
The first function has x in the denominator, so its domain is x != 0.
The second function has x + 4 in the denominator, so its domain is x != -4.
The third function has 1 in the denominator, so its domain is all real numbers.
The fourth function has (x + 4)^2 in the denominator, so its domain is x != -4.
Today is an activity day. The activity for Question 27 comes from Lesson 7-6 of the U of Chicago Geometry text, since this is the lesson on properties of a parallelogram and congruent triangles. The Exploration Question here is on congruent triangles in a regular pentagon. The activity for Question 28 comes from Lesson 13-4 of the U of Chicago Algebra I text, "Function Language," since this lesson discusses the domain and range of a function -- including rational functions like x/(x + 4). The Exploration Question here is about the range of the absolute value function. It's a great question for Common Core, since it involves translating the function up and down.
The guy from the Pre-Calc correctly answers all parts of this question. But the girl marks two boxes for the first part -- x != 0 and x != -4 -- and none for the second part. It could be that she just marks the answer for the second function in the first line instead. I'm not quite sure what would happen on the SBAC interface if a student tries to mark two answers in the same row -- it could be that each row is a radio button where only one box can be marked at a time. This would protect the girl from making this sort of mistake on the computer.
It's also interesting to see the two students' thought process for the fourth function. The guy factors the denominator as (x + 4)^2, then starts to mark x != 4 before crossing it out. On the other hand, the girl doesn't factor at all and just plugs x = -4 into the denominator instead. The guy's method is considered more sound, since the girl is relying on the multiple choice format. She would have had trouble with this question if she'd been asked to state the domain without multiple choice.
Question 27 of the SBAC Practice Exam is on quadrilaterals:
In the given figure, quadrilateral ABCD is a rectangle, and quadrilateral ACED is a parallelogram.
Ted claims that the two shaded triangles [ABC and DCE] are congruent. Is Ted's claim correct? Include all work and/or reasoning necessary to either prove the two triangles congruent or to disprove Ted's claim.
This is a Geometry question. Let's try to come up with a two-column proof of Ted's claim:
Given: Quadrilateral ABCD is a rectangle, and quadrilateral ACED is a parallelogram.
Prove: Triangle ABC = DCE.
Proof:
Statements Reasons
1. bla, bla, bla 1. Given
2. ABCD is a pgram. 2. A rectangle is a parallelogram.
3. AB = DC, AC = DE 3. Opposite sides of a pgram are congruent.
4.
5. Angle BAC = ACD, 5. Alternate Interior Angles (parallel line consequences)
Angle ACD = CDE
6. Angle BAC = CDE 6. Transitive Property of Congruence
7. Triangle ABC = DCE 7. SAS Congruence Theorem [steps 3,6,3]
This proof is sound, but it's tricky to enter a two-column proof into the SBAC. Most likely, the SBAC expects students to enter a paragraph proof into the box. We might try the following:
Paragraph Proof:
Since ABCD is a rectangle, it's also a parallelogram. Since opposite sides of a parallelogram are congruent, we have AB = DC and AC = DE. Since opposite sides of a parallelogram are also parallel by definition,
Triangles ABC and DCE are congruent by SAS. QED
It's interesting to think of this problem in terms of transformations. Notice that if a diagonal divides a parallelogram into two triangles, then a 180-degree rotation about the midpoint of the diagonal maps one triangle to the other. Thus a 180-degree rotation about the midpoint of
Unfortunately, neither the girl nor the guy from the Pre-Calc class supplies the correct proof. The girl marks the congruence relations AD = BC and AB = DC on her paper, then proclaims "Yes correct." The guy doesn't even attempt this problem.
Both of these are Pre-Calc juniors. Neither has sat in a Geometry classroom in two years, since the end of their freshman year. So it's likely that both have forgotten how to write two-column proofs. These students need a refresher on Geometry proofs in order to be successful.
The marks on the girl's paper, by the way, suggests another valid proof based on SSS. If she had drawn in the congruence relations on the sides of parallelogram ACED, she would have had enough information to prove the triangles congruent:
Proof:
Statements Reasons
1. bla, bla, bla 1. Given
2. ABCD is a pgram. 2. A rectangle is a parallelogram.
3. AB = DC, AC = DE, 3. Opposite sides of a pgram are congruent.
BC = AD, AD = CE
4. BC = CE 4. Transitive Property of Congruence
5. Triangle ABC = DCE 5. SSS Congruence Property [steps 3,3,4]
This might be a better proof to show Algebra II or Pre-Calc juniors before the SBAC, since we don't have to remind them what Alternate Interior Angles are. On the other hand, my original proof is excellent for Geometry students before the final, since we want to remind them what Alternate Interior Angles are.
Question 28 of the SBAC Practice Exam is on nonlinear functions:
Choose the domain for which each function is defined.
[The four functions are f (x) = (x + 4)/x, f (x) = x/(x + 4), f (x) = x(x + 4), 4/(x^2 + 8x + 16). The possible domains are all real numbers, x != 0, x != 4, x != -4. Here the "!=" means "does not equal" in ASCII and several computer languages.]
These functions are not linear. It will be hard-pressed to consider this an Algebra I question -- even though many texts (such as the Glencoe Algebra I text) contains a lesson on rational functions (Lesson 11-2), many teachers (including those in our district) skip Chapter 11 altogether. Thus even though students don't have to graph anything, this is mostly likely considered an Algebra II problem.
The first function has x in the denominator, so its domain is x != 0.
The second function has x + 4 in the denominator, so its domain is x != -4.
The third function has 1 in the denominator, so its domain is all real numbers.
The fourth function has (x + 4)^2 in the denominator, so its domain is x != -4.
Today is an activity day. The activity for Question 27 comes from Lesson 7-6 of the U of Chicago Geometry text, since this is the lesson on properties of a parallelogram and congruent triangles. The Exploration Question here is on congruent triangles in a regular pentagon. The activity for Question 28 comes from Lesson 13-4 of the U of Chicago Algebra I text, "Function Language," since this lesson discusses the domain and range of a function -- including rational functions like x/(x + 4). The Exploration Question here is about the range of the absolute value function. It's a great question for Common Core, since it involves translating the function up and down.
The guy from the Pre-Calc correctly answers all parts of this question. But the girl marks two boxes for the first part -- x != 0 and x != -4 -- and none for the second part. It could be that she just marks the answer for the second function in the first line instead. I'm not quite sure what would happen on the SBAC interface if a student tries to mark two answers in the same row -- it could be that each row is a radio button where only one box can be marked at a time. This would protect the girl from making this sort of mistake on the computer.
It's also interesting to see the two students' thought process for the fourth function. The guy factors the denominator as (x + 4)^2, then starts to mark x != 4 before crossing it out. On the other hand, the girl doesn't factor at all and just plugs x = -4 into the denominator instead. The guy's method is considered more sound, since the girl is relying on the multiple choice format. She would have had trouble with this question if she'd been asked to state the domain without multiple choice.
SBAC Practice Exam Question 27
Common Core Standard:
Prove theorems about parallelograms. Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals.
SBAC Practice Exam Question 28
Common Core Standard:
For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship. Key features include: intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; end behavior; and periodicity.
Commentary: The properties of parallelograms and congruent triangles are easily found in Chapter 7 of the U of Chicago Geometry text. Even though rational functions are usually saved for Algebra II, the domain and range of functions -- including rational functions as one of the examples -- appear in many Algebra I texts, including the U of Chicago text.
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