Area of isosceles triangle with rounded corners :S

[High-School-Kid]

Well i got a problem here.
I'm asked to find the area of an isosceles triangle with rounded corners, and I've got no clue how to do so... :?

The corners are like circles and they have a radius of 10 mm
The triangle is 120 mm long

Can anyone please help me?

 

[stapel]

Are the sides of the triangle 120 millimeters before rounding, or after?

Please reply with the full and exact text of the exercise, the complete instructions, and, if possible, a scan of the picture from which you're supposed to be working. When you reply, please include your work and reasoning so far, as this will help us "see" what methods you're supposed to be using (that they gave you in class) and where you're encountering difficulty.

Thank you!

 

[Denis]

Let's see where you're at:
what is the area of isosceles triangle with base=6 and equal sides=5 ?

> The triangle is 120 mm long
That makes no sense. Do you mean the perimeter is 120?

 

[TchrWill]

Well i got a problem here.
I'm asked to find the area of an isosceles triangle with rounded corners, and I've got no clue how to do so... :?
The corners are like circles and they have a radius of 10 mm
The triangle is 120 mm long
Can anyone please help me?

Not having a clear definition as to what "The triangle is 120 mm long" means, consider the geometry of each vertex:

Let A be a vertex.
Let O be the center of the corner radius r
Let B be the tangency points of the corner radius on the 2 adjacent sides.
Let µ be the vertex angle BAB
Draw line AO
We now have two triangle AOB's.
The area of the two triangles is Atn = (r)tan(µ/2)
The total area of the 6 triangles so formed at the 3 vertices is
......At = r(1/tan1 + 1/tan2 + 1/tan3)
The area of the partial circle from B to B is
......Ac = Pi(R^2)2(90 - µ)/360, all three of which sum to Pi(r^2).

The net change in the given isosceles triangle area is therefore
......A - r(1/tan1 + 1/tan2 + 1/tan3) + Pi(r^2).

If the 120mm is meant to be the given triangle's perimeter, you still need the length of the base, the length of one side or one of the three angles in order to determine the initial area of the isosceles triangle using Heron's area formula
......A = sqrt[s(s - a)(s - b)(s - c)] where a, b, and c are the three sides and s = the semi-perimeter = (a + b + c)/2.

If the 120mm is meant to be the given triangle's area (?), then you need only subtract the net change in area as defined above.

Might you be able to clarify what the 120mm represents?

 

[High-School-Kid]

Well it's 120 mm from the base to the top, i added a picture of it.
Thx for the help

 

[stapel]

Well it's 120 mm from the base to the top

Ah. So it's not the sides that have a length of 120 millimeters, but the height. That's quite a bit different. :shock:

Draw the triangle with the pointy corners, and draw a circle inside the peak angle. Draw a radius line from the center of the circle to where the circle meets the triangle. This radius line, by nature, is perpendicular to the side of the triangle.

Draw the altitude line of the triangle, thus splitting the triangle into two 30-60-90 triangles. Note that you now have a right triangle formed by (half of) the peak angle of the original triangle, the corner at the circle's center, and the point where the radius line meets the triangle. Use what you know about 30-60-90 triangles, along with the known value of the circle's radius, to find the length of the segment between the peak of the original angle and the center of the circle.

Since the total height of the rounded-triangle figure is 120, and since the radius of the circle is 10, then what is the height, above the base of the rounded-triangle figure, of the circle's center? Now that you have the length of the segment between the circle's center and the peak of the "regular" triangle, what is the height of the regular triangle?

Using what you know about 30-60-90 triangles, what then must be the length of the base of the regular triangle?

Now that you have the height and base of the regular triangle, find the total area.

Comparing the regular triangle with the rounded one, the area of the rounded triangle will be the area of the regular triangle, less the areas of the three corners (the triangles with bases parallel to the opposite side of the triangle and passing through their circles' centers), plus half the areas of the corners' circles.

See what you can do with this. If you get stuck, please reply showing how far you have gotten.

Thank you!

 

[TchrWill]

Well it's 120 mm from the base to the top, i added a picture of it.

There are an infinite number of these triangles of height 120mm. We need an angle or the base or a side to get the area of a specific triangle.

For example:
Assume an apex angle of 30º
Draw a triangle ABC, A, B and C being the centers of the corner radii.
The altitude of this triangle is 120 - 10 - 10 = 100mm.
The distances AC and AB are 100/cos15º = 103.527mm
The base is 103.527(sin15º)2 = 53.59mm
The area of the triangle is 53.59(100)/2 = 2679.5sq.mm
To this add the 10mm strip that wraps around this triangle
...2(103.527)10 + 53.59(10) + 3.14(10^2) = 2920.6sq.mm
Total area of 120mm high isosceles triangle with 30º apex angle and 10mm corner radii = 5600sq.mm

To work the problem as you define, we need one additional piece of information, such as the base, a side or an angle.

 

[mmm4444bot]

I also wonder about the given information in this exercise.

Are you sure that this is an isosceles triangle with rounded vertices versus an equilateral triangle with rounded vertices?

While Teacher Will was posting, I was drawing an image. As the base of an isosceles triangle increases in length (while the height remains constant), the angles at the base get smaller. This changes the area of the shaded region at each vertex. (See the image below for an example of what happens at each end of the base when the length of the base increases and the height remains constant. Note, too, that the change in shaded area at the top vertex will not be the same change as what occurs at each end of the base, further complicating this senario.)

As Teacher Will stated, there are infinite isosceles triangles with set height. I'll add that there is only one equilateral triangle with the same height.

Please double-check the given information. If you confirm that the object is an isosceles triangle with rounded vertices, and no other measurements are given, then we would need to choose a variable to represent the base (either before rounding the vertices or after), followed by expressing the area symbolically in terms of this variable.

In other words, as posted, there is no way to arrive at a numerical result.

Please confirm also that the height of this object is 120mm after rounding the vertices.

(Double-click the image, to expand.)

[attachment=0:1jae6332]image.JPG[/attachment:1jae6332]

 

[TchrWill]

Re:

As mmm4444bot states,

I also wonder about the given information in this exercise.
In other words, as posted, there is no way to arrive at a numerical result.
Please confirm also that the height of this object is 120mm after roundin
Are you sure that this is an isosceles triangle with rounded vertices versus an equilateral triangle with rounded vertices?

If the given dimension(s) was base to vertex, the first method I posted would have to be used.
If the given dimension was base to rounded vertex, the second method I posted would have to be used.
In either case, one piece of additional information would have to be provided, be it one of the angles, or one of the triangle sides.
The only conditions under which a unique solution is possible, given only the height, are an equilateral triangle or an isosceles right triangle.

 

[High-School-Kid]

the high is 120 from the base to the top after rounding. and it's the drawing i sent in
but i left out an important detail sorry the top is a 40 degree angel
this is the actural drawing from my homework don't mind the circle in the middle unless it's an important factor, whitch i don't think it is.
it's the gray area i need to find well i know how to find the area of the circle so my problem is the triangel
sorry for the mistake

 

[Denis]

Finally got it straight :shock:

Take an isosceles triangle height = 80, angles 20-80-80:
to see where you're at: what is the area of the triangle?

 

[TchrWill]

the high is 120 from the base to the top after rounding. and it's the drawing i sent in
but i left out an important detail sorry the top is a 40 degree angel
this is the actural drawing from my homework don't mind the circle in the middle unless it's an important factor, whitch i don't think it is.
it's the gray area i need to find well i know how to find the area of the circle so my problem is the triangel
sorry for the mistake.

At last. My post of April 25th is what you are looking for. I assumed an angle of 30º, when in fact, it is 40º. I'll let you work out the real solution.