Area question involving trapezoid and parallelogram

[Sophia_123]

Hello all!

Thanks in advance for your help. The original question is as follows:



My first question is about the part that says, "parallelogram with side lengths of a and b." I always thought the side lengths (the sides opposite each other) would be equal in a parallelogram. I am confused about what I am missing here.

As for the trapezoid, I read in someone else's reply that if I draw a line from Y in the upper right corner to point P on the base, I will get a parallelogram on the left. Here is a figure of that:


When I do that, it seems to me that, between the two original figures, I now have two equal parallelograms plus a triangle. Does that mean the trapezoid's area is greater by the area of the triangle? In other words, I am thinking my final answer is just 1/2 base x height or 1/2 ab (choice C)?

Thanks again for your help.

Sophia.

 

[blamocur]

When I do that, it seems to me that, between the two original figures, I now have two equal parallelograms plus a triangle.

Two parallelograms? Where?

Does that mean the trapezoid's area is greater by the area of the triangle?

Yes.

In other words, I am thinking my final answer is just 1/2 base x height or 1/2 ab (choice C)?

What is the base of the triangle? What is the height?

 

[The Highlander]

Thanks in advance for your help. The original question is as follows:

​

My first question is about the part that says, "parallelogram with side lengths of a and b." I always thought the side lengths (the sides opposite each other) would be equal in a parallelogram. I am confused about what I am missing here.?

I too am confused about what you are missing.

Perhaps if you had made a properly labelled diagram, your confusion would not have arisen?

If you draw a line from Y to P such that the measure of ∠YPW is 135°, then |WX| = |PY| = |a| and |XY| = |WP| = |b|, so the opposite sides are equal in length in the newly created parallelogram! See diagram below...

​

Note too that the measure of ∠XYP is 45° (equal & opposite to ∠XWP in the parallelogram).

When I do that, it seems to me that, between the two original figures, I now have two equal parallelograms plus a triangle. Does that mean the trapezoid's area is greater by the area of the triangle?

I have no idea how you see two parallelograms but the area of the trapezium is, clearly, greater than that of the (single) parallelogram by an area equal to that of the triangle formed at the the RHS of the parallelogram.

In other words, I am thinking my final answer is just 1/2 base x height or 1/2 ab (choice C)?

No, your "final answer" is wrong!
Go through the following steps carefully and answer each question before moving on to the next step.

As mentioned above, the measure of ∠XYP is 45°, so what is the measure of ∠PYZ?

Therefore, ΔPYZ is a right-angled, isosceles triangle, isn't it?

If we drop a perpendicular from Y to Q, it will bisect ∠PYZ and create two right angles at Q (as shown below).

​

So what is the measure of ∠PYQ? (And ∠QYZ?)

Therefore, ΔPYQ is also a right-angled, isosceles triangle, isn't it?

I have labelled the height of ΔPYZ as h but, given the geometry here, note which other triangle sides are also h.

You are correct that the area of ΔPYZ will be: "1/2 base x height" but what are the base and the height (in terms of h)?

So what is the area of ΔPYZ in terms of h?

However, none of your "choices" involve h, do they?

But, sin 45° = \(\displaystyle \frac{\color{blue}h}{a}\), doesn't it?

And what is the exact value of sin 45°?

So what does h equal (in terms of a)?

And what can you now say the area of ΔPYZ is?

And, therefore, what is the correct "choice" from the options provided?

Please now come back and show us your (improved) diagram and working (laid out as above) with the correct answers at each stage, finishing off with what you now believe to be the correct choice. Thank you.

Hope that helps.

 

[Sophia_123]

Two parallelograms? Where?

Yes.

What is the base of the triangle? What is the height?

Hello!

The question mentions one parallelogram (the one that is not drawn). That's the first parallelogram. Once I draw the extra line in the trapezoid, I've divided it into a triangle and a parallelogram. That parallelogram is the second one.

"What is the base of the triangle? What is the height?" I see my mistake there. I need to grab the base of the triangle and the height of the triangle.

Thank you!

 

[Sophia_123]

I too am confused about what you are missing.

Perhaps if you had made a properly labelled diagram, your confusion would not have arisen?

If you draw a line from Y to P such that the measure of ∠YPW is 135°, then |WX| = |PY| = |a| and |XY| = |WP| = |b|, so the opposite sides are equal in length in the newly created parallelogram! See diagram below...

View attachment 37857​

Note too that the measure of ∠XYP is 45° (equal & opposite to ∠XWP in the parallelogram).


I have no idea how you see two parallelograms but the area of the trapezium is, clearly, greater than that of the (single) parallelogram by an area equal to that of the triangle formed at the the RHS of the parallelogram.


No, your "final answer" is wrong!
Go through the following steps carefully and answer each question before moving on to the next step.

As mentioned above, the measure of ∠XYP is 45°, so what is the measure of ∠PYZ?

Therefore, ΔPYZ is a right-angled, isosceles triangle, isn't it?

If we drop a perpendicular from Y to Q, it will bisect ∠PYZ and create two right angles at Q (as shown below).

View attachment 37858​

So what is the measure of ∠PYQ? (And ∠QYZ?)

Therefore, ΔPYQ is also a right-angled, isosceles triangle, isn't it?

I have labelled the height of ΔPYZ as h but, given the geometry here, note which other triangle sides are also h.

You are correct that the area of ΔPYZ will be: "1/2 base x height" but what are the base and the height (in terms of h)?

So what is the area of ΔPYZ in terms of h?

However, none of your "choices" involve h, do they?

But, sin 45° = \(\displaystyle \frac{\color{blue}h}{a}\), doesn't it?

And what is the exact value of sin 45°?

So what does h equal (in terms of a)?

And what can you now say the area of ΔPYZ is?

And, therefore, what is the correct "choice" from the options provided?

Please now come back and show us your (improved) diagram and working (laid out as above) with the correct answers at each stage, finishing off with what you now believe to be the correct choice. Thank you.

Hope that helps.

Thank you for your notes. I will look them over.

 

[Sophia_123]

Thank ou for your notes. I will look them over.

I too am confused about what you are missing.

Perhaps if you had made a properly labelled diagram, your confusion would not have arisen?

If you draw a line from Y to P such that the measure of ∠YPW is 135°, then |WX| = |PY| = |a| and |XY| = |WP| = |b|, so the opposite sides are equal in length in the newly created parallelogram! See diagram below...

View attachment 37857​

Note too that the measure of ∠XYP is 45° (equal & opposite to ∠XWP in the parallelogram).


I have no idea how you see two parallelograms but the area of the trapezium is, clearly, greater than that of the (single) parallelogram by an area equal to that of the triangle formed at the the RHS of the parallelogram.


No, your "final answer" is wrong!
Go through the following steps carefully and answer each question before moving on to the next step.

As mentioned above, the measure of ∠XYP is 45°, so what is the measure of ∠PYZ?

Therefore, ΔPYZ is a right-angled, isosceles triangle, isn't it?

If we drop a perpendicular from Y to Q, it will bisect ∠PYZ and create two right angles at Q (as shown below).

View attachment 37858​

So what is the measure of ∠PYQ? (And ∠QYZ?)

Therefore, ΔPYQ is also a right-angled, isosceles triangle, isn't it?

I have labelled the height of ΔPYZ as h but, given the geometry here, note which other triangle sides are also h.

You are correct that the area of ΔPYZ will be: "1/2 base x height" but what are the base and the height (in terms of h)?

So what is the area of ΔPYZ in terms of h?

However, none of your "choices" involve h, do they?

But, sin 45° = \(\displaystyle \frac{\color{blue}h}{a}\), doesn't it?

And what is the exact value of sin 45°?

So what does h equal (in terms of a)?

And what can you now say the area of ΔPYZ is?

And, therefore, what is the correct "choice" from the options provided?

Please now come back and show us your (improved) diagram and working (laid out as above) with the correct answers at each stage, finishing off with what you now believe to be the correct choice. Thank you.

Hope that helps.

Hello Teacher Highlander!

Here are my thoughts now.

I get the two parallelograms this way: the question mentions a parallelogram with base angles of 45 and 135 and sides of a and b. Although there's no figure given for it (I drew it on my own scratch paper), that's the first parallelogram.

The second parallelogram is the one I created when I drew a line in the given trapezoid. I created a triangle on the right and a parallelogram on the left. The 2nd (new) parallelogram has the same angle measurements as the one described in the question.

In the parallelogram embedded in the trapezoid, <WXY is 135 degrees, which means < WPY is 135 degrees. < XWZ is still 45 degrees, and < XYP is 45 degrees. (In figure below.) Because < XYP is 45 degrees, the triangle <PYZ has an angle of 90 degrees (135 - 45 = 90). < YZP is 45 degrees, and so < YPZ also has to be 45 degrees since there are only 180 degrees in a triangle.



To find the difference in size between the trapezoid and the parallelogram, I need the area of the triangle. I see the silly mistake I made before as b is not the base of the triangle.

Since the opposite sides of a parallelogram are equal, line YP must also be a. Since the angle in the triangle across from YP (length of a) is 45 degrees, and the angle across from YZ is also 45 degrees, that makes YZ the same length as YP.

By dropping an altitude from < PYZ to the base, I get this figure below:

I see I have two 45-45-90 triangles because the height cuts the 90-degree angle in two. The ratio of sides of a 45-45-90 triangle is x:x:xrt2. Since the hypotenuse is length a, the base and height of each triangle is a/rt2:


Your explanation included finding information using sine. I understand that sin 45 = opp/hyp or h/a, so h = a/rt2, so sin 45 = a/rt2 divided by a. Simplifying, I get

a 1 1
--- x ---- = ----
rt2 a rt2

Given my gaps in trigonometry, I am not clear on how finding the sine is necessary to move this problem forward. Is there a reason you suggest using sine rather than using the 45-45-90 ratios directly?

In all events, I see I have two 45-45-90 triangles. The hypotenuse of each is a, and given the ratios, the bases and heights of each is a/rt2.

To get the area of one triangle, I use the 1/2 base x height formula, and I get:

1 a a 1 a^2
-- x --- x --- = -- x ----
2 rt2 rt2 2 2

Since I have two triangles, I have to double that result to get the area of the whole triangle. The answer is a^2 / 2 or choice A.


Maybe I am missing something here, but could I also forget about dropping the altitude and using special right triangles if I notice early on that I have a right triangle with equal sides?



YP and YZ can be the base and height, and I can just plug those a's into the 1/2 base x height formula.

1 a^2
-- a x a = --------
2 2

Thanks again for your help. I very much appreciate the time you took on this.

Regards,

Sophia.

 

[The Highlander]

Hello Teacher Highlander!

Here are my thoughts now.

Hi @Sophia_123

I am very pleased to see that you took the time to put all your thoughts down in writing (and added some nice drawings too; they're not quite 100% accurate but are fine for this exercise. A sketch doesn't need to be geometrically perfect as long as it is labelled properly to show what it is meant to represent. )

Thank you for taking the trouble to do all that. It is very rewarding when someone takes the time to respond to the advice one has offered (even if you didn't follow exactly what I suggested, lol).

I noticed that you mention you have "gaps in Trigonometry" and it looks like my suggestion that the Sine Ratio should be used here scared ya, huh? (Did you then draw a complete blank when I talked about an exact value for it too?)

I believe this question was very likely intended to be solved that way, though there's nothing wrong with your approach and you have certainly arrived at the right answer (but there is still a wee bit of wooly thinking going on, so I'll add some comments to your work).

I get the two parallelograms this way: the question mentions a parallelogram with base angles of 45 and 135 and sides of a and b. Although there's no figure given for it (I drew it on my own scratch paper), that's the first parallelogram.
The second parallelogram is the one I created when I drew a line in the given trapezoid. I created a triangle on the right and a parallelogram on the left. The 2nd (new) parallelogram has the same angle measurements as the one described in the question.

There is only one parallelogram! The parallelogram "mentioned" in the question is the very same one you create when you draw a line from Y to P! There is no second parallelogram; the question is just prompting you to create a parallelogram inside the trapezium by drawing a line from Y down to the trapezium's base such that the new vertex created (ie: ∠WPY) has a measure of 135°. And by creating that parallelogram inside the trapezium, you then show that the trapezium's area is greater that that of the parallelogram's by that of the triangle now showing at the RHS. Basically, by drawing the line YP, you have split the trapezium into a parallelogram and a triangle.

In the parallelogram embedded in the trapezoid, <WXY is 135 degrees, which means < WPY is 135 degrees.

No! That is not correct. The ∠WPY is 135° because you made it 135° (when you constructed the line YP) not because ∠WXY is 135° as you have stated. The ∠WPY is the same measure as ∠WXY because you have made it that way so that you would create a parallelogram (with opposite angles and opposite sides in equal measures). You could have made ∠WPY any (other) size you liked but then the quadrilateral created (WXYP) would no longer be a parallelogram! Do you see where the cause and effect lie here?

The remainder of your geometry is fairly sound and gets you right to the point of the exercise (which is finding the area of the triangle).

Given my gaps in trigonometry, I am not clear on how finding the sine is necessary to move this problem forward. Is there a reason you suggest using sine rather than using the 45-45-90 ratios directly?

No, that's just the way I 'saw' it. As a (very old) Maths teacher, I'm just a bit 'beyond' using "special right triangles" (except as a last resort, lol) and I immediately jump to using Trigonometry as my weapon of choice (it's a big gun you see) but there's absolutely nothing wrong with your method. Is that all you've been taught so far? Have you not learned much (or anything) about Trigonometry yet?

Your "special right triangles" and the fact that the ratios of the sides in them remain constant (regardless of the size of the triangles) is, in fact, the very basis of Trigonometry. If you are going to go further in Maths then you will need to learn more about it and (even if you are only just starting in that topic) I think it might be worth it for you to get a copy of a worksheet I give to my pupils that may come in very useful for you in the future (see at the end of the post).

But, in the meantime, to get back to this problem...

In all events, I see I have two 45-45-90 triangles. The hypotenuse of each is a, and given the ratios, the bases and heights of each is a/rt2.
To get the area of one triangle, I use the 1/2 base x height formula, and I get:
\(\displaystyle \frac{1}{2}\times \frac{a}{\sqrt{2}}\times \frac{a}{\sqrt{2}}=\frac{1}{2}\times \frac{a^2}{2}\color{red}=\frac{a^2}{4}\text{ (so your next sentence makes more sense.)}\)
Since I have two triangles, I have to double that result to get the area of the whole triangle. The answer is a^2 / 2 or choice A.

And now you have arrived at the correct answer!

Maybe I am missing something here, but could I also forget about dropping the altitude and using special right triangles if I notice early on that I have a right triangle with equal sides?

Yes, you could indeed; well spotted!

YP and YZ can be the base and height, and I can just plug those a's into the 1/2 base x height formula.

1 a^2
-- a x a = ------
2 2

That works just as well.

Thanks again for your help. I very much appreciate the time you took on this.

Regards,

Sophia.

No problem, you are very welcome.

I promised you a little extra at the end here and it may explain to you why I opted for Trigonometry to solve the problem (I will add my solution at the end).

If you have learned anything about Trigonometry then you should know that it is based on the fact that there is a number (a ratio, in fact) that can be found for any angle and that number (ratio) is specific to that angle.
We can look those numbers up in tables (how I had to do it when I was learning about Trig) or we can get them (more easily) using a scientific calculator. Unfortunately, the modern students miss out on that use of tables (the use of which reinforced the true relationship between angles and their Trig ratios) as just getting them at the press of a button makes it too easy to forget what you are actually doing in their use.

However, regardless of how you obtain a Trig Ratio (whether from tables or electronically), the number you get is almost always just an approximation. In fact, some (indeed many) Trig Ratios cannot be expressed as an accurate number; many are irrational numbers (if that means anything to you).

But there are some Trig Ratios that we can express as exact values and these are very useful so they turn up time and time again both in nature and (especially) in Maths questions!

They are actually based on your very own "special right triangles" and I have produced a worksheet that I hand out to my pupils for them to complete; I will however, give you the sheet with the answers already included. You might wish to download it and store away for safekeeping until the subject comes up?

Here is the Worksheet...      (Have a special look at the Sine (& Cosine) of 45°)

​

Now that you have seen that I can show you the way I attacked the problem...

(But I will do it in a fresh post as this one is becoming so long the forum won't allow anything more)

 

[The Highlander]

Given all the Geometry we've already been through to arrive at my figure below...

​

I then (would have) said...

I have labelled the height of ΔPYZ as h but, given the geometry here, note that there are two other triangle sides which are also h. (Because both ΔPYQ and ΔQYZ are right-angled, isosceles (45-45-90 ) triangles,)

The area of ΔPYZ will be: "1/2 base x height" but the base is h + h = 2h and the height is just h (as shown in the above diagram).

So the area of ΔPYZ is ½ × base × height = ½ × 2h × h = h × h = h²

But, sin 45° = \(\displaystyle \frac{\color{blue}h}{a}\)

And the exact value of sin 45° is: \(\displaystyle \frac{1}{\sqrt{2}}\)

Therefore: \(\displaystyle \frac{\color{blue}h}{a}=\frac{1}{\sqrt{2}}\implies{\color{blue}h}=\frac{a}{\sqrt{2}}\)

And so, the area of ΔPYZ is \(\displaystyle {\color{blue}h}^2=\left(\frac{a}{\sqrt{2}}\right)^2=\frac{a^2}{2}=\frac{1}{2}a^2\)

So Option A) is the correct choice!

PS: Below is another little aide memoire I give my pupils when they are first learning about Trigonometry.
You might want to hold onto a copy of that for future reference too?

Best of luck with your studies.

Take care.

Basic Right-Angled Triangle Trigonometry Aide Memoire...

​

 

[Sophia_123]

Hi @Sophia_123

I am very pleased to see that you took the time to put all your thoughts down in writing (and added some nice drawings too; they're not quite 100% accurate but are fine for this exercise. A sketch doesn't need to be geometrically perfect as long as it is labelled properly to show what it is meant to represent. )

Thank you for taking the trouble to do all that. It is very rewarding when someone takes the time to respond to the advice one has offered (even if you didn't follow exactly what I suggested, lol).

I noticed that you mention you have "gaps in Trigonometry" and it looks like my suggestion that the Sine Ratio should be used here scared ya, huh? (Did you then draw a complete blank when I talked about an exact value for it too?)

I believe this question was very likely intended to be solved that way, though there's nothing wrong with your approach and you have certainly arrived at the right answer (but there is still a wee bit of wooly thinking going on, so I'll add some comments to your work).


There is only one parallelogram! The parallelogram "mentioned" in the question is the very same one you create when you draw a line from Y to P! There is no second parallelogram; the question is just prompting you to create a parallelogram inside the trapezium by drawing a line from Y down to the trapezium's base such that the new vertex created (ie: ∠WPY) has a measure of 135°. And by creating that parallelogram inside the trapezium, you then show that the trapezium's area is greater that that of the parallelogram's by that of the triangle now showing at the RHS. Basically, by drawing the line YP, you have split the trapezium into a parallelogram and a triangle.


No! That is not correct. The ∠WPY is 135° because you made it 135° (when you constructed the line YP) not because ∠WXY is 135° as you have stated. The ∠WPY is the same measure as ∠WXY because you have made it that way so that you would create a parallelogram (with opposite angles and opposite sides in equal measures). You could have made ∠WPY any (other) size you liked but then the quadrilateral created (WXYP) would no longer be a parallelogram! Do you see where the cause and effect lie here?

The remainder of your geometry is fairly sound and gets you right to the point of the exercise (which is finding the area of the triangle).

No, that's just the way I 'saw' it. As a (very old) Maths teacher, I'm just a bit 'beyond' using "special right triangles" (except as a last resort, lol) and I immediately jump to using Trigonometry as my weapon of choice (it's a big gun you see) but there's absolutely nothing wrong with your method. Is that all you've been taught so far? Have you not learned much (or anything) about Trigonometry yet?

Your "special right triangles" and the fact that the ratios of the sides in them remain constant (regardless of the size of the triangles) is, in fact, the very basis of Trigonometry. If you are going to go further in Maths then you will need to learn more about it and (even if you are only just starting in that topic) I think it might be worth it for you to get a copy of a worksheet I give to my pupils that may come in very useful for you in the future (see at the end of the post).

But, in the meantime, to get back to this problem...

And now you have arrived at the correct answer!

Yes, you could indeed; well spotted!

That works just as well.

No problem, you are very welcome.

I promised you a little extra at the end here and it may explain to you why I opted for Trigonometry to solve the problem (I will add my solution at the end).

If you have learned anything about Trigonometry then you should know that it is based on the fact that there is a number (a ratio, in fact) that can be found for any angle and that number (ratio) is specific to that angle.
We can look those numbers up in tables (how I had to do it when I was learning about Trig) or we can get them (more easily) using a scientific calculator. Unfortunately, the modern students miss out on that use of tables (the use of which reinforced the true relationship between angles and their Trig ratios) as just getting them at the press of a button makes it too easy to forget what you are actually doing in their use.

However, regardless of how you obtain a Trig Ratio (whether from tables or electronically), the number you get is almost always just an approximation. In fact, some (indeed many) Trig Ratios cannot be expressed as an accurate number; many are irrational numbers (if that means anything to you).

But there are some Trig Ratios that we can express as exact values and these are very useful so they turn up time and time again both in nature and (especially) in Maths questions!

They are actually based on your very own "special right triangles" and I have produced a worksheet that I hand out to my pupils for them to complete; I will however, give you the sheet with the answers already included. You might wish to download it and store away for safekeeping until the subject comes up?

Here is the Worksheet...      (Have a special look at the Sine (& Cosine) of 45°)

View attachment 37871​

Now that you have seen that I can show you the way I attacked the problem...

(But I will do it in a fresh post as this one is becoming so long the forum won't allow anything more)

Thank you!

-Sophia.

 

[Sophia_123]

Given all the Geometry we've already been through to arrive at my figure below...

View attachment 37873​

I then (would have) said...

I have labelled the height of ΔPYZ as h but, given the geometry here, note that there are two other triangle sides which are also h. (Because both ΔPYQ and ΔQYZ are right-angled, isosceles (45-45-90 ) triangles,)

The area of ΔPYZ will be: "1/2 base x height" but the base is h + h = 2h and the height is just h (as shown in the above diagram).

So the area of ΔPYZ is ½ × base × height = ½ × 2h × h = h × h = h²

But, sin 45° = \(\displaystyle \frac{\color{blue}h}{a}\)

And the exact value of sin 45° is: \(\displaystyle \frac{1}{\sqrt{2}}\)

Therefore: \(\displaystyle \frac{\color{blue}h}{a}=\frac{1}{\sqrt{2}}\implies{\color{blue}h}=\frac{a}{\sqrt{2}}\)

And so, the area of ΔPYZ is \(\displaystyle {\color{blue}h}^2=\left(\frac{a}{\sqrt{2}}\right)^2=\frac{a^2}{2}=\frac{1}{2}a^2\)

So Option A) is the correct choice!

PS: Below is another little aide memoire I give my pupils when they are first learning about Trigonometry.
You might want to hold onto a copy of that for future reference too?

Best of luck with your studies.

Take care.

Basic Right-Angled Triangle Trigonometry Aide Memoire...

​

Thank you!

-Sophia.

 

[The Highlander]

Thank you!

-Sophia.

Thank you!

-Sophia.

You are very Welcome.