Portion of Altitude of a Triangle Inscribed in a Circle

In summary, the length of the base of the red triangle is the same as the length of the base of the other two triangles.
  • #1
zak100
462
11

Homework Statement


In the figure Q image2.jpeg (attached), equilateral triangle ABC is inscribed in circle O, whose radius is 4. Altitude BD is extended until it intersects the circle at E. What is the length of DE?

Solution figure is attached. They formed a right angled triangle & calling it 30-60-90. I know the 90 degree angle but how to find the vertices with 30 degrees & 60 degrees angle?

Homework Equations



Pythagorous thorem:
sqr(a) + sqr(b) = sqr(c)
Sides of a 30-60-90 tirangle are : x, x * sqrt(3), 2 * x (hypotenuse)

The Attempt at a Solution


The solution image is attached. They connected A with O. It then forms a triangle AOD. AO is 4 because its radius (no problem). Therefore 2 * x = 4, so x = 2, so length of smaller side is 2. But which side is smaller. For this we must know the angles associated with the vertices A & O.

Some body please tell me the angle formed by the vertices A & O, how should we find this?

Zulfi.

Zulfi.
Sol image.jpg
.
 

Attachments

  • upload_2016-7-14_22-46-32.png
    upload_2016-7-14_22-46-32.png
    45 KB · Views: 620
Physics news on Phys.org
  • #2
zak100 said:

Homework Statement


In the figure Q image2.jpeg (attached), equilateral triangle ABC is inscribed in circle O, whose radius is 4. Altitude BD is extended until it intersects the circle at E. What is the length of DE?

Solution figure is attached. They formed a right angled triangle & calling it 30-60-90. I know the 90 degree angle but how to find the vertices with 30 degrees & 60 degrees angle?

Homework Equations



Pythagorous thorem:
sqr(a) + sqr(b) = sqr(c)
Sides of a 30-60-90 tirangle are : x, x * sqrt(3), 2 * x (hypotenuse)

The Attempt at a Solution


The solution image is attached. They connected A with O. It then forms a triangle AOD. AO is 4 because its radius (no problem). Therefore 2 * x = 4, so x = 2, so length of smaller side is 2. But which side is smaller. For this we must know the angles associated with the vertices A & O.

Some body please tell me the angle formed by the vertices A & O, how should we find this?

Zulfi.

Zulfi.View attachment 103217 .
If the hypotenuse of the 30-60-90 triangle is 4, and the altitude is 2, what length is the base? Apply the Pythagorean Theorem here.

You know that the center of the circle is at O. What must be the total length of the line BOE which passes thru O? If you know the length of BOE, then you should be able to find the length DE
 
  • #3
Hi,
No altitude 2 is not given. Sorry I have uploaded the solution part of the figure. The figure related to question is:
Q image2.jpg

< If you know the length of BOE,>
According to the question i don't know the length of DE. Actually, i can find the lengths of other 2 sides of ADB if i know the length of hypotenuse using the formulas. But i don't know whether A is 30 or B is 30. This is what i want to know. How can we know which angle is 30 & which angle is 60. For this we must know which side is larger. The angle opposite to the larger side would be 60. We can't believe on the drawing. So without considering the drawing how can we find out which side is larger?

Zulfi.
 
  • #4
zak100 said:
Hi,
No altitude 2 is not given. Sorry I have uploaded the solution part of the figure. The figure related to question is:View attachment 103317
< If you know the length of BOE,>
According to the question i don't know the length of DE. Actually, i can find the lengths of other 2 sides of ADB if i know the length of hypotenuse using the formulas. But i don't know whether A is 30 or B is 30. This is what i want to know. How can we know which angle is 30 & which angle is 60. For this we must know which side is larger. The angle opposite to the larger side would be 60. We can't believe on the drawing. So without considering the drawing how can we find out which side is larger?

Zulfi.

Draw lines AO and OC. This gives you three congruent isosceles triangles AOB, BOC, and COA since they have equal corresponding sides. So their central angles are equal and add to 360 degrees. So each central angle is 120 degrees. Since the triangles are isosceles their smaller angles must all be 30 degrees. Does that help you?
 
  • #5
Hi,
My friend thanks for your guidance. This won't help me at all. Because I have to find DE. If i can find OD, i can subtract BE from BD & this would give me OD. None of these triangles contain the sides DE which is required in the question.

Zulfi.
 
  • #6
upload_2016-7-17_20-6-23.png


The red triangle is equilateral. All the shaded triangles are congruent. Therefore the central angles are 120°. They are isosceles, with sides of length r. The line OE halves the triangle ACO, so OD is perpendicular to AC. What is the length of OD then? OE is one radius of the circle. What is DE?
 
  • #7
Hi,
Thanks for your work. It looks that we now have 30-60-90 triangle. But I have a feeling that 30 degree is not correct. Because its an exterior angle of the triangle and exterior angle is the sum of two opposite interior angles.
Please guide me.

Zulfi.
 
  • #8
zak100 said:
Hi,
Thanks for your work. It looks that we now have 30-60-90 triangle. But I have a feeling that 30 degree is not correct. Because its an exterior angle of the triangle and exterior angle is the sum of two opposite interior angles.
Please guide me.

Zulfi.
Do you see any triangle with exterior angle of the 30° shown?
http://www.mathopenref.com/triangleextangle.html
 
  • #9
Hi,
Thanks for sending me the link. I was wrong. I now understand what is meant by an exterior angle. I also understand why the angle is 120. The angle around O would be 360 & 360 degree divided by 3 is equal to 120 degrees. Line OD halves the triangle so angle AOD in my diagram would be 60 degrees. So larger side would be AD= x * sqrt(3) & x =2 & the length of smaller side OD = x =2 & DE would be 2.

Thanks.
exterior angle.jpg

Sorry for extra white space in the image. When I load image i get this problem.
Zulfi.
 
  • #10
zak100 said:
Hi,
Thanks for sending me the link. I was wrong. I now understand what is meant by an exterior angle. I also understand why the angle is 120. The angle around O would be 360 & 360 degree divided by 3 is equal to 120 degrees. Line OD halves the triangle so angle AOD in my diagram would be 60 degrees. So larger side would be AD= x * sqrt(3) & x =2 & the length of smaller side OD = x =2 & DE would be 2.

Thanks.
View attachment 103422
Sorry for extra white space in the image. When I load image i get this problem.
Zulfi.
Good work!
From the image, select the relevant part. Copy it and paste into your post.
upload_2016-7-18_22-20-58.png
 

1. What is the formula for finding the portion of altitude of a triangle inscribed in a circle?

The formula for finding the portion of altitude of a triangle inscribed in a circle is h = 2r, where h is the altitude of the triangle and r is the radius of the circle.

2. How is the portion of altitude of a triangle inscribed in a circle related to the radius of the circle?

The portion of altitude of a triangle inscribed in a circle is directly proportional to the radius of the circle. This means that as the radius increases, the portion of altitude also increases.

3. Can the portion of altitude of a triangle inscribed in a circle be greater than the radius of the circle?

No, the portion of altitude of a triangle inscribed in a circle cannot be greater than the radius of the circle. This is because the altitude of a triangle cannot be greater than its base, and in this case, the base is the diameter of the circle, which is equal to twice the radius.

4. How does the angle of the inscribed triangle affect the portion of altitude?

The angle of the inscribed triangle does not affect the portion of altitude. As long as the triangle is inscribed in the circle, the portion of altitude will always be equal to 2 times the radius of the circle.

5. Is the portion of altitude of a triangle inscribed in a circle always equal to 2 times the radius of the circle?

Yes, the portion of altitude of a triangle inscribed in a circle is always equal to 2 times the radius of the circle. This is a constant relationship that holds true for all triangles inscribed in a circle.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
14
Views
557
  • Precalculus Mathematics Homework Help
Replies
3
Views
850
  • Precalculus Mathematics Homework Help
Replies
8
Views
3K
  • Precalculus Mathematics Homework Help
Replies
16
Views
2K
  • Precalculus Mathematics Homework Help
Replies
12
Views
2K
  • Precalculus Mathematics Homework Help
Replies
6
Views
2K
Replies
1
Views
899
  • Precalculus Mathematics Homework Help
Replies
2
Views
1K
  • Precalculus Mathematics Homework Help
Replies
10
Views
22K
  • Precalculus Mathematics Homework Help
Replies
5
Views
2K
Back
Top