Olympiad Mathematics | Can you find the Red region area? | (Simple explanation) | #math #maths

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Olympiad Mathematics | Can you find the Red region area? | (Simple explanation) | #math #maths

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[luckshmana]

thank you posting this question sir I am Luckshmana of 10th standard.

[misterenter-izrz]

Note that angle C is a right angle for triangle in a semicircle, so CB^2=144-100=44, then the area is approximately. 😊Actually the area is just equal to the area of the triangle.😮

[MultiYesindeed]

Great video (as always) thank you again

[davidellis]

The area of a semicircle with radius r and diameter d is given by (1/2)pi(r^2) = (1/2)pi(d^2/4) = (1/8)pi(d^2). Since ACB is a right triangle, AC^2+CB^2=AB^2. Multiplying by pi/8, we see that the area of the semicircle on AB equals the sum of the areas of the semicircles on AC and CB, so the red region has the same area as the right triangle.

[marioalb]

That triangle is a right triangle
Pytagorean theorem:
b² = c² - a² = 12²-10²
b= √44 cm

Area = Area of semicircle 1 + Area of semicircle 2 - Area of semicircle 3 + Triangle area

A = ⅛πa²+ ⅛πb² -⅛πc² + ½.a.b
A = ⅛π(a²+b²-c²)+½a.b = 0+½ab
A = ½ a.b = ½.10.√44
A = 33, 166 cm² ( Solved √ )

Red shaded area = Right triangle area

This is the demonstration of "Lune of Hippocrates theorem"

[marioalb]

Lune of Hippocrates theorem :
Red shaded area is equal to triangle area

Right triangle Area
A = ½ b. h
A = ½ 10 √(12²-10²)
A = 33, 166 cm² ( Solved √ )

[HappyFamilyOnline]

Great explanation 👍
Thanks for sharing 😊

[marioalb]

That triangle is a right triangle.
Pytagorean theorem:
b² = c² - a² = 12²-10²
b= √44 cm

Every time we have this configuration,
Red shaded area = Right triangle area

A = ½ a.b = ½.10.√44 = 10√11
A = 33, 166 cm² ( Solved √ )

This is the "Lune of Hippocrates theorem"

[thomast.]

Time travelling to my school time: "Die Möndchen des Hippokrates" ( Lunes of H. de Chios ), but we were asked to show that in all triangles in a semicircles the summe of the 2 lunes is as large as the triangle

[honestadministrator]

∆ AC B being a semi circular triangle angle ACB is right angle
π ( AC^2 + BC^2) /2 + AC*BC/2
- π (AB^2)/2
= AC*BC/2

[marioalb]

That triangle is a right triangle
c² = a² + b²
a² + b² - c² = 0
⅛πa² + ⅛πb² - ⅛πc² = 0

Red shaded area :
A = Aa + Ab - Ac + At
A = ⅛πa² + ⅛πb² - ⅛πc² + ½ab
A = 0 + ½ab = ½.a.b
A = ½ . 10 .(12²-10²)^½
A = 10√11 cm² ( Solved √ )

[mrxmry]

this one is easy. first calculate the length BC. then calculate the areas of the semicircles over the 2 shorter sides, add the area of the triangle and subtract the area of the big semicircle.

[santiagoarosam]

Cateto corto del triángulo rectángulo =a》a^2=12^2 -10^2=44》a=2sqrt11 》Área roja =(Pi/2)[5^2 +(sqrt11)^2 -6^2] +10a/2 =10sqrt11
Sorprendente resultado. Gracias y saludos.

[wyvern]

Love your channel! Quick question: how are chords AC=10 and CB=(eventually)2sqrt11 the respective diametres of the two smaller circles?

[ajbonmg]

Interesting - if you generalise to any lengths, the total red area is always equal to the area of the triangle ABC.

[abstragic]

Thank you sir. My solution was identical with yours this time.

[ScienceNectar]

This is the lune of alhazen so the area of triangle is equal to the area of the red region

[mikeify]

Can problem be solved using area =semi circle.area -semi circle segment area? Thanks

[unknownidentity]

This does not seem to be very complicated. Due to Thales theorem the triangle ABC is a right triangle with ∠ACB=90°. So we can apply the Pythagorean theorem to obtain the missing side length BC:
AB² = AC² + BC²
BC² = AB² − AC² = 12² − 10² = 144 − 100 = 44 ⇒ BC = 2√11

Now we can calculate the red area:
A(red)
= [A(red) + A(white)] − A(white)
= [A(semicircle AC) + A(semicircle BC) + A(triangle ABC)] − A(semicircle AB)

Here we can observe something interesting:
A(semicircle AC) + A(semicircle BC) − A(semicircle AB)
= π(AC/2)²/2 + π(BC/2)²/2 − π(AB/2)²/2
= π[AC²/4 + BC²/4 − AB²/4]/2
= π[AC² + BC² − AB²]/8
= π[AB² − AB²]/8
0

So the red area is exactly as large as the area of the triangle ABC:
A(red) = A(triangle ABC) = (1/2)*AC*BC = (1/2)*10*2√11 = 10√11

Best regards from Germany

[jimlocke]

Your solution is excellent but I would find the video easier to follow if you had presented your plan before doing the solution. Your plan is to find the area comprised of 3 pieces: two smaller semicircles and ΔABC, and then find and deduct the area of the larger semicircle. After you presented your plan, you would introduce Thales' Theorem and use it and the Pythagorean Theorem to find the length of BC, which is needed to find the area of ΔABC and the area of one of the semicircles, and then complete your solution.

The method that you used to solve this problem can be applied to many other geometry problems, including some where it is far less obvious, so it is a very important technique to learn.