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♪ ♪ ♪ ♪ Hello, my name is Lindsay Howard, and I'm a math teacher at Greece.

Olympia.

My name is Erica Georgiadis, also a math teacher at Greece, Olympia.

And we're here today to help you review for a Geometry Regents exam.

Here's an overview of how the Geometry exam looks.

The Geometry Regents is broken up into four parts for a total of 35 questions and 80 total possible points Each.

Each multiple choice question is worth two points, and the open ended questions are worth 2 to 6 points, depending on which section they are in.

The state released this breakdown of topics in the exam, but it is a very broad idea of these concepts.

You can see the breakdown of topics in the middle column and how heavily each topic weighs on your score in the right column.

Every Regents exam is different and it can be challenging to anticipate exactly what you're going to see.

We've looked at all of the exams given between January 2019 and January 2024 and selected topics to review that have consistently been on all of these exams.

Our focus today will be on triangle congruence and transformations, similarity, trigonometry, volume and measurement and quadrilateral proofs.

These topics have been worth a total of 48 to 58 points on the past ten exams, which would have scaled to a score of 77 to 82.

Although these are not all of the topics you will see, focusing your review on these topics will give you a solid foundation in preparation for this exam.

First, we will look at transformations and triangle congruence, which are two topics that are often connected on the Geometry Regents When determining if triangles are congruent, there are five theorems that can be used to prove or justify congruence.

Side side.

Side, side, side, side angle side, angle Side angle angle angle side and hypotenuse leg.

The two combinations that are not sufficient to prove congruent triangles are angle, angle, angle and side side angle.

When performing transformations, congruent triangles are the result of translations, rotations and reflections which are referred to as rigid motions deletions which change the size result in similar triangles, which we'll be looking at in our next section.

But first, we're going to be looking at the types of questions that show up on the Geometry Regents having to do with congruence and rigid motions.

So our first example is going to be looking at our two triangles y IgG and POM that are two distinct non-right triangles such that angle g is congruent to angle N and they did not give me a picture.

So the first thing I'm going to do is draw and label the picture.

They want to know which statement is now sufficient to prove that Y is always congruent to triangle p0m So the first thing I'm going to do is label my option one.

We have angle E congruent to angle O and we have angle y congruent to angle p. And we can see that we have three congruent angles, which would be angle, angle angle, which is not one of our choices.

Let's look at our next option and here they're describing that Y g is congruent to PM and that y e is congruent to P0.

I have two angles in a side this time, but if I look at the order, this one is side side angle.

So that one is also not sufficient to prove that we have triangles congruent.

The remaining two options have rigid motions that they're describing, so we need to understand that rigid motions are really just another way of telling us congruent parts when one thing maps on to another.

So as I read through, it says that I have angle e mapping onto angle.

angle o, that just is another way to say that they're congruent and that y e is congruent to p o.

So now I have two angles in a side and this particular combination angle angle side is one of the options that works.

So three is going to be our correct choice, but we still have to look at this last example.

It looks really similar to number four and maybe that it could be angle side angle, but if you read carefully, they're talking about mapping points and not angles.

So as I look at option four, I can't actually label my point Y.

So as I look at option four, it says that point Y maps onto point P and Y G maps onto PM, so I can label the segments congruent, but I don't actually have another pair of angles and that one will not work.

The next question is where we start to see rigid motions as more of a short answer question.

The question says in the diagram below parallelogram e f g h is mapped onto parallelogram i j k h. After a flexion over line l and we are asked to use properties of rigid motions to prove that these two parallelogram are congruent.

The key word I get in here is reflection.

Reflection is one of our rigid motions, and we know rigid motions preserve size and they preserve shape, which is enough to say that these two parallelograms are, in fact, congruent.

So I can write as follows parallelogram e f g h is congruent to Parallelogram i j k h. Because reflections are rigid motions and rigid motions preserve size and shape.

The next question asked me the opposite Which one does not always preserve distance?

And in the answer choices, you will see that we have this mapping notation.

So I'm going to show us the mapping notations of each transformation.

It is important that you know this prior to taking the exam.

Going back to my question, since it is says the one that does not always preserve distance, I know I'm looking at a dilation, since that is the only transformation that changes size.

The mapping notation for a dilation shows x, y, too k x comma k y where k is just something that you're multiplying by a constant.

In each of my four answer choices, the only one I see that pop up in is answer choice three where you see two x, So that must be my answer.

You will also want to review the mappings and notations for different transformations before taking the exam, because you can be asked to perform transformations rather than identify them.

Our next question asks us to perform a reflection over the line Y equals one.

Since His is a multiple choice question, I would need to know that I need to use the scrap paper from the back of my exam.

On the scrap paper.

I already drew up my x and y axis and I'm just going to plot what they have given me.

So they gave me the point for three, and I'm reflecting that.

Like I said, over the line, Y equals one.

Reflections are all about preserving distance, so I have to see how far away my point is from my line, which is two units and I have to go that much further.

Two more.

This puts me at the point here, which is the point for negative one.

Another really common type of question to see is being asked to identify the rigid motion that's being performed.

So if I look at question 28 here, it says on the set of axes below congruent triangles, ABC, C and D, e, F are graphed.

So that's important.

They're telling me that I have congruent triangles and they're asking me to describe a sequence of rigid motions.

Now, when I see the word sequence, I it sounds like it should be more than one thing, but it really means one or more.

So this might be one transformation or multiple transformations in my answer.

So first thing I want to do is think about all of the different types of transformations it could be.

So I would have dialations, reflections, rotations and translations as my options.

But hopefully you can see it didn't change size and it even says it has to be a rigid motion for my answer so I can cross out dilation right away.

Now I need to really think about the rest and sometimes mapping out how am I?

Triangles match up can be helpful when they don't give me a prime, B prime and C prime as the labels on my second triangle.

So I'm going to use A, b, C mapping on to D,e,f because they told me that the triangles are congruent and that means that D is like a prime E is going to be like B prime and then F is going to be like C prime.

And now I can really think about my reflection.

If it's a reflection, it's going to change direction.

So it's a great way to test it out.

If you don't just instinctively know if it looks like it flipped over a line.

So I've got A, B, C going in a clockwise direction in my original triangle and then a b, C for my primes, my image is still going clockwise and that's how I know it's not a reflection.

So now I'm just really deciding between a translation or a rotation.

So translation is when it just slides and a rotation is if it turns.

If I look at any of the points, I'm going to pick a and see where a prime is.

I can see that A started out on the left.

A prime is now on the bottom.

So it turned.

So this is going to for sure be a rotation.

It's possible then it might be a combination of a rotation and a translation.

So what we really need to do here is perform the transformation and see where it maps.

It moved one quadrant counterclockwise, which means 90 degrees.

And I can see a is the point to five when I'm doing a rotation .25.

When I'm doing a rotation of 90 degrees, the mapping is that x y would become negative y x, so a prime should be the point.

Negative five two.

If I look at where D is, I can see that it is indeed negative five two, which means this is only a rotation.

So my answer to the question is going to be a rotation of 90 degrees.

But I need to be more specific than that because if I don't include the center, I'm not going to get full credit.

So I do need to include that it is centered at the origin.

So our next topic is going to be similar triangles, which is like the result of a dilation when you have the same shape, but the size changes in this diagram, the corresponding angles will be congruent like in congruent triangles.

But now the sides will only be proportional instead of congruent.

In this diagram, you can see the common pictures that will show up with overlapping triangles that create similar triangles.

To understand these pictures better, it could be helpful to separate the picture into multiple triangles.

To stay consistent, we'll be using the big triangle over small triangle model to right proportions.

But this is not the only correct method.

So our first example it says in triangle ABC below D is a point on A, B and E is a point on a C such that we have d, e parallel to b, c. So that's going to be important.

And I see I have my right two triangles that are overlapping, so a great way to start is drawing out those two separate triangles.

So as we're looking at this picture, we can now label our information on the picture that they gave us and our two separate triangles.

So a D is 12.

I can see that that's one of the sides of my smaller triangle.

D B is eight.

I can label that in the original picture, but it's not actually a side of one of my triangles.

E C ten.

And then they're asking me to find the length of a C, so I'm going to call that X a C is the side of my big triangle.

But those are the only two things that they gave me that are actually going to be in my proportion.

What I'm going to have to do to get the rest of my sides in my proportion is add or subtract the information that they gave me.

So I'm going to add up the 12 in the eight to figure out that AB is 20.

I'm going to subtract my X and my ten to figure out that A E is x minus ten.

From here I can now do my big triangle over small triangle proportion.

So I usually like to just really keep track and start with my big triangle and put those two sides on the top.

So I'm going to put the 20 and the X on the top.

But now to make sure that I'm matching up all of the sides correctly in my proportion, I'm going to really think about what the angles are in this problem that have to match.

So as I look, I've got a reflexive angle at angle A, So I know those have to be the same because the reflexive property says anything is congruent to itself and then we have d e parallel to b, c, so that lets me have congruent corresponding angles at D and B and also at E and C. So this lets me finish my proportion correctly because I can see A, B as the 20 that's got to match up with a D which is 12 and then X is a C that's got to match up with a E, which is x minus ten.

From here we're just going to be going through and cross multiplying.

So we are going to have my 20 X and make sure you distribute here -200 equals 12 X and now we're just solving this equation.

So we want to get our X's to be on the same side.

So I'm going to minus my 20 X over and that's going to give me -200 equals a negative eight X and now my last step is just going to be to divide.

When I divide away that negative eight, I get x equals 25.

So that's going to be the answer to the question.

So the next question has a picture of overlapping triangles, just like the last one.

But this time we're going to be asked to determine if the triangles are similar and explain why.

So just like our triangles be incongruent, we have now three options for how we can decide if triangles are similar.

We have angle, angle, side angle, side and side side side.

So notice this time, instead of congruent sides, we actually have proportional side, something that has the same scale factor.

Also, I want to point out we're going to be explaining our reasoning and any time you do that, you're going to have to explain anything you add to the picture to answer the question, in addition to showing that we have some proportional sides.

So let's look at our problem.

It says in Triangle ADC below e It says in Triangle ADC below eb is drawn such that and then they gave me all of the numbers for the sides that they've already labeled for us.

And the question at the bottom says is a B similar to a I'm sorry, ab E similar to ADC and explain why.

So the first thing I want to do is again draw out my two separate triangles.

So now I'm going to label the information that they gave me just like I did in the last one.

I know my small triangle has a 5.

6 and a 4.

1, but for my big triangle, once again, I'm going to have to add the sides.

So I have a D is 9.02 and I have a C is 12.32.

So I have two sides.

So I know I'm going to be testing outside angle side, but we still need to have our angle.

So remember, just like in our last problem, we have a reflexive angle up at a that I can label and now I have my angle between the two sides.

The only thing I have to do now is do my big triangle over small triangle proportion this time to match up the sides.

We can do a mapping because they gave me the order of the triangles in the question.

I want to know if abe is similar to adC, and that's how I can make sure I'm matching my triangles up correctly.

So I've got my big triangle over small triangle.

My two big triangle sides are the 9.02 and the 12.32.

I look at 9.02.

That is HD.

So if I look over at my mapping, a D has to match a B, so that's how I know I'm going to be putting the 4.1 under the 9. putting the 4.1 under the 9.02.

And likewise we have a C matching a E, So that's how I 5.6 under that 12.32.

So it is possible to cross multiply here, but sometimes you test out three fractions, which is why I prefer to just type those fractions into my calculator to see if they're equivalent.

So when I type each of those fractions in, I'm going to get 2.2, which means that the sides are proportional.

So now we can just say that we have our similar triangles.

But remember we had that angle, a congruent angle A because of the reflexive property.

So first we're going to have to explain that as part of our answer.

So I'm going to say I've got those congruent angles and then I can say that I have my similar triangles because of side angle side similarity.

This is another common similarity question.

When a right triangle has an altitude drawn that creates three similar triangles.

So our question here says in the diagram below of right triangle ACB altitude cd is drawn to hypotenuse ab.

We're told a D equals two and a C equals six.

So I'm going to use this picture to draw out a big medium and small right triangle.

I'm going to start with labeling my right angles from the picture.

So I know in my big triangle that angle C is my right angle.

And then in the medium and small triangle, I know it is angle D because of the altitude.

Next I'm going to label my reflexive angles.

So I see in the small triangle and the big triangle I have angle a in the big and medium triangle, I have angle B and then I'm just going to fill in the rest with C, So now I have them separated and I'm going to fill in the sides.

So wherever I see in a D, I'm going to put a two and wherever I see an AC, I'm going to put a 6.

I am asked to solve for aB, so I'm going to let that be X.

There are no labels on the medium triangle, so I'm not going to be using that one.

I will be looking at my big triangle and my small triangle.

So I'm going to use the big over small, set up my proportion.

So in my big triangle I have an X and I have and my big triangle, I have an X and I have a six.

NOW I'M GOING TO GET THE PIECES FROM MY SMALL TRIANGLE SO I KNOW X IS MY HYPOTENUSE OF MY BIG TRIANGLE.

SO I WILL USE MY HYPOTENUSE OF MY SMALL TRIANGLE TO GO WITH THAT ONE.

AND THEN SIX IS MY BOTTOM LEG ON MY BIG TRIANGLE.

SO I'LL USE THE TWO BECAUSE THAT IS THE BOTTOM LEG OF MY SMALL TRIANGLE AS WE'VE BEEN DOING.

I CAN CROSS MULTIPLY HERE TO SOLVE AND I GET TO X EQUALS 36.

2 X EQUALS 36.

I WANT TO GET X ALONE.

SO I HAVE TO DIVIDE BY TWO AND I GET THAT X IS 18.

WE HAVE TO MAKE SURE OUR X IS WHAT WE'RE LOOKING FOR.

SINCE I LABELED A B AS X, I CAN JUST GO IN AND SAY THAT THAT MEANS THAT A B IS EQUAL TO 18.

WHEN DEALING WITH RIGHT TRIANGLES, KNOWING TWO SIDES AND NEEDING THE THIRD SIDE MEANS TO USE THE PYTHAGOREAN THEOREM.

BUT WHEN YOU HAVE QUESTIONS THAT INVOLVE SIDES AND ACUTE ANGLES OF RIGHT TRIANGLES, THAT'S WHEN YOU KNOW THAT YOU NEED TO DO TRIGONOMETRY AND USE SIGN COSINE OR TANGENT.

YOU WILL NEED TO KNOW HOW TO LABEL THE OPPOSITE ADJACENT AND HYPOTENUSE OF YOUR TRIANGLE BASED ON THE ACUTE ANGLE THAT YOU'RE TALKING ABOUT.

SO IF WE'RE TALKING ABOUT A IN THIS PICTURE THAT WE HAVE, MY SIDE THAT'S ACROSS FROM ANGLE A IS ALWAYS GOING TO BE THE OPPOSITE SIDE.

SO I'M GOING TO CALL THAT, O, WE KNOW THAT THE SIDE ACROSS FROM MY RIGHT ANGLE IS ALWAYS THE HYPOTENUSE, SO I'M GOING TO CALL THAT H, AND THEN THE REMAINING SIDE IS GOING TO BE A I USUALLY LIKE TO WRITE OUT.

SOA CAH TOA AS A REMINDER AS TO WHAT SINE, COSINE AND TANGENT ARE, BECAUSE THEY DO NOT GIVE YOU THAT ON THE REGENTS.

OUR FIRST EXAMPLE COMBINES THE IDEA OF SIMILAR TRIANGLES WITH TRIGONOMETRY.

THE QUESTION SAYS IN DIAGRAM BELOW TRIANGLE SPECIES IS SIMILAR TO TRIANGLES CMJ AND THE COSINE OF J IS EQUAL TO 3/5 AND ASSESS TO DETERMINE AND STATE THE MEASURE OF ANGLE AS TO THE NEAREST DEGREE.

THAT PIECE TELLS ME THAT I NEED TO MAKE SURE MY CALCULATOR IS IN DEGREE MODE, OTHERWISE I AM NOT GOING TO GET THE CORRECT ANSWER.

I'M GOING TO START OFF WITH MY MAPPING BECAUSE I KNOW THAT THESE ARE SIMILAR TRIANGLES.

SO FROM LOOKING AT THIS, I KNOW THAT ANGLES C AND ANGLE J ARE GOING TO BE CONGRUENT.

SO THE FIRST THING I NEED TO DO IS FIGURE OUT WHAT ANGLE J IS FROM MY GIVEN INFORMATION.

SO I'M TOLD THAT THE COSIGN OF J IS 3/5.

SO TO FIND THE VALUE OF ANGLED J, I NEED TO USE MY INVERSE TRIG 22 COSINE INVERSE OF THREE FITS.

WHEN YOU PLUG THAT INTO YOUR CALCULATOR, IF YOU ARE IN DEGREE MODE, THAT SHOULD GIVE YOU 53 DEGREES.

SO I KNOW THAT ANGLE J IS 53 DEGREES AND I ALSO KNOW THAT ANGLE C IS 53 DEGREES.

THIS QUESTION ASKED ME FOR THE MEASURE OF ANGLE S WHICH I STILL DO NOT HAVE.

I NEED TO USE THE FACT THAT THE ANGLES IN A TRIANGLE ADD UP TO 180 DEGREES TO HELP ME FIND THIS MISSING ANGLE.

SO IF I JUST DO 180 MINUS THE 90 DEGREES FROM MY RIGHT ANGLE P AND ALSO MINUS THE 53 DEGREES I JUST CALCULATED FOR ANGLE.

I AM LEFT WITH A TOTAL OF 37 DEGREES, WHICH IS MY LEFT OVER AND MUST BE THE VALUE OF ANGLE S. SO THIS QUESTION GAVE US A PICTURE.

SOMETIMES WE'RE GOING TO NEED TO DRAW A PICTURE FROM GIVING INFORMATION.

THE NEXT QUESTION HAD A PICTURE AND WE REMOVED IT SO WE CAN GET PRACTICE WITH DRAWING THESE PICTURES ON OUR OWN.

THE QUESTIONS HAVE SAID IN THE DIAGRAM BELOW A WINDOW OF A HOUSE IS 15 FEET ABOVE THE GROUND.

A LADDER IS PLACED AGAINST THE HOUSE WITH ITS BASE AT AN ANGLE OF 75 DEGREES, WITH THE GROUND DETERMINED IN STATE THE LENGTH OF THE LADDER TO THE NEAREST 10TH OF A FOOT.

WHEN I START DRAWING A RIGHT TRIANGLE.

MY RIGHT ANGLE IS HERE.

SINCE IT IS A WINDOW OF A HOUSE THAT'S A 15 FEET ABOVE THE GROUND, THAT'S A HEIGHT.

SO I KNOW THAT THAT'S GOING TO GO OVER HERE.

AND I'M TOLD THAT THE LADDER IS PLACED AGAINST THE HOUSE MAKES AN 75 DEGREE ANGLE WITH THE GROUND.

SO I KNOW THAT THAT ANGLE IS GOING TO GO OVER HERE, SINCE THE OTHER ANGLE WITH THE GROUND IS 90 DEGREES.

THE LADDER ITSELF IS LEANING ON THE HOUSE.

SO THAT MEANS THAT MY UNKNOWN OR MY X IS THIS SIDE RIGHT HERE.

SO I'M GOING TO START WITH WRITING OUT MY SOA COH TOA TOWER AND LABELING MY OPPOSITE ADJACENT AND HYPOTENUSE.

SO LOOKING HERE NOW FROM MY 75 DEGREES, IF I DRAW A CROSS, I GET TO MY 15, WHICH IS NOW MY OPPOSITE.

AND IF YOU DRAW A CROSS FROM YOUR RIGHT ANGLE, YOU GET TO YOUR X, WHICH IS YOUR HYPOTENUSE.

SO I HAVE INFORMATION ON MY OPPOSITE AND I'M LOOKING FOR MY HYPOTENUSE, WHICH TELLS ME THAT I AM GOING TO BE USING SIGN.

SO I'M GOING TO SET UP MY PROPORTION.

SO I KNOW THAT'S A SIGN OF 75 DEGREES IS EQUAL TO MY OPPOSITE 15 OVER MY HYPOTENUSE X, WHICH I DO NOT KNOW.

WHAT I CAN DO NOW IS THROW THE SIGN OF 75 OVER ONE AND THEN WE CAN CROSS MULTIPLY.

WHEN I CROSS MULTIPLY I GET X TIMES SIGN OF 75.

EQUALS 15 TIMES ONE, WHICH IS JUST 15.

NOW I WANT TO GET X COMPLETELY ALONE, SO I'M GOING TO DIVIDE BOTH SIDES BY SINE OF 75.

AND NOW I PUT INTO MY CALCULATOR AND I WAS TOLD TO ROUND TO THE NEAREST 10TH OF A FOOT.

AND WHEN I DO SO I SHOULD GET 15.5 FEET AND THAT WOULD BE MY LENGTH OF THE LADDER.

SO OUR LAST QUESTION IS A DOUBLE TRIANGLE PROBLEM.

THIS IS A REALLY COMMON FOR POINT QUESTION TO SEE ON THE REGENTS EXAM WHERE YOU HAVE OVERLAPPING RIGHT TRIANGLES.

BECAUSE REMEMBER SOKOTO YOU CAN ONLY DO WITH RIGHT TRIANGLES.

SO IF I LOOK AT THIS QUESTION, IT SAYS CAPE CANAVERAL, FLORIDA, IS WHERE NASA LAUNCHES ROCKETS INTO SPACE AS MODELED IN THE DIAGRAM BELOW, A PERSON VIEWS THE LAUNCH OF A ROCKET FROM OBSERVATION AREA, A 3280 FEET AWAY FROM THE LAUNCH PAD B AFTER LAUNCH, THE ROCKET WAS SITED AT C WITH AN ANGLE OF ELEVATION OF 15 DEGREES AND THEN LATER SITED AT D WITH AN ANGLE OF ELEVATION OF 31 DEGREES.

WHAT THEY'RE ASKING US TO FIND IS TO THE NEAREST FOOT, THE DISTANCE THE ROCKET TRAVELED BETWEEN C AND D. SO THIS IS REALLY WHAT THEY WANT US TO FIND.

I'M GOING TO CALL IT H BECAUSE IT'S THE HEIGHT OF THE ROCKET BETWEEN THOSE TWO POINTS.

THE PROBLEM IS THAT IS PART OF A TRIANGLE THAT'S NOT ACTUALLY A RIGHT TRIANGLE.

SO I NEED TO FIND MY TWO RIGHT TRIANGLES IN THIS PICTURE SO I CAN SEE THAT I HAVE A BIG RIGHT TRIANGLE THAT HAD THE ANGLE OF ELEVATION OF 31 AND THE HEIGHT OF THAT I'M GOING TO CALL X. I CAN ALSO SEE THAT I HAVE THIS SMALLER RIGHT TRIANGLE SITTING ON TOP OF IT WITH THE ANGLE OF ELEVATION OF 15.

I'M GOING TO CALL THAT HEIGHT Y.

SO THESE ARE GOING TO BE THE TWO TRIANGLES WE ACTUALLY HAVE TO LOOK AT AND USE TRIG WITH TO FIND THE ANSWER.

AND WE'RE GOING TO BE ABLE TO FIND THE HEIGHT BY JUST SUBTRACTING THE TALLER ONE, WHICH IS X MINUS THAT SHORTER ONE, WHICH IS Y.

SO WE'VE GOT OUR GAME PLAN.

I'M GOING TO START BY DRAWING OUT MY TWO SEPARATE TRIANGLES AND LABELING THE INFORMATION.

SO WE HAVE OUR TWO TRIANGLES.

I'VE GOT MY 31 DEGREE ANGLE OF ELEVATION AND MY FIRST ONE, AND THAT'S THE ONE THAT I CALLED X AS THE HEIGHT.

AND THEN WE HAVE THAT 3280 THAT'S GOING TO BE AT THE BOTTOM FOR BOTH.

SAME IDEA WITH MY SECOND TRIANGLE, MY ANGLE OF ELEVATION IS NOW 15.

THE HEIGHT OF THIS ONE IS Y, AND THEN I STILL HAVE THAT 30 TO 80 THE BOTTOM.

SO WE'RE GOING TO HAVE TO DO TRIG TWO TIMES.

SO WE'RE STILL THINKING SOACAHTOA, WE'RE STILL GOING TO HAVE TO LABEL OUR OPPOSITE ADJACENT AND HYPOTENUSE.

AND FOR BOTH OF THEM IT'S GOING TO BE FROM THAT ANGLE OF ELEVATION.

SO FROM A WITH THE 31 DEGREE ANGLE AXIS, MY OPPOSITE AND THE HYPOTENUSE IS REALLY THE BLANK SIDE THIS TIME THAT WE'RE NOT GOING TO BE USING, WHICH LEAVES 3280 TO BE THE ADJACENT.

MY SECOND TRIANGLE IS GOING TO BE LABELED EXACTLY THE SAME.

THE HEIGHT IS GOING TO BE THE OPPOSITE AND THE ADJACENT IS ON THE GROUND.

THE HYPOTENUSE IS ACROSS FROM THAT RIGHT ANGLE.

SO I'M JUST GOING TO HAVE TO WRITE OUT TWO DIFFERENT EQUATIONS.

AND BECAUSE WE'RE DEALING WITH THE OPPOSITE AND ADJACENT, THIS TIME IT'S GOING TO BE A TANGENT EQUATION.

THESE TWO TANGENT EQUATIONS ARE GOING TO LOOK ALMOST IDENTICAL.

MY FIRST ONE IS GOING TO HAVE 31 AS MY ANGLE AND THEN OPPOSITE OVER ADJACENT.

THIS TIME IS X OVER THAT 30 TO 80.

MY SECOND EQUATION IS GOING TO BE TANGENT OF 15 EQUALS Y OVER 30 TO 80, AND THEN WE'RE JUST CROSS MULTIPLYING TO SOLVE FOR BOTH.

THIS TIME X IS ON THE TOP.

SO WHEN I DO X TIMES ONE, I JUST GET X AND THEN WHAT I'M GOING TO HAVE TO TYPE IN MY CALCULATOR IS THE 30 TO 80 TIMES THE TANGENT OF 31.

SAME IDEA FOR MY NEXT ONE WE ARE GOING TO END UP WITH Y EQUALS 32 80 TIMES THE TANGENT OF 15.

SO WHEN I PUT THOSE IN MY CALCULATOR, X IS GOING TO GIVE ME 1,970.

ME 1,970.8 AND Y IS GOING TO GIVE ME 878.9.

SO WE'RE ALMOST DONE.

WE JUST HAVE ONE LAST STEP.

REMEMBER, IF WE GO BACK TO THAT ORIGINAL PICTURE, THE HEIGHT WAS JUST SUBTRACTING THOSE TWO VALUES.

SO WHEN I SUBTRACT, I'M GOING TO PLUG IN MY H WAS THAT 19 70.

8 AND MY Y WAS THAT EIGHT 98.

9, I'M SORRY, POINT NINE.

AND THEN WE'RE ROUNDING, REMEMBER, TO THE NEAREST FOOT.

SO MY FINAL ANSWER IS GOING TO BE 1092.

NEXT WE'RE GOING TO BE TALKING ABOUT MEASUREMENT AND VOLUME OF 3D FIGURES.

THESE PROBLEMS WILL REQUIRE THE FORMULAS FROM THE REFERENCE SHEET, AND THEY OFTEN HAVE MORE THAN ONE STEP, EVEN IF IT'S MULTIPLE CHOICE.

SO IT'S REALLY IMPORTANT TO READ CAREFULLY AND MAKE SURE YOU'RE USING ALL PARTS OF THE QUESTION TO ANSWER THE PROBLEM.

AS I LOOK AT MY FIRST EXAMPLE, IT SAYS A SMALL TOWN IS INSTALLING A WATER STORAGE TANK IN THE SHAPE OF A CYLINDER.

AS SOON AS I SEE CYLINDER, I KNOW I'M GOING TO BE WRITING DOWN THE VOLUME OF A CYLINDER FORMULA.

THEN IT SAYS THE TANK MUST BE ABLE TO HOLD AT LEAST 100,000 GALLONS OF WATER.

SO THAT IS THE VOLUME OF THE TANK.

I USUALLY LIKE TO JUST KEEP TRACK OF THE INFORMATION THAT THEY'RE GIVING ME AS I READ THROUGH.

OKAY.

SO MY VOLUME WAS THAT 100,000 GALLONS AND THEN THEY'RE TELLING ME THAT MY HEIGHT HAS TO BE 30 FEET.

SO WE'VE GOT A LITTLE PROBLEM HERE BECAUSE I CAN SEE THAT MY FIRST NUMBER IS MEASURED IN GALLONS.

MY SECOND NUMBER IS MEASURED IN FEET.

SO BEFORE WE CAN ACTUALLY PLUG INTO THE VOLUME FORMULA, WE'RE GOING TO HAVE TO MAKE SURE THAT THEY MATCH UP.

NOW, FORTUNATELY, THEY'VE GIVEN ME A CONVERSION RIGHT IN THE QUESTION.

THIS TIME THEY'RE TELLING ME ONE CUBIC FOOT HOLDS 7.

CUBIC FOOT HOLDS 7.48 GALLONS OF WATER.

SOMETIMES YOU NEED TO GO TO THE REFERENCE SHEET TO FIND THE CONVERSION AND YOU JUST HAVE TO NOTICE THE MISMATCHED UNITS ON YOUR OWN.

BUT THIS TIME THEY KIND OF HELPED ME OUT.

SO I KNOW THAT I'M GOING TO HAVE TO START OUT BY DOING A CONVERSION.

I'M GOING TO USE DIMENSIONAL ANALYSIS, BUT THERE'S OTHER WAYS OF CONVERTING IN DIMENSIONAL ANALYSIS.

WE START WITH WHAT WE'RE GOING TO CHANGE LATER ON AND THE QUESTION THEY'RE ASKING FOR OUR ANSWER TO BE IN FEET, WHICH IS HOW I KNOW THAT I NEED TO GET RID OF THE GALLONS.

SO I'M GOING TO START WITH MY 100,000 GALLONS AND PUT IT OVER ONE.

NOW I'M JUST GOING TO MULTIPLY BY A FRACTION THAT HAS THE TWO PARTS OF MY CONVERSION IN IT.

SO I HAVE GALLONS I'M GETTING RID OF.

SO NOW I NEED TO PUT THEM IN THE BOTTOM AND CUBIC FEET THAT I'M TRYING TO TURN IT INTO.

SO NOW I'M GOING TO PUT THAT CUBIC FEET IN THE TOP AND I JUST MATCH UP MY NUMBERS.

ONE GOES WITH THE CUBIC FEET AND THEN MY 7.48 GOES WITH THE GALLONS.

SO THIS JUST GETS TYPED INTO THE CALCULATOR AND WE END UP GETTING A VOLUME OF 13,368.

98 AND THE NUMBER KEEPS GOING.

A VOLUME OF 13,368.98 AND THE NUMBER KEEPS GOING.

IF YOU HAVE A CALCULATOR THAT YOU CAN USE THE UNROUNDED ANSWERS TO MOVE FORWARD, THAT'S ALWAYS YOUR BEST BET.

SO NOW WE HAVE THE RIGHT UNITS THAT ARE MATCHING BECAUSE THESE ARE CUBIC FEET AND WE CAN PLUG INTO OUR VOLUME FORMULA AND FINISH ANSWERING THE QUESTION.

SO WHAT THEY'RE ACTUALLY ASKING ME FOR IS THE MINIMUM DIAMETER OF THE TANK.

SO WE HAVE TO THINK ABOUT ONE MORE THING.

OUR EQUATION HAS A RADIUS IN IT, BUT WE NEED TO HAVE THE DIAMETER AS OUR FINAL ANSWER.

SO JUST REMEMBER THAT A DIAMETER IS JUST TWO TIMES THE RADIUS, AND IT'S REALLY IMPORTANT TO MAKE SURE THAT WE INCLUDE THAT BEFORE WE CHOOSE AN ANSWER.

SO LET'S PLUG IN AND SEE WHAT WE GOT.

MY VOLUME WAS THAT 13,000 NUMBER.

I NEED TO HAVE MY PI.

THE RADIUS IS WHAT I DON'T KNOW.

AND THEN I'M PLUGGING IN 30 FOR THE HEIGHT, SO I'M GOING TO BE SOLVING FOR MY RADIUS.

SO EVERYTHING HERE IS BEING MULTIPLIED.

SO I JUST HAVE TO DIVIDE AWAY THE 30 PI THEY CANCEL AND ANYTHING I DO TO ONE SIDE, I JUST DO TO THE OTHER.

SO WE END UP GETTING THAT MY RADIUS SQUARED IS 141, BUT WE'RE STILL NOT DONE BECAUSE WE HAVE TO SQUARE ROOT TO GET THAT RADIUS.

WHEN I DO THE SQUARE ROOT OF 141, I GET 11.9.

REMEMBER, THAT'S MY RADIUS.

AND WHAT THEY'RE ACTUALLY ASKING ME FOR IS THE DIAMETER.

BUT NOTICE THAT OPTION ONE IS 12.

WE WANT TO REJECT THAT ONE BECAUSE IT'S NOT THE DIAMETER.

THAT ONE IS THE RADIUS.

WE HAVE TO FIRST MULTIPLY BY TWO AND WE COME UP WITH A DIAMETER OF 24.

SO OUR NEXT QUESTION IS ANOTHER WORD PROBLEM.

BUT THIS TIME WE WILL SEE DENSITY, WHICH IS A FORMULA THAT IS NOT GIVEN TO YOU ON YOUR REFERENCE SHEET.

SO YOU HAVE TO MAKE SURE YOU KNOW IT.

SO THIS ONE SAYS A REGULAR PYRAMID.

SO AGAIN, AS SOON AS I SEE THAT, I KNOW I'M USING MY PYRAMID FORMULA FROM THE REFERENCE SHEET HAS A SQUARE BASE AND IT'S MADE OF SOLID GLASS, IT HAS A BASE AREA.

SO EVEN THOUGH THEY TOLD ME THAT I HAD A SQUARE BASE, I DON'T ACTUALLY NEED THAT BECAUSE THEY'RE JUST TELLING ME THAT THE BIG B THE AREA OF THE BASE IS 36 AND MY HEIGHT IS TEN.

I KEEP READING AND NOW THEY'RE TALKING ABOUT DENSITY.

SO NOW I NEED TO HAVE ANOTHER FORMULA.

AS SOON AS I SEE THE WORD DENSITY, I NEED TO KNOW THAT THAT'S MASS OVER VOLUME.

AND THEY'RE TELLING ME THAT THE DENSITY OF THE GLASS IS THIS 2.7.

GOING TO BE THE 2.7 WHEN I SOLVE, AND THEN THEY'RE ASKING ME TO FIND THE MASS OF THE PYRAMID, AND I CAN SEE THAT THAT IS IN THAT DENSITY FORMULA.

SO LET'S PLUG IN.

WE'RE GOING TO HAVE TO START WITH OUR VOLUME FORMULA SO THAT WE HAVE VOLUME TO PLUG INTO THE DENSITY FORMULA.

SO WE'VE GOT V EQUALS ONE THIRD.

MY BIG B IS 36 AND MY HEIGHT IS TEN.

WHEN I PUT THAT IN THE CALCULATOR, I'M GOING TO GET A VOLUME OF 120.

NOW I'M READY TO PLUG INTO MY DENSITY FORMULA.

SO REMEMBER THE D IS 2.

7 M IS WHAT WE'RE SOLVING FOR AND WE JUST CALCULATED THE VOLUME TO BE 120.

SO ONCE AGAIN WE'RE GOING TO CROSS MULTIPLY AND TIMES ONE IS M AND THEN MY 2.

7 TIMES MY 120 GIVES ME 324, WHICH IS OPTION TWO.

BUT NOTICE OPTION ONE AGAIN WAS 120.

SO IF I WASN'T READING CAREFULLY, IT WAS EASY TO MAYBE STOP EARLIER ON IN THE QUESTION.

OUR NEXT QUESTION, WE SEE A PICTURE OF A RIGHT CIRCULAR CONE AND WE'RE ASKED TO FIND ITS VOLUME.

WE'RE THAT IT HAS A DIAMETER OF TEN AND A SLANT HEIGHT OF 13.

THE VOLUME FORMULA FOR MY CONE FROM MY REFERENCE SHEET IS ONE THIRD PI TIMES RADIUS SQUARED TIMES THE HEIGHT.

SO I'M GIVEN MY DIAMETER IS TEN, SO I KNOW I NEED TO CHANGE THAT TO BE MY RADIUS.

MY RADIUS IS GOING TO JUST BE HALF OF THAT.

SO IT'S GOING TO BE FIVE.

I'M GIVING MY SLANT HEIGHT OF 13, BUT THE HEIGHT I NEED TO PLUG INTO MY FORMULA IS THIS ONE.

ONCE I DRAW THAT IN, I CAN SEE I CREATED A RIGHT TRIANGLE WITH TWO SIDES THAT I ALREADY KNOW THE FIVE AND THE 13.

IN THESE CASES, I CAN GO RIGHT TO USING PYTHAGOREAN THEOREM TO FIND MY THIRD SIDE OR A SQUARED PLUS B SQUARED EQUALS C SQUARED IN THE IN THE PYTHAGOREAN THEOREM, C IS YOUR HYPOTENUSE, WHICH IS, LIKE WE SAID BEFORE, ACROSS FROM THE RIGHT ANGLE.

SO I KNOW 13 IS GOING TO BE MY C, SO I'M GOING TO PLUG THIS IN.

SO I HAVE H SQUARED PLUS FIVE SQUARED EQUALS 13 SQUARED.

I WANT TO GET H ALONE.

SO I'M GOING TO SUBTRACT FIVE SQUARED FROM BOTH SIDES.

SO I'M LEFT WITH H SQUARED EQUALS AND WHEN I PLUG 13 SQUARED MINUS FIVE SQUARE TO MY CALCULATOR I GET 144.

BUT NOW I JUST NEED TO GET THAT H TO BE COMPLETELY ALONE, TO UNDO THE SQUARE.

I HAVE TO TAKE THE SQUARE ROOT OF BOTH SIDES AND I'LL GET H IS EQUAL TO 12.

SO NOW I HAVE SOMETHING TO PLUG IN FOR MY HEIGHT.

SO I'M GOING TO PLUG MY RADIUS OF FIVE AND MY HEIGHT OF 12 INTO MY VOLUME FORMULA.

AND BEFORE I PUT THIS INTO MY CALCULATOR, I NEED TO FINISH READING THE QUESTION.

IT SAYS TO HAVE THIS IN TERMS OF PI, WHICH IS BASICALLY JUST A WAY OF TELLING ME THAT THEY DON'T WANT ME TO PLUG THE PI INTO MY CALCULATOR, SO I'M JUST GOING TO PLUG THE ONE THIRD TIMES FIVE SQUARED TIMES 12 INTO MY CALCULATOR, WHICH WOULD GIVE ME 100.

SO MY FINAL ANSWER WOULD BE 100 PI.

OUR NEXT QUESTION IS A LENGTHY WORD PROBLEM, SO LET'S READ IT CAREFULLY.

IT SAYS JOSH IS MAKING A SQUARE BASED FIRE-PIT OUT OF CONCRETE FOR HIS BACKYARD AS MODELED BY THE WRIGHT PRISON.

BELOW, HE PLANS TO MAKE THE OUTSIDE WALLS OF THE FIRE-PIT 3.

5 FEET ON EACH SIDE, WHICH YOU CAN SEE LABELED WITH A HEIGHT OF 1.5 FEET THAT YOU CAN ALSO SEE LABELED THE CONCRETE WALLS OF THE FIRE-PIT GOING TO BE NINE INCHES THICK.

I'M ASKED IF A BAG OF CONCRETE MIX WILL FILL POINT SIX FEET CUBED, DETERMINE AND STATE THE MINIMUM NUMBER OF BAGS NEEDED TO BUILD THE FIRE-PIT.

SINCE THIS IS GIVEN TO ME AND FEET CUBED, I KNOW I'M GOING TO BE LOOKING AT MY VOLUME.

THIS VOLUME IS A LARGER RECTANGULAR PRISM, WHICH I'LL CALL V ONE THE OUTSIDE.

WITH A SMALLER ONE, I'LL CALL V 2 CUT OUT OF IT FOR WHERE THE FIRE-PIT IS GOING TO GO ON OUR LARGER OR V ONE.

I KNOW I'M GOING TO HAVE THE LENGTH OF 3.5 AND I'M GOING TO HAVE THE WIDTH OF ALSO 3.5.

SINCE THIS IS A SQUARE BASE BEFORE I CAN PLUG IT IN, YOU EITHER NEED TO LOOK AT YOUR REFERENCE SHEET TO HAVE THE AREA OF THE BASE TIMES THE HEIGHT OR SINCE THIS IS A RECTANGULAR PRISM, WE CAN JUST USE LENGTH TIMES WITH TIMES HEIGHT, WHICH IS WHAT I'M GOING TO BE USING.

SO LIKE I SAID BEFORE, WE KNOW WE HAVE 3.

5 TIMES 3.

5 TIMES 3.5 TIMES OUR HEIGHT OF 1.5.

AND IF YOU PLUG THAT INTO YOUR CALCULATOR, YOU GET 18.375.

NOW, I NEED TO FIGURE OUT WHAT'S ON THE INSIDE TO TAKE AWAY.

SO I'M TOLD THAT THE WALLS ARE GOING TO BE NINE INCHES THICK, SO I NEED TO HAVE NINE INCHES ON BOTH SIDES.

SO IF I DO NINE TIMES TWO, THAT GIVES ME 18 INCHES I HAVE TO TAKE AWAY.

NOW THIS IS IN INCHES WHEN EVERYTHING ELSE IS GIVEN IN FEET.

SO I HAVE TO DIVIDE THAT 18 BY TO GIVE ME 1.5 FEET.

NOW THAT'S GOING TO BE WHAT I'M TAKING AWAY TO CREATE THAT SMALLER HOLE.

SO MY ORIGINAL ONE WAS 3.5 FEET.

SO I HAVE TO TAKE AWAY THAT 1.

5 FEET AND I WILL BE LEFT WITH TWO FEET FOR A LENGTH AND A WIDTH ON MY SMALLER RECTANGLE FOR V TWO.

SO FOR MY SECOND VOLUME I'LL JUST CHANGE THE 3.

JUST CHANGE THE 3.5 TO 2 IS THE HEIGHT IS STILL 1.

HEIGHT IS STILL 1.5.

WHEN I PLUG THAT INTO MY CALCULATOR, I GET SIX.

SO NOW TO FIND THE VOLUME OF THE ACTUAL CONCRETE FIRE-PIT, I NEED TO DO THE BIGGER VOLUME MINUS THE SMALLER VOLUME.

SO I NEED TO DO THE 18.375 MINUS SIX, WHICH GIVES ME 12.375 AS MY VOLUME.

NOW I NEED TO USE THAT VOLUME TO FIGURE OUT HOW MANY BAGS OF CONCRETE I ACTUALLY NEED.

SO I NEED TO DO THAT 12.375 DIVIDED BY THE 0.6 AND THAT WILL GIVE ME 20.625.

NOW, SINCE I'M PURCHASING BAGS OF CONCRETE, I CAN'T PURCHASE A FRACTION OF IT, SO I NEED TO PURCHASE 21 BAGS.

OUR LAST TOPIC WILL BE QUADRILATERAL PROOFS, WHICH IS THE MOST COMMON TYPE OF TO SEE ON THE REGENTS EXAM.

OUR FIRST EXAMPLE WILL BE A COORDINATE PROOF WHERE WE WILL NEED TO CALCULATE SLOPES AND DISTANCES TO PROVE DIFFERENT SHAPES.

SO YOU CAN SEE IN THE CHART THAT WE HAVE, YOU HAVE TO USE SLOPES TO PROVE THINGS LIKE PARALLEL AND PERPENDICULAR SEGMENTS AND THEN THE DISTANCE FORMULA OR PYTHAGOREAN THEOREM TO PROVE CONGRUENT SEGMENTS.

YOU CAN BE ASKED TO PROVE ANY OF THE SPECIAL QUADRILATERAL.

SO YOU NEED TO BE PREPARED TO PROVE ANY SHAPE.

HERE ARE THE DIFFERENT METHODS THAT YOU CAN USE TO PROVE THE DIFFERENT QUADRILATERAL THAT YOU MIGHT SEE ON THE EXAM.

IN A COORDINATE PROOF, YOU HAVE SOME CHOICES.

BUT OUR SECOND EXAMPLE IS GOING TO BE A STATEMENT RECENT PROOF, AND YOU REALLY HAVE A LITTLE BIT LESS CHOICE BECAUSE YOU'RE GOING TO BE USING THE DEFINITIONS OF GIVEN VOCABULARY INSTEAD OF THINGS LIKE SLOPES AND DISTANCES TO PROVE WHAT YOU NEED FOR EACH SHAPE.

SO YOU REALLY NEED TO BE PREPARED FOR ANY OF THESE DIFFERENT OPTIONS TO SHOW UP FOR A STATEMENT REASON PROOF.

WE'RE GOING TO BE FOCUSING IN ON PARALLELOGRAM BECAUSE THAT'S THE ONE THAT REALLY SHOWS UP THE MOST.

WE'LL BE STARTING OFF WITH OUR COORDINATE PROOF.

YOU CAN SEE YOU GAVE US A QUADRILATERAL ABCD WITH THE VERTICES GIVEN AND I'M ASKED TO PROVE THAT ABCD IS A PARALLELOGRAM BUT NOT A RECTANGLE.

THE FIRST THING YOU NEED TO DO WHEN YOU ARE GIVEN A COORDINATE PROOF IS PLOT.

PROOF IS PLOT THE POINTS WHICH I HAVE ALREADY DONE AND SINCE WE ARE ON A GRAPH, THE PREFERRED METHOD TO PROVE THAT THIS IS A PARALLELOGRAM WOULD BE TO SHOW THE SLOPE OF BOTH SIDES.

THE PREFERRED METHOD WOULD BE TO SHOW THAT THE BOTH PAIRS OF OPPOSITE SIDES ARE PARALLEL.

SO I'M GOING TO GO IN AND FIND THE SLOPES OF ALL FOUR SIDES AND FOCUS ON THE OPPOSITE PAIRS.

SINCE IT'S ON GRAPH PAPER, I CAN JUST USE RISE OVER RUN TO FIND ALL OF MY SLOPES.

SO IF I START WITH AB AND I GO UP AND OVER, YOU'LL SEE WE GO UP SO POSITIVE FOUR AND TO THE RIGHT THREE FOR A POSITIVE THREE.

SO MY SLOPE OF AB IS 4/3.

I CAN DO THE SAME THING FOR DC, IT'S OPPOSITE SIDE.

I WOULD JUST GO UP FOUR AND TO THE RIGHT THREE AGAIN, WHICH ALSO GIVES ME A SLOPE OF 4/3.

SO I CAN ALREADY SEE THAT THOSE TWO SIDES HAVE EQUAL SLOPES.

I JUST HAVE TO CHECK THE OTHER TWO.

SO NOW IF I START WITH A D, I SEE THAT I HAVE TO GO DOWN FIVE AND TO THE RIGHT FIVE, WHICH IS NOW A NEGATIVE FIVE OVER FIVE, WHICH WOULD REDUCE TO A NEGATIVE ONE AND B C I ALSO GO DOWN FIVE AND TO THE RIGHT FIVE, WHICH IS ANOTHER NEGATIVE FIVE OVER POSITIVE FIVE, WHICH REDUCES TO A NEGATIVE ONE.

SO FROM HERE YOU CAN ALREADY SEE THAT TWO PAIRS OF OPPOSITE SIDES HAVE EQUAL SLOPES, WHICH IS ENOUGH TO TELL ME THAT THEY ARE PARALLEL.

SO TO DRAW THAT FIRST CONCLUSION, I CAN SAY A, B, AND C, D ARE PARALLEL AND B, C AND A, D ARE PARALLEL BECAUSE THEY HAVE EQUAL SLOPES, WHICH THEN TELLS ME THAT ABCD IS A PARALLELOGRAM BECAUSE BOTH PAIRS OF OPPOSITE SIDES ARE PARALLEL.

NOW I HAVE TO LOOK AT THE SECOND HALF OF THAT QUESTION WHICH ASKS ME TO SHOW THAT IT'S NOT A RECTANGLE.

TO SHOW THAT SOMETHING IS NOT A RECTANGLE, WE NEED TO SHOW THAT THERE ARE NO RIGHT ANGLES SO YOU CAN PICK ANY TWO SIDES THAT TOUCH LIKE A, B AND B, C AND SHOW THAT THEY DO NOT HAVE NEGATIVE RECIPROCAL SLOPES.

FOR A NEGATIVE RECIPROCAL SLOPE IT JUST MEANS THAT YOU IT JUST MEANS THAT YOU CHANGE THE SINE AND FLIP THE FRACTION.

SO FOR EXAMPLE, IF I DID HAVE THE NEGATIVE SLOPE, NEGATIVE RECIPROCAL SLOPE OF 4/3, THAT WOULD BECOME NEGATIVE THREE FOURTHS.

WHAT YOU CAN CLEARLY SEE IS NOT THE NEGATIVE ONE THAT WE HAVE.

SO BECAUSE THEY ARE NOT NEGATIVE RECIPROCAL SLOPES, THEY ARE NOT PERPENDICULAR, WHICH MEANS THAT ANGLE B IS NOT A RIGHT ANGLE.

SO I CAN GO IN AND SAY THAT THAT A, B AND B, C ARE NOT PERPENDICULAR BECAUSE THEY ARE NOT NEGATIVE RECIPROCAL SLOPES, WHICH MEANS THAT ANGLE B IS NOT A RIGHT ANGLE.

SO THEREFORE ABCD IS NOT A RECTANGLE BECAUSE ANGLE B IS NOT A RIGHT ANGLE.

SO OUR LAST EXAMPLE IS AN EXAMPLE OF A STATEMENT.

REASON QUADRILATERAL PROOF.

IT SAYS IN THE DIAGRAM OF QUADRILATERAL ABCD BELOW A B IS CONGRUENT TO CD AND A B IS PARALLEL TO CD, SO THE SAME PAIR OF SIDES ARE PARALLEL AND CONGRUENT.

THERE ALSO TELLING ME THAT SEGMENT C AND A F ARE DRAWN AND THAT B IS CONGRUENT TO F. WHAT THEY'RE ASKING ME TO PROVE AT THE END IS THAT C E IS CONGRUENT TO AF.

USUALLY WHEN WE HAVE SEGMENTS THAT WE'RE PROVING CONGRUENT, WE FIRST HAVE TO PROVE THAT TRIANGLES ARE CONGRUENT AND THEN WE CAN USE SOMETHING CALLED CP C CPCTC TO PROVE THAT THOSE CORRESPONDING PARTS OF CONGRUENT TRIANGLES ARE CONGRUENT.

WE REALLY HAVE A COUPLE OF OPTIONS FOR HOW WE CAN DO THAT, BUT BECAUSE WE REALLY STARTED OUT KNOWING THAT WE HAVE THAT SAME PAIR OF SIDES PARALLEL AND CONGRUENT, I CAN ACTUALLY SAY THAT I HAVE A PARALLELOGRAM JUST WITH THOSE TWO PIECES OF INFORMATION.

SO IF I START LABELING WHAT THEY GAVE ME, THEY SAID A B WAS CONGRUENT TO CD AND PARALLEL TO CD.

AND THEN THE OTHER THING THAT THEY GAVE ME WAS B E CONGRUENT TO DFE.

TO DF.

SO EVEN THOUGH I MIGHT BE THINKING THAT I NEED TO START WITH THOSE LEFT AND RIGHT TRIANGLES, IT'S ACTUALLY GOING TO BE EASIER IN THIS PROOF TO PROVE THAT THE TOP TRIANGLE AND BOTTOM TRIANGLE ARE CONGRUENT BECAUSE THEY GAVE ME PARTS OF THOSE TRIANGLES CONGRUENT AS WELL.

SO I'M GOING TO START OUT MY PROOF BY JUST SAYING THAT I HAVE MY GIVEN BECAUSE THAT'S HOW WE ARE.

THAT'S HOW WE ALWAYS START.

WE START ANY PROOF AND THEN I'M GOING TO GO STRAIGHT INTO SAYING THAT I HAVE A PARALLELOGRAM, BECAUSE WHEN I SAY I HAVE A PARALLELOGRAM, IT JUST OPENS UP WAY MORE OPTIONS FOR WHAT I CAN USE IN THE PROOF TO LABEL MORE CONGRUENT PARTS.

SO IT'S A PARALLELOGRAM BECAUSE I HAVE ONE PAIR OF SIDES OPPOSITE THAT ARE PARALLEL AND CONGRUENT.

NOW THAT I HAVE THAT I CAN LABEL, MY OTHER PAIR OF SIDES ARE CONGRUENT AND PARALLEL BECAUSE THOSE ARE PROPERTIES OF A PARALLELOGRAM.

SO MY NEXT STEP IN MY PROOF WOULD BE SAYING THAT B, C, AND SO MY NEXT STEP IN MY PROOF WOULD BE SAYING THAT B, C, AND AD.

ARE PARALLEL AND CONGRUENT BECAUSE OF PARALLELOGRAM PROPERTIES.

SO IF I REALLY TAKE INVENTORY OF WHAT I'VE GOT HERE, I'VE GOT TWO PAIRS OF CONGRUENT SEGMENTS THAT I'VE ALREADY TALKED ABOUT IN MY PROOF.

BUT IN ORDER TO PROVE TRIANGLES ARE CONGRUENT, IF YOU THINK ALL THE WAY BACK TO THE BEGINNING OF OUR REVIEW, WE ALWAYS NEED THREE THINGS.

SO WE'RE GOING TO NEED A THIRD THING STILL, AND IT'S GOING TO BE A PAIR OF ALTERNATE INTERIOR ANGLES THAT WE CAN LABEL BASED OFF OF THOSE PARALLEL LINES.

SO I WOULD HAVE ANGLE B E CONGRUENT TO ANGLE ADF BECAUSE I HAVE ALTERNATE INTERIOR ANGLES THAT ARE CONGRUENT.

SO MY NEXT STEP IN MY PROOF WOULD BE SAYING THAT THE ANGLES ARE CONGRUENT BECAUSE I HAVE THOSE ALTERNATE INTERIOR ANGLES THAT ARE CONGRUENT.

SO NOW I HAVE ALL THREE PARTS TO BE ABLE TO SAY THAT I HAVE TRIANGLES CONGRUENT.

AND IF YOU LOOK BACK AT OUR PICTURE, YOU CAN SEE THAT WE KNOW THE ANGLE BETWEEN THE TWO SIDES.

SO THIS IS GOING TO BE SIDE, ANGLE SIDE.

SO MY STATEMENT IS THAT THE TRIANGLES ARE CONGRUENT.

AND THEN THE REASON IS GOING TO BE SIDE ANGLE SIDE.

SO WE'RE ALMOST THERE.

THERE'S JUST ONE MORE STEP AFTER SAYING MY TRIANGLES ARE CONGRUENT, WE HAVE TO MAKE SURE WE REMEMBER THAT WHAT THEY ASKED FOR WAS C, E, CONGRUENT TO AF, CPCTC.

SO THE LAST STEP OF MY PROOF IS GOING TO BE SAYING EXACTLY THAT.

THAT'S WRAPS UP OUR GEOMETRY REGENT'S EXAM.

JOINING US TODAY WAS A GOOD FIRST STEP TO GET READY FOR THE EXAM BUT DON'T STOP HERE.

CHECK OUT ADDITIONAL RESOURCES ONLINE AND PUT IN THE WORK TO FEEL PREPARED FOR YOUR TEST.

THANKS FOR JOINING US TODAY AND GOOD LUCK ON YOUR GEOMETRY REGENT'S EXAM.

IF YOU'RE LOOKING FOR ADDITIONAL REGENTS EXAM INFO, THEN JUMP ON TO YOUR COMPUTER AND LOG ON TO REGENTSREVIEW.

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