Can you find area of the Green shaded region? | (Fun Geometry Problem) |

[@wackojacko3962]

Basic concept reviews are so revealing to solving. I absolutely love it. 🙂

[@PreMath]

Glad it was helpful! Thanks for the feedback ❤️

[@jimlocke9320]

Critique: At 0;45, a strategy is introduced, but it is to find the radius. I feel that the strategy should be to divide the figure into regions for which we know how to compute the areas, such as squares, triangles and circle sectors, and add and subtract the areas to obtain the area of the green region. The first step might be to construct the diagonals AC and BD of the square, which intersect at O. We observe two quarter circles, one with minor arc AB and the other with minor arc CD, and three triangles, ΔBOC, ΔAOD and the white triangle. We need the radius of the circle and then we have enough information to solve. We note that square ABCD has been divided into 4 congruent isosceles right triangles, ΔAOB, ΔBOC, ΔCOD and ΔAOD. We can take two of them and place them next to each other, with a common hypotenuse. The resulting square has the radius of the circle as a side, so its area is r² and half the area of square ABCD, or 338/2 = 169. So, r = √(169) = 13. So, we have 2 quarter circles with area (1/4)πr² = (1/4)π(13)², or 169π/2 for both, and 2 isosceles right triangles with sides r, so (1/2)(r)(r) for each, or r² = 169 for both. Now, we need to add these two areas and deduct the area of the white triangle. To compute the white triangle's area, we need the length of the diagonals of the congruent isosceles right triangles, which are equal to the sides of ABCD, and are 13√2. So, the white triangle has a base of 13√2 and height of 13√2, or area (1/2)(13√2)(13√2) = 169. So, blue area = area of 2 quarter circles + area of two isosceles right triangles - area of white triangle = 169π/2 + 169 - 169 = 169π/2, as PreMath also found. Note that, at 7:45, a shortcut is shown. The white triangle's area is equal to the sum of areas of 2 of the isosceles right triangles, so we really only need to sum the areas of two quarter circles.

[@yalchingedikgedik8007]

That’s very nice Thanks Sir Thanks PreMath Very nice method of solve With glades

[@unknownidentity2846]

Let's find the area: . .. ... .... ..... For reasons of symmetrie the green area outside the square has exactly the same size than the white area outside the square. The area of the white triangle within the square is exactly half of the area of the square, so its area has exactly the same size than the green area within the square. Therefore the green area within the circle has the same size than the white area within the circle and we can conclude: A(green) = A(white) = A(circle)/2 Since ABCD is a square and A, B, C and D are located on the circle, the triangle OAD is a right isosceles triangle. With R being the radius of the circle we obtain by applying the Pythagorean theorem: OA² + OD² = AD² R² + R² = A(blue) R² = A(blue)/2 Now we can finally calculate the size of the green area: A(green) = A(circle)/2 = πR²/2 = π*A(blue)/4 = π*338/4 = 84.5*π ≈ 265.46 Best regards from Germany

[@phungpham1725]

Thank you for a new approach!

[@unknownidentity2846]

@@phungpham1725 Thanks a lot for your kind feedback and best regards from Germany.

[@Mediterranean81]

area of the green area Circle area -(circular segments +unshaded triangle)-# circular segment area 1/2r(@ rads-sin@) if we draw two diagonals it will pass through the center and bisect the 90º angles to 45º lets focus on AOD triangle angle sum is 180º so 180º-45º-45º=angle AOD=90º=Pi/2 @ is angle AOD we have segments BC and AD are congruent since they got a common segment x so the area of the segments is 2*1/2r^2(Pi/2-1) =r^2(Pi-2/2) lets focus on the unshaded triangle now its area is the area of the square minus the shaded part let's call BE=y (E is the unlabeled vertex of the triangle) so EC= x-y area of CDE = x(x-y)/2=(x^2-xy)/2 area of ABE = xy/2 area of the square = x^2 so the unshaded triangle area is : x^2-(xy/2+(x^2-xy)/2) x^2-(xy+x^2-xy)/2 x^2-(x^2/2) = x^2/2 2r is the diagonal of the square so x=2r/sqrt(2) x= r*sqrt2 so the area of the triangle is (r*sqrt2)^2/2 2r^2/2 =r^2 go back to equation # circular segments = r^2(Pi-2)/2 triangle area = r^2 circle area = Pi*r^2 so green shaded area Pi*r^2-r^2/2(Pi-2)-r^2 factor by r^2 r^2(Pi-1/2Pi+1-1) r^2*1/2Pi 1/2Pi*r^2-& so the area of the shaded region is half of the circle let's calculate r we have the area of the square is 338 so the side is x= sqrt 338 =13sqrt2 the diagonal is 2r and also x*sqrt2 so 2r=13*sqrt2*sqrt2 2r=13*2 r=13 lets plug 13 in equation & 1/2Pi*r^2=1/2Pi*13^2=169/2 Pi=84.5Pi

[@marioalb9726]

Area = Half square area + 2x Circular segment area Asq = 338 cm² = ½d² d = 26 cm, R = 13 cm A = ½ Asq + 2 . Acs A = 338/2 + 2.[ ½ R² (α - sin.α)] A = 169 + 13² (90°-sin90°) A = 265,46 cm² ( Solved √ )

[@laxmikantbondre338]

One Step missing. When you connect points A and C, you need to mention why it will pass through point O. The reason is as ABCD is a square angle B is a right angle. So according to Inverse of Thale's Theorum, AC is a Diameter and hence it passes through center of the circle i.e. O.

[@quigonkenny]

As ABCD is given to be a square and is fully inscribed in circle O, then AB = BC = CD = DA = s, ∠AOB = ∠BOC = ∠COD = ∠DOA = 90°, and (as both a diagonal of the square and diameter of the circle) AC = DB = √2s = 2r. The green shaded region of the figure will be as follows: [1] The area inside the blue square will be equal to half the area of the blue square, as the area of a triangle with the same base and height as the height and width of a rectangle is half that of the rectangle. A1 = s²/2 [2] The area outside the blue square will be equal the area of two circular segments, each covering 90° of the circle. This latter amount will be equal to two 90° sectors, or quarter circles, which is equal to half of the circle, minus two isosceles right triangles, or quarters of the square, which is half the area of the square. A2 = 2πr²/4 - 2s²/4 A2 = πr²/2 - s²/2 A1 + A2 = s²/2 + πr²/2 - s²/2 A(total) = πr²/2 This means that the shaded area inside the square from [1] cancels out with the triangular value from the circular segment calculation from [2], leaving the total area as one half of the circular area. Thus we need only find the value of r to calculate the area. 2r = √2s (2r)² = (√2s)² 4r² = 2s² 4r² = 2(338) = 676 r² = 676/4 = 169 A(total) = πr²/2 = 169π/2 sq units

[@MrPaulc222]

The green area that lies within the square is half the square's area, so 169 sq un. Square's side lengths are sqrt(338), so the diagonal is sqrt(676), which is 26. Half of that is 13, which is also the radius. Area of full circle is 169pi. Areas of sectors OCD + OAB is the equivalent of a semicircle, so (169/2)*pi. As half the square is green, and half of the circle's area outside the square is also green, (169/2)*pi is the green area. This is approx 265.46 sq un. Yes, I see we got their in different ways, but the essentials were the same.

[@phungpham1725]

1/ By observation we can see that the area of the green region inside the sqare = 1/2 that of the square. Just draw 2 diagonals, we find out that the sum of the green areas = 2 x( 90 degree sector of the circle of which the diameter is the diagonal of the square. So, area of the green= 1/2 area of the circle The side of the square= 13 sqrt2--> the diagolnal= 13sqrt2 x sqrt2=13x2 Area of the green= 1/2 pi sq 13= 1/2 pi . 169 sq units

[@srirajan1933]

Alternate way to compute green area (apologies if already mentioned below, did a quick check but couldn’t easily find): after computing the radius as the professor had done, instead of reshaping the green triangles into right triangle equivalents at the circle’s center to form part of the sector: subtract the scalene white triangle area (1/2 * base AD * height AB) + (circle area - square area)*1/2 (for just 2 white chords on left and right sides) from the circle area. Essentially, leaving just the 3 white portions remaining

[@srirajan1933]

I do like the Professor’s area reshaping area technique!! He is fantastic with all his creative problems!! Thanks for all you do!!!

[@robertlynch7520]

I'm kind of amazed at how much math y'all did! WAY too much for me. Taking a close look at the diagram, I see that it is perfectly symmetric because of the square. This means that the upper and lower (green) lenses are half of the total of four of them. Moreover, the triangle cutting the square into parts does so into exactly half. So… the green area equals the white area. And either area would thus be ½ of the area of the circle. Then to figure its area, the square is 2 × 13 × 13 square units. (338 u²) Thus each side must be 13√2 units. And the diagonal across the square is (13√2)√2 or 26 units. And the radius is half that … or 13 units. Well, half the area of the now well-defined circle is [1.1]  ½π13² = 169π/2 ≈ 265.5 u² And that's that! ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅

[@gelbkehlchen]

Solution: a = side of the square = √338, r = radius of the circle = √[2*(√338/2)²] = √[2*338/4] = √[338/2] = √169 = 13. Green area = area of the square - area of the white triangle + 2*area of a section of a quarter circle = a²-a²/2+2*(π*r²/4-r²/2) = a²/2+π*r²/2-r² = 338/2+π*13²/2-13² = 169+π*169 /2-169 = 84.5π ≈ 265.4646

[@marcgriselhubert3915]

side length of the square: c = sqrt(338) = 13.sqrt(2),; radius of the circle: R = (sqrt(2)/2).c = 13. Area of the circle: 169.Pi; Area of the circle outside the square: 169.Pi - 332; Area of the green region outside the square: (1/2).(169.Pi - 332) Area of the green region inside the square: (1/2).332. Finally, green area: (1/2).(169.Pi - 332) +(1/2).332 = (1/2).(169.Pi) = (169/2).Pi.

[@santiagoarosam430]

Área ABCD =338=(2r)²/2→ r²=338/2=169. Área verde =(338/2)+[(πr²-338)/2]=πr²/2 =169π/2 =265,4645........ Gracias y un saludo cordial.

[@himo3485]

338=√338*√338 r=√338/2*√2=√676/2=26/2=13 13*13*π - 338 = 169π - 338 area of Green shaded region : (169π - 338)*2/4 + √338*√338/2 = 169π/2 - 169 + 169 = 169π/2

[@bernhardmuller4436]

r2 = 338 /2

[@OnlengKu-oj9bu]

即求半圓面積！

[@giuseppemalaguti435]

Ag=π169/2

[@kalavenkataraman4445]

169 pi/2

[@GillesF31]

Very nice problem I tried to solve with one formula: The answer is: ■ green area: (338·(π/2) - 338)/2 + 338/2 = 265.46 cm² ~~~~~~~~~~~~~~~~~~~~~~~~~~ Explanations (step by step): • 338·(π/2) is the circle area (*) • 338·(π/2) - 338 is the area of the 4 circular segments • (338·(π/2) - 338)/2 is the area of 2 circular segments • 338/2 is the half of the square area (**) • (338·(π/2) - 338)/2 + 338/2 is the green area (final answer) ■ green area = 265.46 cm² ----- notes ----- (*) property: • the area of a circle is π/2 (≈ 1,570796) multiplied by the area of its inscribed square (**) property: • the area of ​​a triangle inscribed in a square, having as base one of the sides of the square, is equal to half the area of ​​the square 🙂

[@LuisdeBritoCamacho]

STEP-BY-STEP RESOLUTION : 1) Blue Square [ABCD] Area = 338 Square Units 2) Side of Blue Square = sqrt(338) = 13*sqrt(2) Linear Units 3) Diagonal of Blue Square = 26 Linear Units 4) Radius of Circle = R = 26/2 ; R = 13 5) Area of Circle = 169*Pi Square Units ~ 531 Square Units 6) Area of the 4 Minor Segments = 531 - 338 = 193 Square Units 7) Area of each Minor Segment = 48,25 Square Units 8) Area of Half Square = 169 Square Units 9) Area of Green Shade Region = 169 + (48,25 * 2) = 169 + 96,5 = 265,5 Square Units 10) ANSWER : The Area of the Green Shaded Region is equal to approx. 265,5 Square Units.

[@sorourhashemi3249]

I didn't get it

[@sergeyvinns931]

a=13\/2; r=13. Circle area = r^2*pi. Area of two green segmenfs (530,66-338)/2=96,33. Area triangle =a^2/2=169, Area of the Green shaded region = 169+96,33 = 265,33!

[@erwinkurniadi1850]

Ini jawaban yang benar.

[@alexreid-wh9gq]

Must have missed something on this one. The triangle is not an Isosceles triangle. The 2 green areas are not symmetrical.

[@srirajan1933]

He’s reshaping the triangle area equivalents to form 4 equal right triangle parts such that 2 right triangles equals 169 square units

[@waheisel]

You are correct that it isn't isosceles. And PreMath's proof that A (the white part of the square) = A1+A2 (the 2 green triangles) is incomplete. But it is nonetheless true. Here is a proof: Let x be the length of the sides of the square. Then the area of the white triangle will be (x^2)/2. Let E be any point on CB. The area of the upper triangle will be EC*x/2, and for the lower triangle is is EB*x/2. Add those two areas, EC*x/2 + EB*x/2 = (EC+EB)*x/2 EC+EB is CB, a side of the square, or x. So the sum of the green triangles, like the white triangle, equals (x^2)/2, each half of the square's total area. The proof also works for non-square rectangles.