Inscribed circle in a triangle with known sides

[Ahatmose]

Hi I am having a devil of a time figuring this out and every explanation doesn't seem to work.

It seems simple enough. we have a triangle with base of 439.823 units and slope sides of 356.035 units. What is the diameter or radius of the largest circle that perfectly fits inside this triangle. Thank you in advance.

 

[Dr.Peterson]

Hi I am having a devil of a time figuring this out and every explanation doesn't seem to work.

It seems simple enough. we have a triangle with base of 439.823 units and slope sides of 356.035 units. What is the diameter or radius of the largest circle that perfectly fits inside this triangle. Thank you in advance.

Please show us what didn't work for you (your attempts, and whatever explanation you are referring to), so we don't just do the same thing again.

Are you talking about an isosceles triangle?

 

[Ahatmose]

I am referring to The Great Pyramid and I am trying to calculate what the largest circle would be that rests on the base and exactly touches both sides (slopes) I have it somewhere between about 213.9 and 214.9 cubits using my measurements as cubits.

Here is the only explanation I have found. Inscribed circle in a triangle

regards
db

 

[The Highlander]

Hi I am having a devil of a time figuring this out and every explanation doesn't seem to work.

It seems simple enough. we have a triangle with base of 439.823 units and slope sides of 356.035 units. What is the diameter or radius of the largest circle that perfectly fits inside this triangle. Thank you in advance.

I am referring to The Great Pyramid and I am trying to calculate what the largest circle would be that rests on the base and exactly touches both sides (slopes) I have it somewhere between about 213.9 and 214.9 cubits using my measurements as cubits.

Here is the only explanation I have found. Inscribed circle in a triangle

regards
db

Since the two "slope" sides (plural) of the triangle are given as "356.035 units" then (surely?) it is safe to assume that it is isosceles.

Read through the following and you should then know where to go next (Google can help you with that. ?)
Hint: Pay particular attention to point number 16.

Isosceles Triangle: A triangle having two sides of equal length is called the Isosceles triangle. In an isosceles triangle, the angles opposite to the equal sides are equal. In the triangle given below, two sides are of 5inches and one side is 3 inches. Thus, it is an Isosceles Triangle.

Properties:-

• Two sides are congruent to each other.
• The unequal side of an isosceles triangle is called a base.
• The two angles opposite to the equal sides are congruent to each other. Thus it has two congruent base angles.
• Apex angle is the angle that is not congruent to the two congruent base angles.
• The height drawn from the apex of an isosceles triangle divides the base into two equal parts and also divides the apex angle into two equal angles.
• Area of Isosceles triangle = ½ × base × height
• The perimeter of an Isosceles triangle = sum of all the three sides
• The third unequal angle of an isosceles can be acute or obtuse.
• The circumcenter of an isosceles triangle lies inside the triangle if all the three angles of the three triangles are acute.
• The sides of the triangle are the chords of the circumcircle.
• If one of the angles is 90 degrees, then the circumcenter lies outside the triangle.
• The centroid is the intersection of the medians of the Isosceles triangle.
• The median drawn from Apex divides the triangle at right angles.
• The perpendicular bisectors of an isosceles triangle intersect at its circumcenter.
• The angle bisectors of an isosceles triangle intersect at the incenter.
• The circle drawn with the incenter touches the three sides of the triangle internally.
• Each median divides the isosceles triangle into two equal triangles having the same area.
• The area of the triangle can be estimated:
• If the measure of one angle and one side are given
• If three sides of the triangle are given.
• If two sides of an isosceles triangle and their included angles are given.
• Joining the midpoint of three sides divides the triangle into 4 smaller triangles of the same area.
• When a circle with a diameter equal to the base is drawn:-
  • For an obtuse-angled isosceles triangle, the apex lies inside the circle.
  • For a right-angled isosceles triangle, the apex lies on the circumference.
  • For an acute-angled isosceles triangle, the apex lies outside the triangle.
  • When the midpoint apex is taken as a radius and a circle is drawn with the midpoint of the base as the centre
  • For acute-angled isosceles, the base vertices lie inside the circle.
  • For a right-angled isosceles the base vertices lie on the circumference
  • For an obtuse-angled isosceles triangle, the base vertices lie outside the circle.

 

[Ahatmose]

Hi I appreciate the help but I am 73 years old and I have gone over all these things and I can easily mark the points and draw the circle but I can't figure out exactly what the size is. I am sorry I didn't realize this was school. and yes I came here to be spoon fed the solution. Now I remember this from the way back days.

I will keep on trying but what you posted makes no sense to me, sorry but thanks anyway..

regards
db

 

[The Highlander]

Hi I appreciate the help but I am 73 years old and I have gone over all these things and I can easily mark the points and draw the circle but I can't figure out exactly what the size is. I am sorry I didn't realize this was school. and yes I came here to be spoon fed the solution. Now I remember this from the way back days.

I will keep on trying but what you posted makes no sense to me, sorry but thanks anyway..

regards
db

Getting my plastic spoon out, @Ahatmose, ?

You want the incentre. There is a calculator here.

Assuming this to be an isosceles triangle, I calculated the altitude to be: 279.99997 cubits (using Pythagoras).

Then placing the triangle at the origin (and its base on the x-axis), its vertices' coordinates become: (0, 0), (219.9115, 279.99997) & (439.823, 0).

I get the radius of your circle to be: 106.911 cubits. (and I'm 70! ?)

Your age should not be a barrier to further learning; keeping the brain ticking over helps you stay young. ?

 

[Ahatmose]

No need I did it myself, thank you. I found a site that explained it very nicely.

Here is my solution (Gosh I am so proud of myself ! )


with a little, well a lot of help from this website but I have it at last. Solution to inscribed circle

 

[Ahatmose]

Hi I don't know why they had to make it so complicated. Here is a formula I came up with that makes it quite easy to remember and solve.

 

[Dr.Peterson]

Hi I don't know why they had to make it so complicated. Here is a formula I came up with that makes it quite easy to remember and solve.

View attachment 35330

That does work (though it gives the radius, not the diameter) if you happen to know the area. But you weren't given that! That's why they got a more complicated formula.

(In this case, the Pythagorean Theorem provides a way to find the height, and thence the area. This is, in fact, what they did for their specific problem, not the formula they show. There is also a formula for the area of any triangle, given the sides, but it is not needed here.)

The general formula for the radius of an inscribed circle is r = A/s, where s is the semiperimeter. In this case, we have A/((B + 2S)/2), which immediately gives your formula.

I'm curious to see how you arrived at this formula, and whether you have used it for your problem.

I should also add that the word "slope" doesn't mean what you are using it to mean. The slope tells how steep something is; your S is the sloped side, or, better, the leg of the isosceles triangle.

 

[Ahatmose]

Hi firstly my formula yields the diameter and not the radius. But to answer your question I originally was looking for the diameter not the radius and I noticed in the formula that everything was divided by 2. Thus I simply took away the divided by 2 and was left with this. As to the area of the triangle all solutions given assume you know it is base times height divided by 2. And to label 1/2 base + slope or as you choose to cal it "the leg of the triangle" although why slope is not correct I will never know, remember I was talking about the Great Pyramid at Giza) and "s" as " the semi perimeter" well that is a term I have never heard before and finally your formula "The general formula for the radius of an inscribed circle is r = A/s, where s is the semi perimeter. In this case, we have A/((B + 2S)/2), which immediately gives your formula." actually yields the diameter not the radius. You have to divide it all by 2 to get the radius.

Proof: In The Great Pyramid simplified size is base of 440 cubits and height is 280 cubits slope is 356.09 ( sq rt of 220 squared + 280 squared)

Area is 440 x 280 = 123,200 1/2 base = 220 + Slope or 356.09 total = 576.09

123,200 / 576.09 = 213.85547397108090749709246819073 which is THE DIAMETER.

OOOPS !

Better change that to 2A, or base x height , sorry ... silly me

THANKS FOR THE HEADS UP CORRECTION !!!!!!

Cheers
db

Well at least we got a simple formula for the radius

 

[Ahatmose]

My humbly corrected diagram.

 

[Dr.Peterson]

Proof: In The Great Pyramid simplified size is base of 440 cubits and height is 280 cubits slope is 356.09 ( sq rt of 220 squared + 280 squared)

Area is 440 x 280 = 123,200 1/2 base = 220 + Slope or 356.09 total = 576.09

123,200 / 576.09 = 213.85547397108090749709246819073 which is THE DIAMETER.

No, the area of a triangle with base 440 and height 280 is (440*280)/2 = 61,600. You accidentally used twice the area (even though you correctly stated the formula), so you got twice the radius, which is the diameter.

The formula does give the radius, when used correctly.

I see you corrected your claims before I got around to sending this response.

although why slope is not correct I will never know

As for the meaning of "slope", you might want to look it up, rather than just declaring your ignorance.

 

[Ahatmose]

I did:




... okay I won't use it for mathematics anymore but remember I was talking about The Great Pyramid