MATH DESTROYER Triangle inside ellipse problem

[SN1]

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Practice test #2
question #6.

An ellipse is circumscribed about a triangle, whose base of length 8 is the line connecting the foci of the ellipse. If the height of the triangle is 3, find the area of the shaded region.

picture


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I know that I need to find the area of the ellipse and then subtract that by area of the triangle to find the area of the shaded region.

Area of triangle = 1/2 x 8 x 3 = 12

I know that the height of the triangle, which is 3 is also the semi-minor axis of the ellipse. My question is that how in the hell are we supposed to find the semi-major axis? The solution says to use phytagorean theorem to find the major semi-major axis. But isn't semi-major axis the horizontal distance from the center of the ellipse to one of the ends of the ellipse?

If I tried to use the phythagorean theorem, am I not finding the hypotenuse of the triangle? how is the hypotenuse of the triangle the same as the semi-major axis?

The solution use the P-theorem (a=hypotenuse, b= height of triangle, c = 1/2 base of triangle)
a^2 = b^2 +c^2
a^2= 16 + 9
a= 5
how is this "a" value the semi-major axis? isn't this the hypotenuse of the triangle?

to find the area of ellipse:
Pi * semi major * semi minor

I labeled the diagram. Please let me know if I am wrong


Uploaded with ImageShack.us




I also attached a clean picture in the beginning in case you would like to label something.


thanks.

 

[wired202808]

Practice test #2
question #6.

An ellipse is circumscribed about a triangle, whose base of length 8 is the line connecting the foci of the ellipse. If the height of the triangle is 3, find the area of the shaded region.

picture


Uploaded with ImageShack.us

I know that I need to find the area of the ellipse and then subtract that by area of the triangle to find the area of the shaded region.

Area of triangle = 1/2 x 8 x 3 = 12

I know that the height of the triangle, which is 3 is also the semi-minor axis of the ellipse. My question is that how in the hell are we supposed to find the semi-major axis? The solution says to use phytagorean theorem to find the major semi-major axis. But isn't semi-major axis the horizontal distance from the center of the ellipse to one of the ends of the ellipse?

If I tried to use the phythagorean theorem, am I not finding the hypotenuse of the triangle? how is the hypotenuse of the triangle the same as the semi-major axis?

The solution use the P-theorem (a=hypotenuse, b= height of triangle, c = 1/2 base of triangle)
a^2 = b^2 +c^2
a^2= 16 + 9
a= 5
how is this "a" value the semi-major axis? isn't this the hypotenuse of the triangle?

to find the area of ellipse:
Pi * semi major * semi minor

I labeled the diagram. Please let me know if I am wrong


Uploaded with ImageShack.us




I also attached a clean picture in the beginning in case you would like to label something.


thanks.

hey so while I cannot provide an answer to this, I can tell you that I have never ever heard of anyone having an ellipse problem on the real exam. So in all honesty I would not worry about it. I took the DAT and I cannot confirm or deny what I had on my exam, but I know from hearing other peoples stories, there have never been 1 ellipse problem that anyone ever mentioned to be on the real DAT.

Hopefully someone can explain the problem to you, so you feel comfy, but I wouldnt spend hours on it.

 

[Demps]

Where major axis and minor axis meet is the dead center. In your picture where you wrote 8 in pencil on the first picture.

From there the height is given as 3.

And I think you realize that from the center to where triangle ends is 4.

Eclipse area is given as pi*a*b

b = 3

a = 4 + x

a^2 = 3^2 + 4^2

Hence a = 5

5x3xpi - triangle area which is 12

so choice A is your answer

 

[Demps]

I think you understand it, but overanalyzing it.

Look at the "Important Formulas" section in the beginning

If you look at the picture on page 10. I think you can clarify yourself

 

[SN1]

Where major axis and minor axis meet is the dead center. In your picture where you wrote 8 in pencil on the first picture.

From there the height is given as 3.

And I think you realize that from the center to where triangle ends is 4.

Eclipse area is given as pi*a*b

b = 3

a = 4 + x

a^2 = 3^2 + 4^2

Hence a = 5

5x3xpi - triangle area which is 12

so choice A is your answer

I looked at the formula page but still don't see the relationship.
How in the world is c^2= a^2-b^2?

Isn't the pythaghorean suppose to involve a hypotenuse? based on the formula in the book neither a, b, or c is a hypotenuse.

 

[Demps]

I looked at the formula page but still don't see the relationship.
How in the world is c^2= a^2-b^2?

Isn't the pythaghorean suppose to involve a hypotenuse? based on the formula in the book neither a, b, or c is a hypotenuse.

What you just wrote is the same thing

if you rearrange the equation

A^2 (The longest segment extending middle to X-axis)
= b^2 + c^2

I think it is just easier to think that the longest length^2 = sum of the two shorter segment squared.

Or I would just memorize 🙂

 

[Demps]

I never noticed but,

I guess the long segment on ellipse is equal to the hypotnuse of the triangle formed

 

[SN1]

I never noticed but,

I guess the long segment on ellipse is equal to the hypotnuse of the triangle formed

Exactly. That's what the formula is basically saying. But I just don't see it lol.

 

[wired202808]

Exactly. That's what the formula is basically saying. But I just don't see it lol.

man you should focus on rate problems instead. your more likely to ever see that vs any ellipse problem lol id make a bet with you for $100 that you will never see one, use the time you have wisely, unless you're taking it in January.

 

[DentistDMD]

Practice test #2
question #6.

An ellipse is circumscribed about a triangle, whose base of length 8 is the line connecting the foci of the ellipse. If the height of the triangle is 3, find the area of the shaded region.

picture


Uploaded with ImageShack.us

I know that I need to find the area of the ellipse and then subtract that by area of the triangle to find the area of the shaded region.

Area of triangle = 1/2 x 8 x 3 = 12

I know that the height of the triangle, which is 3 is also the semi-minor axis of the ellipse. My question is that how in the hell are we supposed to find the semi-major axis? The solution says to use phytagorean theorem to find the major semi-major axis. But isn't semi-major axis the horizontal distance from the center of the ellipse to one of the ends of the ellipse?

If I tried to use the phythagorean theorem, am I not finding the hypotenuse of the triangle? how is the hypotenuse of the triangle the same as the semi-major axis?

The solution use the P-theorem (a=hypotenuse, b= height of triangle, c = 1/2 base of triangle)
a^2 = b^2 +c^2
a^2= 16 + 9
a= 5
how is this "a" value the semi-major axis? isn't this the hypotenuse of the triangle?

to find the area of ellipse:
Pi * semi major * semi minor

I labeled the diagram. Please let me know if I am wrong


Uploaded with ImageShack.us




I also attached a clean picture in the beginning in case you would like to label something.


thanks.

Since it's not clear whether the top vertex lies exactly on the minor axis, I would just assume that it does. In such case, the triangle is an isosceles one, meaning the two sides are equal. Let left side of the triangle be d1 and the other side d2.

Now from here you can use the property of an ellipse d1 + d2 = 2a, where a is the length of a semi major axis. Since the triangle is isosceles, d1 = d2 and consequently d1=d2=a.

Hope this helps.

Note d1 represents the distance from one focus to the point that lie on the ellipse called the vertex, and d2 represents the distance from the other focus to the vertex.

 

[SN1]

man you should focus on rate problems instead. your more likely to ever see that vs any ellipse problem lol id make a bet with you for $100 that you will never see one, use the time you have wisely, unless you're taking it in January.

I'm taking the dat 1 year from now lol.