I haven't watched the video yet but my answer right now to the question in the thumbnail; "Easy or Not?" is DEFINITELY NOT. I did get the answer, I think, but it took 3 pages of algebra, most of which was to come up with 625x^2-5050x+10,201=0 where x is the distance from the apex of the triangle to the tangent on the 13 length side. Since you asked the question, and since the video is only 6:30 I think the answer is supposed to be easy. So while I got the answer I am going to keep looking for the easy solution before I play the video. Thanks Phantom for the puzzle!
So I found a simpler solution (not quite as elegant as the one in the video). And I found that my original difficult solution was a little off. Now I get to figure out how I messed up!
I do like the Ancient. I look for all the right angles. The three center points of a triangle is often a good start. Bisecting angles and sides and in this case finding the altitudes since the area is known from the circumface. Then also the close sibling of the triangle, the circle that is also defined by three points. Tangents, chords, secants et c. It's a set of visual tricks to be combined. Avoid algebra!
It's more fun to put the semicircle on the length 14 side. Then r = 6, and the shaded area becomes 84 - 36 * pi. That could be a mental math problem for some. As an alternative method of determining the area of the full triangle, note that (5,12,13) and (9,12,15) are Pythagorean triples, establishing the height of the triangle (to the side of length of 14) as 12, so the area is 1/2 * 14 * 12 = 84. That's evens simpler than Heron's formula.
Area of triangle: A = ½b.h + ½b.h = ½(13.r)+½(14.r) A = 13,5 . r Area of triangle, Heron's formula: A²= s(s-a)(s-b)(s-c) A²=21(21-13)(21-14)(21-15) A = 84 cm² = 13,5.r --> r=56/9cm Área of semicircle: A = ½πr² = 60,815 cm² Red shaded area: A = A₁- A₂ = 23,185 cm² (Solved √)
After going on the obvious tangent and Heron (our hero) I did use (deduced) the altitude from C to F on base 14, and from A to G on base 13. Altitude and one side is known on each of these right triangles. And there are congruent triangles with the side r. But that was less helpful in finding the radius, than what this method as shown was. I thought B was less useful, but heck, two tangents converge there, duh!
All good - but, unlike the opening description, the requested area is NOT the intersection between the triangle and the semicircle, that would be just the semicircle .... but we know what you mean!
Let the vertices of the triangle be A, B, and C, clockwise from the top. Let the center of the semicircle be O. We can determine the area of the triangle by Heron's Formula: s = (13+14+15)/2 = 42/2 = 21 A = √(s(s-a)(s-b)(s-c)) A = √(21(21-14)(21-14)(21-15)) A = √(21(8)(7)(6)) A = √(3•7•2³•7•2•3) A = √(3²•7²•2⁴) A = 3•7•2² = 84 Let the points of tangency between semicircle O and AB be M and between semicircle O and NC be N. As OM and ON are radii of length r and AB and BC are tangent to semicircle O at M and N respectively, ∠AMO = ∠ONC = 90°. Draw OB. This creates two triangles ∆BOA and ∆COB with height r and bases of 13 and 14 respectively. As their combined areas are the same as the area of triangle ∆ABC, we can solve for r. [ABC] = [BOA] + [COB] 84 = 13r/2 + 14r/2 = 27r/2 27r = 84(2) = 168 r = 168/27 = 56/9 Semicircle O: A = πr²/2 = π(56/9)²/2 A = π(3136/162) A = 1568π/81 ≈ 60.815 Shaded area: A = 84 - 1568π/81 ≈ 23.185 sq units
ABC=(14r/2)+(13r/2)=27r/2 Perímetro de ABC =13+14+15=42→ Semiperímetro =42/2=21 → Según fórmula de Herón: ABC =√[21*(21-13)*(21-14)*(21-15)=84. 27r/2=84→ r=56/9→ Área sombreada =84-(πr²/2)=84-(1568π/81) =23,18497... Gracias y un saludo.
I haven't watched the video yet but my answer right now to the question in the thumbnail; "Easy or Not?" is DEFINITELY NOT. I did get the answer, I think, but it took 3 pages of algebra, most of which was to come up with 625x^2-5050x+10,201=0 where x is the distance from the apex of the triangle to the tangent on the 13 length side. Since you asked the question, and since the video is only 6:30 I think the answer is supposed to be easy. So while I got the answer I am going to keep looking for the easy solution before I play the video. Thanks Phantom for the puzzle!
So I found a simpler solution (not quite as elegant as the one in the video). And I found that my original difficult solution was a little off. Now I get to figure out how I messed up!
@@waheisel You really did amazing brainstorming job! Well done!
I do like the Ancient. I look for all the right angles. The three center points of a triangle is often a good start. Bisecting angles and sides and in this case finding the altitudes since the area is known from the circumface. Then also the close sibling of the triangle, the circle that is also defined by three points. Tangents, chords, secants et c. It's a set of visual tricks to be combined. Avoid algebra!
It's more fun to put the semicircle on the length 14 side. Then r = 6, and the shaded area becomes 84 - 36 * pi. That could be a mental math problem for some.
As an alternative method of determining the area of the full triangle, note that (5,12,13) and (9,12,15) are Pythagorean triples, establishing the height of the triangle (to the side of length of 14) as 12, so the area is 1/2 * 14 * 12 = 84. That's evens simpler than Heron's formula.
Area of triangle:
A = ½b.h + ½b.h = ½(13.r)+½(14.r)
A = 13,5 . r
Area of triangle, Heron's formula:
A²= s(s-a)(s-b)(s-c)
A²=21(21-13)(21-14)(21-15)
A = 84 cm² = 13,5.r --> r=56/9cm
Área of semicircle:
A = ½πr² = 60,815 cm²
Red shaded area:
A = A₁- A₂ = 23,185 cm² (Solved √)
Good video sir
Thanks a lot.
Ty sir!
After going on the obvious tangent and Heron (our hero) I did use (deduced) the altitude from C to F on base 14, and from A to G on base 13. Altitude and one side is known on each of these right triangles. And there are congruent triangles with the side r. But that was less helpful in finding the radius, than what this method as shown was. I thought B was less useful, but heck, two tangents converge there, duh!
Nice. My first thought was to mirror the triangle along the 15 line, but it does not seem to help much.
Yeah, I know. Brain wants to do that!
All good - but, unlike the opening description, the requested area is NOT the intersection between the triangle and the semicircle, that would be just the semicircle .... but we know what you mean!
@@bonaldisillico Yeah, you're right... I meant 'difference area,' but thankfully, you got the point, and that's what matters most.
Area=84-1568π/81=23.18.❤
Let the vertices of the triangle be A, B, and C, clockwise from the top. Let the center of the semicircle be O. We can determine the area of the triangle by Heron's Formula:
s = (13+14+15)/2 = 42/2 = 21
A = √(s(s-a)(s-b)(s-c))
A = √(21(21-14)(21-14)(21-15))
A = √(21(8)(7)(6))
A = √(3•7•2³•7•2•3)
A = √(3²•7²•2⁴)
A = 3•7•2² = 84
Let the points of tangency between semicircle O and AB be M and between semicircle O and NC be N. As OM and ON are radii of length r and AB and BC are tangent to semicircle O at M and N respectively, ∠AMO = ∠ONC = 90°.
Draw OB. This creates two triangles ∆BOA and ∆COB with height r and bases of 13 and 14 respectively. As their combined areas are the same as the area of triangle ∆ABC, we can solve for r.
[ABC] = [BOA] + [COB]
84 = 13r/2 + 14r/2 = 27r/2
27r = 84(2) = 168
r = 168/27 = 56/9
Semicircle O:
A = πr²/2 = π(56/9)²/2
A = π(3136/162)
A = 1568π/81 ≈ 60.815
Shaded area:
A = 84 - 1568π/81 ≈ 23.185 sq units
@@quigonkenny Nice! Great job!
ABC=(14r/2)+(13r/2)=27r/2
Perímetro de ABC =13+14+15=42→ Semiperímetro =42/2=21 → Según fórmula de Herón: ABC =√[21*(21-13)*(21-14)*(21-15)=84.
27r/2=84→ r=56/9→ Área sombreada =84-(πr²/2)=84-(1568π/81) =23,18497...
Gracias y un saludo.
Nice!
How do you know that the circle is exactly a 1/2 circle???
Because that's the condition of the problem. Check 0:12 please :)