Calculate area of the Blue shaded Rectangle | Semicircle | Important Geometry skills explained

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Calculate area of the Blue shaded Rectangle | Semicircle | Important Geometry skills explained

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[procash]

Fantastic. What a seemingly unsolvable problem and what an amazing solution

Thanks PreMath Guru Ji

[ayushshinde]

you can construct a perpendicular to the chord of EC and OC would be length of the rectangle since its the radius of the circle and CD would be breadth and then use similarities
We get,
L/6√3=12√3/B
LB=6√3*12√3=216
(I have taken BC as length and AB as breadth)

[marioalb]

c = 12 √3 = 20, 785 cm

Similarly of triangles:
(c/2) / R = (R+x) / c
(R+x). R = (c/2). c = Area
(R+x). R = c² / 2
(R+x). R = 20, 785² / 2
Area = 216 cm² ( Solved √ )

The area of this rectangle with base R+x and height R
is equal to:
The area of a square with diagonal 12√3

[solimana-soli]

Call G the opposite of C along the diameter in the semicircle, then I noticed a similitude between GEC and DEC triangles (because they are rectangles and have an angle in common)
GC : EC = EC : DC
2r : 12sqrt(3) = 12sqrt(3) : DC
and we get
DC = 216/r
Area = r*216/r = 216

[vara]

Wish you were my school teacher. Thank you for the way the problem was solved. Very methodical.

[montynorth]

Interestingly, point E can be moved to coincide with point F and we then have a square FBCO with the diagonal of 12 root 3.
The area of that square is given by the formula 1/2 x diagonal squared.
1/2 x (12 root 3 )^2 = 216.

[michaelkouzmin]

I dare say there is some room left for optimization: if we leave aside EA and use only DE for our workout we will have a slight simplified process:
1) Let h= DE;
2) r^2= = h^2+x^2 => h^2= r^2-x^2;
3) (12*sqrt(3))^2= (r+x)^2+ h^2;
4) lets substitute h^2:
432= (r+x)^2+ (r^2-x^2);
432= r^2+x^2+2rx+r^2-x^2;
432=2r^2+2rx
AreaBlue = r^2+rx= 432/2=216.

[nareshkange]

The area of the rectangle is(1/2)*semicircle chord square. We can solve it by shortly if CF is the radius of your drawn figure of semicircle, then by tangent triangle are ∆ EDC & ∆ FEC are congruent triangles, so we have EC/FC=DC/EC, EC^2=FC*DC, 12√3 * 12√3= 2r * (x+r), 432= 2r * (x+r), 216= r * (x+r) = area of rectangle

[spiderjump]

Let r be the radius of the semicircle. And OD be a.
DE^2= r^2 —a^2
By Pythagorean theorem,
DE^2 + CD^2= 144•3
r^2 -a^2 +(r+a)^2=144•3
2r^2 + 2ar=144•3
r(r+a)=216
Area of blue rectangle = r(r+a)=216

[bsmith]

Lets call the other end of the semicircle's diameter point G. Now draw EG.
Triangle EGC is a right triangle from angle GEC subtending the diameter GC.
Triangle EDC is given as a right triangle.
Triangles EDC and GEC share common angle ECD/GCD.
Then triangles EDC and GEC are similar triangles.

From similar triangles we have GC/CE = CE/ED.
CE = 12*sqrt(3) is given.
Let the radius be r, then diameter GC equals 2*r and the height of the rectangle is BC=r.
CD is the length of the rectangle, call that x.
The area of the rectangle can be expressed as CD*BC = r*x.
Now just substitute and simplify: GC/CE = CE/ED -> (2*r)/(12*sqrt(3)) = (12*sqrt(3))/x -> r*x = 216.

The area of the rectangle is 216.

[vierinkivi]

otetaan taas yksinkertaisin ehdot täyttävä tapaus. Piste E siirretään puoliympyrän vasempaan nurkkaan . Ala on 12*sqrt3 * 6*sqrt3.

[rabotaakk-nwnm]

Интересно, что к этому времени только три пользователя догадались, что задачу можно решить простым и быстрым способом, основанном на том, что результат не зависит от положения точки E на стороне AD (угол DCE может изменятся от 0 до 45°, соотношение сторон ABCD - от 1:1 до 2:1).
Все они рассмотрели один крайний случай: E совпадает с A (угол DCE=45°, ABCD - квадрат).
Мне остаётся только рассмотреть другой крайний случай: E совпадает с D (угол DCE=0, ABCD - прямоугольник 2:1).
[ABCD]=12sqrt3×6sqrt3=216
Что и требовалось доказать!

[bigm]

Your approach to these minimal information projects is fascinating!🥂😀❤

[wackojacko]

Awesome! I think I'm ready now too tackle a Millineum Prize Prize Problem . I need the money!🙂

[misterenter-izrz]

Seems to be a interesting puzzle.😃 First let r be the radius, OD be x, so the answer is (r+x)r. Now DE^2=r^2-x^2, so 144x3=432=(r^2-x^2)+(r+x)^2=r^2-x^2+r^2+2rx+x^2=2r^2+2rx=2r(r+x), therefore my answer is 432/2=216.🙂

[vidyadharjoshi]

Draw a perpendicular OG to EC. Join OE. By calculations of angles. OD = OG. So DE = 6*sqrt3. Angle OCE = 30.
So DC = 18, OC = 12. So Area ABCD = 12*18 = 216

[ybodoN]

Generalization: the area of the blue rectangle is CE² / 2
Trivia: the area of this blue rectangle is 3³ + 4³ + 5³ = 6³

[sebastianoardita]

Ho ottenuto lo stesso risultato, molto più velocemente, con le proporzioni tra il rettangolo DCE e EC/2 OC . DC x OC = 6√3 x 12√3 = 216

[theoyanto]

Nice problem, square root of 3 was my homing beacon, then a spoonful of Thales and alakazam 🤓

[EPaozi]

On peut librement faire varier l'angle DCE. Prenons 45°. OCBF (ou ABCD, c'est le même) est alors un carré de coté c et dont la diagonale est 12.3^0, 5 . Ce qui donne : 2c^2=(12.3^0, 5)^2 c^2=216