You could also do it by scale drawing! 😃Once you bisect the three angles the bisectors would intersect at the centre of the circle. The perpendiculars to the sides from the centre of the circle would then give you the radius. At school we did it by drawing before we did the mathematical proof as it was a good method of reinforcing the theorems and trigonometrical identities at play.
@@konradyearwood5845 Good approach to get a close estimate...as long as you remember it's an estimate--not exact. But using it to verify the math derivation or proof is a great idea.
Using Heron's formula for the triangle - A=sqrt{s*(s-a)*(s-b)*(s-c)}, where s= (a+b+c)/2, we have A =sqrt(21*8*7*6)=84 sq units. Now adding the areas of the three triangles, we get A=1/2*R*(13+14+15)=21R. Hence 21R=84 or R=4. Hence area is 16 pi sq units.
By breaking the triangle into 3 "kites", you can quickly show that 7r + 6r + 8r = 21r is the same as the area of the triangle, which Heron's formula gives as 84. Hence, r = 4. This way you don't have to bring angles into it at all (other than recognizing the right angle where each radius meets the sides of the triangle - and thereby forms two sides of each 4-sided kite).
Where do you get the 7r and 6r? The only way to know for sure is if the triangle were drawn to scale. Being that the sides are 14,13, and 15 where could you draw the kite?
Freeze the video at 6:42. Apply x = 6 to the diagram and you can see where the 7 and 8 come from. Break the triangle into 3 right-angled kites (draw the perpendicular from the incenter to each side of the triangle - Andy did this for just one of them). The top kite has sides 6, r, r, and 6, which means since it is a right-angled kite, its area is 6r. Similarly, the kite on the left has sides 7, r, r, 7, with area 7r, and the kite on the right has sides 8, r, r, 8, with area 8r. Thus the area of the triangle is the combined area of the 3 kites, or 21r.
I’m a retired airline pilot educated in Aerospace Science with a BS. 77 years old. I haven’t been doing math since college. I got the first few of your videos wrong. That was a wake up call for my brain and an ego killer. Ouch! Now, I’m finally getting most of them right although it takes me awhile to get the cobwebs out of my head. Love your videos man. How exciting! lol
That's a wonderful approach, Andy. I especially like the use of trig to find sides and side ratios, not angles. And really, no calculator needed. How exciting indeed!
What about using Heron’s formula? Semiperimeter (s): General formula: s = (a + b + c) / 2. With values: s = (13 + 14 + 15) / 2 = 21. Area (A) of the triangle (using Heron’s formula): General formula: A = sqrt(s * (s - a) * (s - b) * (s - c)). With values: A = sqrt(21 * (21 - 13) * (21 - 14) * (21 - 15)) = sqrt(21 * 8 * 7 * 6). Radius (r) of the incircle: General formula: r = A / s. Substituting the area and semiperimeter, we find r that simplifies to the formula for calculating the area of the circle. Area of the circle: General formula: π * r^2. With the calculated r, the area of the circle simplifies to 16π square units.
With Heron's formula the derivation is more straithforward, no need to use the law of cosines: A² = s(s-a)(s-b)(s-c) where s = (a+b+c)/2. Also A = rs and equating both and multiplying both sides by 𝜋, gives the area of the circle which is r²𝜋 = 𝜋(s-a)(s-b)(s-c)/s, and s = 21 ⇒ r²𝜋 = 16𝜋. End of story.
This problem takes a lot of concepts from Geometry to solve, requiring a pretty thorough understanding of the subject. Great job! You made me want to touch up on my Geometry skills 😁
@@xarlixe7565 That is easy but a long method. So I can't explain in comments. But another shorter method is by using Heron's formula. Divide the circle into three triangles and the height of all those triangles would be equal to radius of circle. Equate the sum of area of these three small triangle equal to larger traingle whose area would be found by Heron's formula. Then you would get radius and then you can find area of circle. If you would know all of this, the answer can be found under 1 minute.
I used to love Math very much since i was a kid, however stress of college life and other stuff make me forgot those feeling. Thanks for reminding me how fun math is to me. 😊
This was a fun one. I originally went down the trig rabbit hole, but ended up using the Base*Height/2 formula. Computed the Height of the whole triangle to find its area. The big triangle is also made up of 3 smaller triangles of total area 13r/2 + 14r/2 + 15r/2. Then just solve for r.
I recently discovered your videos and love it. Can you give us a sense of what grade (if any) each problem rises too? I have a couple elementary age kids so I don't want to scare them too soon!
Very nice problem and solution. I like that you always go straight to the point instead of fluffing around repeating the same thing 4 or more times, as other people do here in RU-vid. Heck, there are people who would had made a 1 hour long video for this problem!
YES!!! I tried it before watching the video and found a much faster solution: just find the average value of sides of the triangle (14) and calculate it as if the triangle had all sides 14 (it’s just a simple tan30=r/7) result is the same and I feel good about myself :)
If you have all 3 sides of the triangle just use Herons formula to find the area of the triangle. Then the radius of the inscribed circle is 2S/(a+b+c). No need for all the trigonometry. Heron did it for us 2000 years ago.
There is a much simpler way to do this with a bit of calculation First devide the traingle into 3 more parts such that each smaller triangle has the radius of the circle as it's height ( r is perpendicular to t). Then find the area of the triangle using herons formula. Find the area of each smaller triangle in terms of r (1/2base x height) and add them Substitute the values and find r
heron formula to find the area of the triangle gives Area = 84. Area is also equal to (half the perimeter) * r, from which we find r = 4, therefore Area of the circle is = 16 Pi
Is it purely accidental that the numeric values of the sides and area of the inscribed circle are 13,14,15,16? Or is this repeated in other triangles? Sure i could do the arithmetic and trig myself (proof might be harder to do though) but it's late here and I'm tired! ;-) (it clearly doesn't for the 3-4-5 triangle btw: 3, 4, 5, Pi.)
Interestingly, this problem can be solved in a pretty clean manner. You can actually break a 13 14 15 triangle into a 9 12 15 right triangle and a 5 12 13 right triangle. Then, you can find that the area of the triangle is 84 since bh/2 = (12)(14)/2 = 84. Since inradius * semiperimeter = Area of the triangle, (21)r=84 and r=4. Therefore, the area of the circle is 16(pi).
It depend on what you know oc. If you know the formula for the radius of the incircle in terms of the sides, it's just a computation :). If you know Heron's formula, then you can see how the incircle helps cutting the triangle into 3 pieces with height R and base = the sides. If you don't know Heron's formula but you can solve this problem, then you're on par with the ancient Greek mathematicians.
To be honest I don’t think this is a hard problem but indeed interesting. I just use heron formula and this formula I don’t really know what it call “s = (r*(a + b + c))/2” where r is the radius of incircle of a triangle, s is the surface area of the triangle that has the incircle that we are dealing with, a b and c are the length of each side of the triangle. So the problem ask us to find the radius of incircle of the triangle which we can use the second formula but first we have to rearrange it so s = (r*(a + b + c))/2 therefore r = 2*s/(a + b + c) but we don’t know the surface area of the triangle yet but in the problem they tell the length of each side of the triangle so we can use heron formula to calculate which s = squarerootof((semi_perimeter_of_triangle)*(semi_perimeter_of_triangle-a)*(semi_perimeter_of_triangle-b)*(semi_perimeter_of_triangle-c)) so s = squarerootof(((13+14+15)/2)*(((13+14+15)/2)-13)*(((13+14+15)/2)-14)*(((13+14+15)/2)-15)) so s = 84 unit square so plug numbers into the second formula we got r = 2*(84)/(13+14+15) so r is 4 unit length. Edit: My bad this problem actually ask for the surface area of the incircle but since we already figured out the radius of the circle we just need to use formula circle_surface_area = pi*(radius’s_length)^2 where pi is approximately 3.14 so circle_surface_area (approximately) = 3.14*(4)^2 = 3.14*16 = 50.24 square unit.
There's another reasonably interesting way to do this, though a bit off-beat as far as RU-vid geometers go. Strategy • find height of the triangle • use that to find θ (say on left corner) and φ (right corner); • tangent of ½θ is the slope of a line from corner to center of incircle… likewise × -tangent of ½φ is the 'other slope' of the incircle center to right corner • mathematically cross 'em, to find 𝒙 • and use that times the first slope to find 𝒓, the radius. Doing it: [1.1] 𝒉² = 𝒂² - 𝒔² … where 𝒔 is an unknown bit of 𝒃 [1.2] 𝒉² = 𝒄² - (𝒃 - 𝒔)² … the other half of 𝒃 Set those two to each other, and expanding, moving things around, solve [1.3] 𝒔 = (𝒂² - 𝒄² + 𝒃²) ÷ 2𝒃 the height then follows [2.1] 𝒉 = √(𝒂² - 𝒔²) Having that, we can now find θ [3.1] θ = arctan( 𝒉 / 𝒔 ) And the slope of the incircle-center line from corner is [4.1] 𝒎₁ = tan( ½θ ) … for the line [4.2] f(𝒙) = 𝒎₁𝒙 ⊕ 0 The exact same logic can be used on the right side finding φ and tan( ½φ ) to give a slope. [5.1] 𝒎₂ = -tan( ½φ ) [5.2] 𝒃₂ = -𝒎₂ • 𝒃 … intercept; [5.3] g(𝒙) = 𝒎₂𝒙 + 𝒃₂ Now we have formulæ for lines that can be mathematically crossed [6.1] 𝒙 = 𝒃₂ / (𝒎₁ - 𝒎₂) and of course, the radius is [6.1] times 𝒎₁ ⊕ 0. In PERL: (just a convenient (for me) calculator) --------- CODE --------------------------------------------------------------------- my $a = 13; my $b = 15; my ¢ = 14; my $h; my $x; my $r; my $s; $s = ($a •• 2 - ¢ •• 2 + $b •• 2) / (2 × $b); $h = √($a •• 2 - $s •• 2); $x = $b • tan( ½ • atan2( $h, $b - $s ) ); $x /= ( tan( ½*atan2( $h, $s ) ) + tan( ½*atan2( $h, $b - $s ) ) ); $r = $x • tan( ½ • atan2( $h, $s ) ) ⊕ 0; --------- OUTPUT --------------------------------------------------------------------- a = 13 b = 15 c = 14 s = 6.6 h = 11.2 x = 7 r = 4
We had this question in class 10 its easy not to prolong but you can just the use the formula area of the triangle = perimeter * radius of circle /2 to find the radius and then find area
Long, complicated way to do it. I broke it up into 3 pairs of congruent right triangles and set the sum of those areas equal to the total area of the triangle. From that, I solved for the radius of the circle. r/2(2a + 2b + 2c) = r(a + b + c) = sqrt[s(s-2a)(s-2b)(s-2c)] = rs: s = a + b + c. Square both sides, divide by s and multiply by pi. Done.
Since the 345 triangle is involved in squaring the Circle and since that uses the golden ratio, the ratio of the circumference to diameter Hass to be congruent with the golden number. Pi is 3.144... or 4/phi squared
You could take radius perpendicular to Side and dividing the whole triangle into 3 triangle then by herons formula, you could solve the problem It would take comparatively less time
Found an easeir way. First find the included angle of side 13 and 15 which is 59.48.... Using the laws of cosine then usinf the trignometic ratio of sin find the shortest distance of triangle (the height) which is sin59.48 x 13 and the shortest distance i.e the perpendicular will be 11.2. Then find the included angle of 13 and 14 which would be 67.38... Then divide this by two as we know from cirlce properties (as told by andy) that tangents from same points have every angle and side same thus the angle wil be 33.69... Then as seen in the picture we can subract the radius,r, from from 11.2 i.e 11.2 -r to get the hypotnease of the triangle of the congurent triangles and then using the trignometic ratio of sin on the congurent triangles we can do sin33.69..=r/11.2-r which will give us the radius 3.99 rounded of to 4 and usinf the area formula we will get the answer 16pi
my journey in solving this problem is even longer. i first figured out what the two lengths of a side are considering the three altitudes intersecting them, then i found the measure of the angles and the lengths of the altitudes, then I figured out the lengths of the angle bisectors, then i converted the angle bisectors into equations considering the lower left corner as (0,0), then i found the intersection, then i plugged the y value into the area of a circle. this took me roughly 2 hours. didnt know heron's formula, didnt remember law of cosines or double angle formula
I see i'm not the only one who found the solution with heron's formula. I actually didn't know about that formula, and had to look it up. I also wouldn't have found the solution without the first note about tangent lines in the beginning.
You don't need any law of cosinus. If you do the second part, calculating the length of parts divided by r in each side (6,7,8), and also calculate the area of the whole trangle by adding up the area of the small triangles 6r+7r+8r=21r, then apply the general rule for area of a triangle, here: 15 x height of the triangle/2. The height of the triangle = r+ square root of( 36+r2). Then you have two eguation: 21r= r+ square root of( 36+r2) x 15/2. Which also gives r=4
The radius of a circle inscribed in a triangle is two times the area of the triangle divided by the perimeter of the triangle. In your case r = 2 x 84 / 42 = 4.
Is it possible to make out of the 14 a 13.5 and out of the 13 a 13.5? And then out of the 15 a 14 and the one we took away there can't I add 0.5 to the 13.5 of both sides now so that all sides are now 14?
i created a spread sheet after watching your videos. I just need to enter any triangle data and it gives me all the other information. Triangle 13, 14, 15 = inside circle area: 50.265
a better method let one coordinate be (0,0) and one be (15,0) so applying distance formula we can get the 3rd coordinate as x^2 + y^2 = 169 (x-15)^2 + y^2 = 196 solving this we get 3rd coordinate as (33/5 , 56/5) now apply in-centre formula I =[(ax1 + bx2 + cx3)/(a+b+c) , (ay1 + by2 + cy3)/(a+b+c)] so coordinate of in-centre comes out to be I=(7,4) since we took one coordinate as (0,0) ie the origin and other as (15,0) the line joining this is the x axis the distance between the in-centre and x axis is the radius here that is = 4 (by simple observation) hence area = pi(4)^2 = 16pi
i don’t understand, so when you got r/x=2/3, why isn’t r=2 and x=3? that seemed to make the most sense to me as you can see the values reflected in the expression.
We can find the radius of the circle by applying herons formula to the sides of the triangle and also finding the area of the triangle by using 1/2.b.h formula to all the three triangles inside it and finding the total area of the triangle and then find r. how exciting...
A similar question is in the NCERT Book for class 10 maths. I was just scrolling through yt one day before my maths board exam and saw this diagram. I didn't instantly remember how to do it but still screamed 16π.
We can use the fact that the (Radius of Circle)*(Semi-perimeter of Triangle) = Area of Triangle We can find the Area using Heron's formula to be 84 cm^2, and the Semi-Perimeter is 21 cm So Radius of Circle is 4 cm, and Area is 16pi cm^2
Another way to solve it: The area of the triangle A = 13*r/2 + 14*r/2 + 15*r/2 = 21r; Also, A = SQRT(s*(s-a)*(s-b)*s-c)). This is Heron's formula. Where a, b, c are the triangle sides and s = (a+b+c)/2; Than s = (13+14+15)/2 = 21; A = SQRT(s*(s-a)*(s-b)*s-c)) = SQRT(21*(21-13)*(21-14)*21-15)) = SQRT(21*8*7*6) = 84; Than 21r=84 than r = 4; The circle area = PI*4*4 = 16*PI
idk if this is legal but i gave angles the value of their opposing side as in 15k, 14k and 13k, then added them up and solved for k; 42k=180 and then i figured the true angle size, then i used the sin theorem and got the value of the outside radius R and then got the area of tge triangle with A=abc/4R, and at last i used the formula A=s×r with r being the inner radius and then i got the area of the inner circle wich came up as right! Please excuse me for my english and correct me if i did smt wrong!
I solved it in a totally different way but had the same result.. what i did is I divided the shape into smaller squares that had smaller quarter of circles and then solved it
U can solve this in less than a minute just by taking half the base of the triangle and u will get an approximate number to the diameter of the circle wich will lead to the area (its not precise but efficient )
If half angle theta(t) at vertex with sides of length 13 and 15 and half angle alpha(a) at vertex with sides of length 15 and 14, tan(t) = r/7; tan(a) = r/8. Here, r is radius of incircle. Also, tan(90-t-a) = r/6. Thus, 6 = r*tan*t+a) = r * (tan(t) +tan(a))/(1-tan(t)tan(a)). This leads to 6 = (15*r^2)/(56-r^2). This leads to r = 4. Thus area is 16pi.
You could EASILY solve this in only two steps. • Find the area of the triangle • Use the following formula: S (area of the triangle) = P (½ of a perimeter) × r (radius of the circle inscribed within a triangle) Easy
