Can you find area of the Green shaded triangle? | (Semicircle) |

Learn how to find the area of the Green shaded triangle in the semicircle. Important Geometry and Algebra skills are also explained: Pythagorean theorem; area of the triangle; area of the rectangle formula; similar triangles. Step-by-step tutorial by PreMath.com
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Can you find area of the Green shaded triangle? | (Semicircle) | #math #maths #geometry
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[@parthtomar6987]

Very nice and easy solution sir with no mistake thx🎉

[@PreMath]

Glad to hear that! You are very welcome! Thanks ❤️

[@jimlocke9320]

Very well done! A minor shortcut. At 8:20, we can treat CD as the base of ΔBCD, then the length of CF or DP becomes the height. Applying the area of a triangle formula, A = (1/2)bh, (1/2)(2√(34))(√(30)) = (√(34))(√(30)) = √(1020) = 2√(255), as PreMath also found. When taking the area of a right triangle, we are used to taking the product of the sides and dividing by 2. When the hypotenuse and the distance from the right angle's vertex are known, we may overlook the fact that we can treat them as base and height in the area formula!

[@PreMath]

Excellent! Thanks ❤️

[@KipIngram]

Note to channel - this is a really nice problem. I actually had to think about this one for a while. -- Kip --- Let points C and D be at locations (x, y) and (-x, y), respectively. Momentarily consider point B the origin of a 2D vector space. Consider the vectors BC and BD. These are perpendicular, so their dot product must be zero. Let's write that out and simplify: [ (x+2)*i + y*j ] dot [ -(x-2)*i + y*j ] = 0 -(x-2)*(x+2) + y^2 = 0 -(x^2-4) + y^2 = 0 -x^2 + 4 + y^2 = 0 x^2 - y^2 = 4 We also know that x^2 + y^2 = 64, so: x^2 + y^2 = 64 x^2 - y^2 = 4 2*x^2 = 68 x = sqrt(34), y = sqrt(30) x = sqrt(2*17), y = sqrt(2*15) Now, the area is just 2*x*y / 2 = x*y, so Green_Area = 2*sqrt(255) Q.E.D.

[@hookahsupplier.5155]

Your solution escorted with unmatched clarity and honed problem solving skills are amazing and pretty inspirational!

[@JLvatron]

Wow!

[@DB-lg5sq]

شكرا لكم على المجهودات يمكن استعمال a=COE 180-a=COB 180-2a=COD x=BC y=BD x^2=8^2+2^2-2×2×cos(180-a) x^2=67+32cosa y^2=8^2+2^2-2×2×8cosa y^2=68-32cosa x^2+y^2=136 DC^2=136 DC^2=8^2+8^2-2×8×8cos(180-2a) cos2a=1/16 cosa=(جذر34)/8 x^2=68+4(34جذر) y^2=68-4(34جذر) x^2 × y^2=4080 xy=4(34جذر) S=1/2 xy S=2(255جذر)

[@PreMath]

Thanks ❤️

[@tombufford136]

Taking a second glance, A very fast technique for this one, draw a rectangle DCGF, where G and F are the perpendicular intersections from D and C to AE. Reflect the rectangle about axis AE to form a square, noting DC is parallel with AE with D and C on the circumference of the semicircle radius, 16/2. The green shaded area is half the area of the rectangle = 1/4 the area of the square. Then the green triangle area = 0.25 * (Diagonal^2) /( sqrt(2)^2 )= 0.25 * 16^2/2 = 32 square units

[@soli9mana-soli4953]

Being AP=FE=x and applying Euclid theorem on ADE we have: DP²=x*(16-x) [1] On DBC we have: DP² =(6-x)*(10-x) [2] Comparing [1] with [2] you find x 16 - 2x is the hypotenuse of the green triangle √ x*(16-x) is its height

[@PreMath]

Thanks ❤️

[@limfilms1089]

AP = FE = x. (a) Δ DBC ~ Δ BPD ~ Δ BFC. Take ratios y^2= (10-x)(6-x) (i) (b) Draw right triangle Δ ACE. Δ ACE ~ Δ AFC ~ Δ EFC. Take ratios y^2= x(16-x) (ii). (c) thus (i) and (ii): (10-x)(6-x)=x(16-x), so x= 2.07 approx. (d) thus y=5.36 approx. (e) DC= 16-2x = 11.86 approx. (f) A = (1/2)(11.86)(5.36)= 31.8 aprox.

[@ybodoN]

Draw the chord GBH perpendicular to AE. Then GB = BH = √(6 · 10) = √60 and DP = CF = √(60 / 2) = √30. Since the radius of the semicircle is (10 + 6) / 2 = 8 then OP = OF = √(64 − 30) = √34 and PF = 2√34. Therefore, the area of the green triangle is ½ 2√34 √30 = √1020 = 2√255 ≈ 31.937 square units.

[@PreMath]

Thanks ❤️

[@LuisdeBritoCamacho]

I solved this problem geometrically. Let the coordinates of point A be the Origin (0 ; 0) Let the coordinates of point B equal (0 ; 6) The Equation of the given Circle is: (x - 8)^2 + y^2 = 64 The coordinates of point D are the interception of (x - 8) ^2 + y^2 = 64 and x^2 +y^2 = 36 Solution: x = 9/4 and y = (3*sqrt55))/4 Drawing a Straight Line, y = 3*sqrt(55)/4, I can calculate the Distance between points D and C; wich is 11,5 lin un (13,75 - 2,25) Now, knowing the Base of triangle [BCD] and the height = (3*sqrt(55))/4 I can calculate the Area of triangle [BCD] Area = [11,5 * (3*sqrt(55))/4] / 2 = (34,5 * sqrt(55)) / 8 ~ 255,859 / 8 ~ 31,982 sq un Answer: The Green Triangle Area is equal to approx. 31,982 Square Units.

[@PreMath]

Thanks ❤️

[@marcgriselhubert3915]

Let's use an adapted orthonormal as often, center O, first axis (AB), the radius of the semi circle is 8, its equation is x^2 + y^2 = 64. Let's nama a the distance from D or C to (AB), then C(sqrt(64 -a^2); a) and D(-sqrt(64 -a^2), and B(-2; 0) Then VectorBC(sqrt(64 -a^2) +2; a) and VectorBD(-sqrt(64 -a^2) +2; a). These vectors are orthogonal, so: (sqrt(64 - a^2) +2).(-sqrt(64 -a^2) +2) + a.a = 0 or 4 - (64 -a^2) + a^2 = 0, so 2.(a^2) = 60 and a= sqrt(30) Then the length of the green triangle is DC = 2.sqrt(64 -a^2) = 2.sqrt(34) and its height is a = sqrt(30) and finally its area is (1/2).2.sqrt(34).sqrt(30) = sqrt(1020) = 2.sqrt(255)

[@PreMath]

Excellent! Thanks ❤️

[@aryankushwaha7028]

We can easily do it with symmetry and chrod intersection property

[@aryankushwaha7028]

Do this thinking is correct please , think about it ,

[@PreMath]

Thanks ❤️

[@LENAKSOY]

Awesome🎉

[@PreMath]

Thanks ❤️

[@tombufford136]

At a quick glance, Rotate Triangle BDC about the line AE to form a square in the Circle with diagonals 2 * radius = 10+ 6 = 16. Then dc = 16 /SQRT(2) and height = 8/sqrt(2) Then area of Green shaded triangle = 0.5 / 2 * 16 * 8 = 32 square units.

[@ybodoN]

To get a square, B and O should be the same point 🧐 (since DCFP should be a domino, implying OF = OP)

[@tombufford136]

Thanks for your comment, In reply , The area of a Triangle is half * Base * Height. DC is parallel to AE and the height and area does not alter if you move the B along AE. To form a square and circle by rotating DC and the semicircle, 180 degrees about axis AB The height is half DC to form a square with sides DC in length @@ybodoN

[@ybodoN]

@@tombufford136 DP = CF = √30 and OP = OF = √34. So, by rotating, you get a rectangle which is 2√30 × 2√34. It is very close to 2√32 × 2√32 but the green area is 2√255 *≈* 31.937, not 2√256 *=* 32 😉

[@PreMath]

Thanks ❤️

[@PreMath]

Thanks ❤️

[@ybodoN]

In any similar construction where a rectangle is divided into three similar right triangles, DP = CF = √AB × √BE / √2 🤩

[@PreMath]

Thanks ❤️

[@arthurschwieger82]

Is there a rule that says two parallel chords of a circle are equally divided by a perpendicular line that runs through the center of the circle?

[@johnwilson839]

I don't understand where the fact that dc is parallel to ae comes from?

[@jamestalbott4499]

Thank you!

[@PreMath]

You are very welcome! Thanks ❤️

[@ap3xmath123]

Brilliant absolutely brilliant

[@PreMath]

Glad to hear that! Thanks ❤️

[@Atharva-jf7cq]

Thanks for giving us knowledge by tricky question which were asked in competitive exams

[@PreMath]

It's my pleasure! Glad to hear that! Thanks dear ❤️

[@dirksteele1477]

Why are DP and CF equal in length? Was it stated that CD and AE are parallel?

[@Abby-hi4sf]

Yes , it was stated parallel, by the arrow sign

[@孫西萍]

Why does angle FCB equals angle PDB, and angle CBF equals angle BDP?

[@Teamstudy4595]

1st View ❤

[@PreMath]

Super Thanks ❤️

[@StephenRayWesley]

(6)^2 =36 ,(10)^2 =100 (3x(15°)=45°x 3x(15°)=45°x (90°+45°x+45°x)=180°x^2 (36+100)=136 (136-180°x^2)=√44x^2 4^√11x^2 4^√11^1x^2 √4^1^1x^2 √2^√2√1^√1x^2 √1^1x^2 1x^2 (x+1x-2)

[@PreMath]

Thanks ❤️

[@plenin6072]

How can you assume that AB and DC are parallel? The problem description doesn't mention anything of that sort.

[@PreMath]

Yes , it was stated parallel, by the arrow sign

[@sergeyvinns931]

S = BD*DC*sin90/2 = OD*OC*sin90/2 = 8^2/2 = 32! sin90 = 1!

[@PreMath]

Thanks ❤️

[@misterenter-iz7rz]

x^2+y^2=8^2=64, ((x-2)^2+y^2)+((x+2)^2+y^2)=4x^2, x^2-y^2=4, so x^2=34, y^2=30, therefore the area is 1/2×(2x)×y=xy=sqrt(30×34)=2sqrt(15×17)=2sqrt(255)=31.9 approximately 😅

[@PreMath]

Thanks ❤️

[@wackojacko3962]

🙂

[@PreMath]

😀 Thanks ❤️

[@giuseppemalaguti435]

r=8...Ag=√(8^2-h^2)h....dai calcoli risulta h^2=30....Ag=2√255

[@PreMath]

Excellent! Thanks ❤️

[@s900991]

it is 5^21 ?

[@Teamstudy4595]

Ans 32

[@PreMath]

Thanks ❤️

[@prossvay8744]

Area of the green triangle=1/2(2√17+√34)(2√17-√34))=2√255=31.94 square units.❤❤❤ ThAnks.

[@PreMath]

Great! You are very welcome! Thanks ❤️

[@jessewallis6589]

Since the height and width will stay the same no matter where the right angle is on the diameter, you can just move the vertex to the center giving you a 45-45-90 triangle with 8 as each leg. So the area is 32.

[@ybodoN]

If you move the vertex at the center, ∠DOC will be approximately 93.58° 🧐

[@PreMath]

Thanks ❤️

[@misterenter-iz7rz]

My method is clumsy for neglecting a pair of similar triangles in the figure.

[@PreMath]

No worries! Thanks ❤️

[@misterenter-iz7rz]

It is very clumsy and difficult puzzle. 😅😅😅

[@PreMath]

Thanks ❤️

[@pralhadraochavan5179]

Good night sir

[@PreMath]

Thanks dear ❤️

[@EPaozi]

DC // AE ?????????

[@brudo5056]

at the start it's 'said' to be parallel ...