Can You Find the Area of the Red Rectangle?

[@sorourhashemi3249]

Thanks. Connect B to O and have triangle BCO in which it is similar to triangle ABC.( two angled and side BC), and mark CD as X. So : 2R-X/2=R-X/R===> R =1 and area of rectangke 2×1=2

[@jamesfaizi7771]

All four methods are correct and very interesting I came out with the fifth method that I saw many other people have the same idea as you see in comments because it is the shortest way to find the area of the rectangle so every time this chord touches the side of rectangle where intersect with The Arc of semicircle no matter how big or smal itl is . just square the chord and divided by 2 and that would be the area of the rectangle 😃

[@txikitofandango]

circle is centered at origin. Its equation is x² + y² = r². If the coordinates of the intersection of the semicircle and right edge of the rectangle are (x,y), then by the Pythagorean Theorem we know (r+x)² + y² = 2², and expanding yields r² + 2rx + x² + y² = 4. Substitute in r² from the equation of the circle to get r² + 2rx + r² = 4. Simplify to get 2r(r+x) = 4. Divide by 2 to get r(r+x) = 2. And this expression r(r+x) is the area of the rectangle. Therefore, the area is 2. No radicals, no needing to solve for x.

[@nandisaand5287]

3* different methods, and I couldn't come up with a single one. SAD! I love that you show multiple ways to attack problems. * I don't think Method #4 holds up. If you make AB=10: Using the Limit Method, b=10, a=5 Area=½5•10=25 Using the Intersecting Cords Method, 2ab=100 Area=ab=½•100=50. Unless I'm doing something wrong, #4 just happens to work when AB=2, not with other #s.

[@oscarcastaneda5310]

It can be shown that the rectangle's area is "Half the Square of the given length."

[@santiagoarosam430]

Si desplazas el extremo derecho de la cuerda propuesta hasta el extremo del radio vertical del semicírculo, el rectángulo se transforma en un cuadrado cuya diagonal mide 2 ud y la superficie buscada es 2²/2=2, quedando probada tu afirmación.

[@ThePhantomoftheMath]

@@oscarcastaneda5310 Yes! Check Method 4 in the video.

[@ludosmets2018]

I am not good at geometry but immediately saw the fourth method (limit case). I am quite proud of myself.

[@ThePhantomoftheMath]

@@ludosmets2018 You should be. Not a lot of people see that.

[@nenetstree914]

Using Multiple-angle formulae is an another way 😀

[@rgcriu2530]

Magnifica edición 👍👍

[@KipIngram]

Well, the point on the circle where the diagonal line intersects it is not specified. So the answer must not depend on that. That means we're free to put it anywhere. Let's solve it two different ways. First, let's lie the line flat, so it becomes a diameter of the circle. That means D=2, R=1. So the rectangle has width 2 and height 1 and the area is 2. Now let's let the line intersect the circle at the top. That means two is the diagonal of the square, and so the side length is 2/sqrt(2) or sqrt(2). And once again the areas is 2. So, the area is 2 independent of the diagonal line angle. Q.E.D.

[@santiagoarosam430]

Respetando las condiciones del trazado de la figura, podemos desplazar el extremo de la cuerda propuesta hasta el extremo del radio vertical del semicírculo; con esta transformación, el rectángulo se convierte en un cuadrado cuya diagonal mide 2 ud y su superficie es equivalente a la del rectángulo original → Área del rectángulo rojo =2²/2 =2 ud². Interesante problema. Gracias y un cordial saludo.

[@jimlocke9320]

Let's use the designations for points used in Method #1. In addition, let's designate the point of tangency between the half circle and CD as point F and the other end of the diameter chord as G. The problem statement implies that point E lies on the arc between F and G and that the area of the rectangle is the same regardless of where point E is located. Method #4 covers the limit case where points B, E and G coincide (although it uses different designations for the points). There is a second limit case where E and F coincide and O and B also coincide. ABCD becomes a square with diagonal length 2. The sides of the square have length √2. The area of the square is (√2)(√2) = 2, same as found by the other methods. We could plot the area of the square on an x-y coordinate system as y where x is the arc length from F to E. If area does not change, as implied by the problem statement, the plot is a horizontal line segment. We do not expect a discontinuity between x = 0 and x greater than 0, nor a discontinuity where the arc length approaches π/2. So, the computed areas at these points should equal the area computed for points in between. We can prove that there is no discontinuity by solving the general case, but there should be other ways.

[@Celanish]

A(adcb) = (AB).(AD) (AE)^2 = 2(AO).(AB)= 2(AD).(AB)=4=2(abcd) !

[@JoeCarsto]

Pythagorean Theorem for the win! 🥇

[@drJavi]

This was fun to solve. Thanks for sharing.

[@ThePhantomoftheMath]

Thanks for watching! :)

[@colina64]

👍

[@khalidaladaili7331]

Another cheating method is if we select the arbitrary point to make it square, then the radius becomes the square root of 2 => Area =2

[@jamesfaizi7771]

The square root of 2 is 1.4142....not 2 but if you square the square root of 2 and then divide it by 2 you will get 1 which will be the area of square not rectangle only if the radius is 1 in this case the radius is actually more than 1😉

[@Eyalkamitchi1]

How does the third method work? Didn't you change the area of the rectangle??

[@ThePhantomoftheMath]

@@Eyalkamitchi1 Hi. This is just a mathematical trick to create a full circle, which allows us to use the intersecting chord theorem. While it's true that we doubled the area of the rectangle, this doesn't affect the outcome, as we're only interested in the product of the rectangle's sides. Switching to a full circle simply made it easier to find that product.

[@dakcom-mk6mp]

Nice

[@motogee3796]

Easy

[@Bayerwaldler]

You could rule out the last, „cheating“ solution by setting that line segment equal to „D“ and then say „compute the area of the rectangle as a function of D and the radius R of the semi circle“. That will cause a little surprise at the end. 😊 Nice problem nevertheless!

[@nenetstree914]

2

[@gelbkehlchen]

The 2nd method is faster.

[@khalidalward3483]

The quadratic solution proves that r = 1 & 2r = 2 So the area = 2 And this is the area of a rectangle whose shape is larger than the rectangle of the question, and whose area was found in the previous solutions, namely = 2 So there is an error in the fourth solution

[@jamesfaizi7771]

You're right this rectangle area is more than 2 units squared but he says what if the chord and diameter of the semicircle are same 2 units each 🙂

[@julianarce3070]

This cannot be solved

[@vivektulja4516]

You never encounter such problems in real life. That's why people who do well in Math Olympiads don't do well in real life.

[@MarieAnne.]

Really? You've got proof of that?

[@vivektulja4516]

@MarieAnne. I am speaking from personal experience. I am an aerospace engineer, and have dealt with the most complex geometry (and math) problems but never one like this problem.