Circle Overlap Challenge: Find the Hidden Area of This Red Rectangle

[@nandisaand5287]

I took the analytical geometry approach and got there, but because life just isnt hard enough, I made O(0,0) the intersection of AB and CD: Sm Circle: X²+(Y-K1)²=9 Lg Circle: X²+(Y-K2)²=25 K1-K2=6 {Lots of overly difficult Algebra later...} 12K2-12Y=(-52) Since @ O, Y=0: 12K2=(-52) K2=(-52/12)=(-13/3) From there, I was able plug in as you show and get the answer.

[@RAG981]

Method 1. Rather simpler to find area of triangle BAD by Heron's formula = Root( 7x4x2x1)= rt56. Area BAD also = half of 6 times x, so x = rt56/3 = 2rt14/3, width rect = 4rt14/3, and area = 8rt14. Method 2 works fine. Whole thing is a simple problem dressed up.

[@bjorntorlarsson]

Even simpler after Heron: Area of triangle BAD = rt(7x8) is also 1/4 of the area of the rectangle.

[@PrithwirajSen-nj6qq]

I like to say to take ADBC as a KITE. +++++--++++++++++ Being a kite, the diagonals intersect each other perpendicularly and the longer diagonal bisects the shorter diagonal. +++++++++ Then the proof will be shorter. We only have to evaluate the area of 🔺 ADB using Heron formula ++++++++++++++ and multiply the area with 4 to have the area of the rectangle.

[@babisstafulas7874]

IF K1 and K2 are the centers of two circles and A,B the common points of them then we have a 4 sides K1AK2B WITH 3,5,5,3 and diagonios 6. 2) Area of one triange is A=sq(τ(τ-α)(τ-β)(τ-γ))= sq(7*1*2*4)=2*sq(14) 3) From the type of area base*height/2 we have h=2*sq(14)/3 for K1AK2 triangle (base 6) 4) The other diagonios of K1AK2B is 4*sq(14)/3 5) The asking area is 6*4*sq(14)/3=8*sq(14)~29,93 s.u. Thank's NGE iF you can answer from what country are you from? Again thank's!

[@MelcomMulder]

South Africa

[@bpark10001]

There is another way to solve. Use Heron's formula to find triangle ABD area & divide by base to get height. Area is √(7*4*2*1) = 2√14. Height OD is (2√14)/3. CD = (4√14)/3. Area rectangle = ((4√14)/3)(6) = 8√14. Problems should be calculated with exact answers. The exact answer is 8 √14.

[@sorourhashemi3249]

Interesting. The center of small circle marks as O and big one M. Connect o to Q and Q to M. Focus on triangle OQM . OQ=3, QM=5 and OM=6. By Heron formula that area is 7.48. And the height of this triangle is ~2.5. 2.5×2=5 which is the width of t Rectangle . and area is 5×6=30

[@bjorntorlarsson]

Instead of calulating the height of the triangle, one can just multiply the area of triangle OQM by 4 to get the area of the rectangle. Move one part of the triangle, and put it along the diagonal of the rectangle to make it obvious. One doesn't need to use the height (or the half-width of the rectangle).

[@marioalb9726]

Triangle ABC, sides & semi perimeter : R= 5cm ; r = 3cm ; b=6cm ; s=7cm (¼A)²=s(s-a)(s-b)(s-c)=7(7-3)(7-5)(7-6) A = 8√14 = 29,93cm² ( Solved √ )

[@marioalb9726]

A₁=πR² --> R²=A₁/π --> R= 5cm A₂=πr² --> r²=A₂/π --> r = 3cm Triangle ABC: A₃²=s(s-a)(s-b)(s-c) A₃²=7(7-3)(7-5)(7-6) = √56 = 2√14 cm² Red shaded rectangle : A = 4A₃ = 8√14 = 29,93cm² ( Solved √ )

[@vishalmishra3046]

From area of the 2 circles, the radii = 3 and 5. So, area of triangle ABC = √ [ s (s-a) (s-b) (s-c) ] where s = (3+5+6)/2 = 7, so Area Δ = √ [ 7 x 4 x 2 x 1 ] = 2 x √14 Area of rectangle = 2 x each half = 2 x (2 x Area ΔABC) = 4 x 2 x √14 = 8 √14 = 8 x 4 x √(1 - 2/16) =~ 32 (1 - 1/16) = 32 - 2 = 30 (using binomial approximation of square root). So, Answer = 8 √14 =~ 30 square units.

[@concerned9890]

I just used Desmos distance formulas lol

[@gelbkehlchen]

Lösung: A = Mittelpunkt des großen Kreises, B = Mittelpunkt des kleinen Kreises, R = Radius des großen Kreises, r = Radius des kleinen Kreises, S = linker Schnittpunkt der beiden Kreise, T = rechter Schnittpunkt der beiden Kreise, M = Mittelpunkt der Strecke ST. Für die Fläche des großen Kreises gilt: π*R² = 25π |/π ⟹ R² = 25 |√() ⟹ R = 5 Für die Fläche des kleinen Kreises gilt: π*r² = 9π |/π ⟹ r² = 9 |√() ⟹ r = 3 Pythagoras: (1) AM²+SM² = R² (2) BM²+SM² = r² ⟹ (1a) AM²+SM² = 5² (2a) (6-AM)²+SM² = 3² ⟹ (2b) 36-12*AM+AM²+SM² = 3² ⟹ (1a) - (2b) = (3) -36+12*AM = 25-9 = 16 |+36 ⟹ (3a) 12*AM = 52 |/12 ⟹ (3b) AM = 52/12 = 13/3 |in (1a) ⟹ (1b) (13/3)²+SM² = 5² |-(13/3)² ⟹ (1c) SM² = 5²-(13/3)² = 25-169/9 = (225-169)/9 = 56/9 |√() ⟹ (1d) SM = √56/3 ⟹ Fläche des roten Rechtecks = 2*SM*6 = 2*√56/3*6 = 4*√(4*14) = 8*√14 ≈ 29,9333 Solution: A = center of the large circle, B = center of the small circle, R = radius of the large circle, r = radius of the small circle, S = left intersection point of the two circles, T = right intersection point of the two circles, M = center of the line ST. The following applies to the area of ​​the large circle: π*R² = 25π |/π ⟹ R² = 25 |√() ⟹ R = 5 The following applies to the area of ​​the small circle: π*r² = 9π |/π ⟹ r² = 9 |√() ⟹ r = 3 Pythagoras: (1) AM²+SM² = R² (2) BM²+SM² = r² ⟹ (1a) AM²+SM² = 5² (2a) (6-AM)²+SM² = 3² ⟹ (2b) 36-12*AM+AM²+SM² = 3² ⟹ (1a) - (2b) = (3) -36+12*AM = 25-9 = 16 |+36 ⟹ (3a) 12*AM = 52 |/12 ⟹ (3b) AM = 52/12 = 13/3 |in (1a) ⟹ (1b) (13/3)²+SM² = 5² |-(13/3)² ⟹ (1c) SM² = 5²-(13/3)² = 25-169/9 = (225-169)/9 = 56/9 |√() ⟹ (1d) SM = √56/3 ⟹ Area of ​​the red rectangle = 2*SM*6 = 2*√56/3*6 = 4*√(4*14) = 8*√14 ≈ 29.9333

[@marioalb9726]

A₁=πR² --> R²=A₁/π --> R= 5cm A₂=πr² --> r²=A₂/π --> r = 3cm Triangle ABC: A₃²=s(s-a)(s-b)(s-c) A₃²=7(7-3)(7-5)(7-6) = √56 = 2√14 cm² Red shaded rectangle : A = 4A₃ = 8√14 = 29,93cm² ( Solved √ )

[@marioalb9726]

A₁=πR² --> R²=A₁/π --> R= 5cm A₂=πr² --> r²=A₂/π --> r = 3cm Triangle ABC: A²=s(s-a)(s-b)(s-c)=7(7-3)(7-5)(7-6) A = √56 = 2√14 cm² = ½b.x x = 2A/b = 2*2√14/6 = 2/3 √14 cm Red shaded rectangle : h = 2x =4/3 √14 = 4,989 cm A = b.h = 6*4,989 = 29,93cm²