I'm still lost in translation, because I can't figure out what "esquare root" is...

Call angle MDC θ, this angle is congruent to angle PAB. Call DC = AB = w. We want wcos(θ)) + 4 which is 6 + 4 = 10. To help "picture" this ... From C draw the perpendicular to side x : )

Drop a perpendicular line from D to line BC. Call that E. Draw a perpendicular line from A to line DC. Call that F. Now you have two triangles with the same length hypotenuse and opposite angle of 90°. 7X + 90 + 2X = 180 9X = 90 X = 10

10, Just draw a line parallel to the base passing through C; a triangle equal to APB is formed, with a height of 6, then add 4 of CQ. The max height of a rotating rectangle is always equal to the sum of height of A and C, it is obvious.

Completing the rectangle with base PQ and height 10. Checking out that D will be along the top edge, and the area [ABCD] is 5.PQ At first this is just a likely guess, which would give DM to equal 10 X=10? Seems a very quick answer. A horizontal line , EJ, across half way between the height of A and the height of C will start on AP and will meet QC produced at J. The very small right-angled triangles EAF and JCH are congruent, with F where EJ crosses AB, and H where EJ crosses CD because AB and CD are parallel Their sides are 1/6 of the corresponding sides of the triangle PAB . A triangle congruent to PAB is GCD which is in the top right corner of the larger rectangle So this shows that the top edge of the rectangle which is 10 units high does pass though D. X = 10 , answer.

別解法　欠けた部分に直角三角形を足すと、長方形になる。三角形の相似、合同から長方形の縦の長さは４＋６＝１０となる。

Tive que assistir esse segundo método duas vezes para entender ! Que maravilha !

By the theorem of the chord, any chord AB of a circumference of a determined radius r is equivalent to its diameter multiplied for the sinus of one of the angles at the circumference pointed to the chord: AB=2r*sin(a) Since you have calculated x and the sinus of theta, then you will have x=2r*sin(theta), so r=x/(2*sin(theta))

I like that you had a second different solution ready for it, I know the 1st solution is faster, simpler, and more intuitive than the second (at least for me and many others), but seeing a second solution made me feel better. I do suggest that you try generalizing the answers though, like, in this case it'd be fun to show that x is gonna be 10 even if we change the size of the rectangle, or even the ratio of its dimensions, as long as the 6 and 4 remain. Of course, animating that would be cool, but since that might be a bit of a drag, and takes a lot of extra time, and you also might not have the knowledge or software to do that (no offense there ✋, I know many don't, me included), but I believe that'd attract ALOT of extra traffic to your videos (views, likes, even subscriptions)

Extremely easy : x = 4+6 = 10 cm ( Solved √ )

Area = 20😅

20😂

Trazo AT//PQ. AT=PM. Ángulo TAD=ángulo QBC, correspondientes entre paralelas. AD=BC, lados del rectángulo. Entonces: triángulo ATD congruente con triángulo BQC (LAL). Luego: TD=CQ= 4 u. AP=TM= 6u. X= TD+TM= 4+6= 10 u.

You're kidding ? it took me 0.15143 second to find the answer ...

La paralela a PQ por D, corta las verticales PA y QC en E y F respectivamente ---> AED y CQB son congruentes ---> MD=PE=PA+AE=PA+CQ =6+4=10. Gracias y saludos

Here is sothing else: Be t= angleQBC and BC'D'A' a rectangle of same dimensions than BCDA with C' on (BQ). Then BCDA is the image of BC'D'A' in the rotation center B and angle t, whose analytic expression is x' = cos(t).x - sin(t).y with y' = sin(t).x + cos(t).y. Be L = BC = BC' and l = BA = BA' the side lengthes of the two rectangles. We have C'(L; 0), so C(L.cos(t); Lsin(t)) with L.sin(t) = 4, we also have A'(0; l), so A(-l.sin(t); l.cos(t)) with l.cos(t) = 6, Finally we have D'(L; l), so D(L.cos(t) - l.sin(t); L.sin(t) + l.cos(t)) with X = L.sin(t) + L.cos(t) = 4 + 6 = 10.

It seems so simple! Draw a line from [c] to line [dm], parallel to the [pq] base line. Call the intersection point [F]. The triangle [DFC] is congruent to triangle [APB] because the rectangle's parallel sides give the same angles and lengths. Therefore, line [FD] is 6 units long. By the same reasoning, the segment [FM] is 4 units long because it too is by definition parallel to the [QC] segment. Therefore X = 6 + 4 = 10 units. No sines, cosines, wiggly piggly geometry required. Just an intersection point and a pair of congruent triangles.

drawing the perpendicular CH to line DM we get right triangle DHC that is congruent with APB (note infact that are AP // DM and AB // DC so angles in A and in D are equal), then DH = 6 and HM = 4 DM = 6 + 4 = 10

This can also be solved by drawing line to left from point C to meet DM at point N'. N'M = 4. Triangle DN"C is congruent to triangle APB so N'D = 6. X = 4 + 6 = 10.

Its very simple to calculate. just consider the rectangle is at angle 'k' with the base and smaller side of rectangle is 'a'. Hence cosk=(x-4)/a and in other side cosk=6/a. So, (x-4)/a = 6/a, hence x=10

х=10 - от AD легко построить ещё треугольник и доказать его равенство.

∎PQRS → PS = PA + AS = 6 + 4 = QC + CR = PA + QC = 6 + 4 → x = 10 en detail: φ = 30φ; AB = CD = a; BC = AD = b; AC = BD = y; DM = x = ? PBA = CBQ = δ; PQ = k = PB + QB = m + (k - m); PA = 6; QC = 4; sin⁡(ABC) = 1 → 1/6 + 1/4 = 5/12 = 1/m → m = 12/5 → tan⁡(δ) = 2/5 = 4/(k - m) = 20/(5k - 12) → k = 62/5 → k - m = 10; ABD = β; DBC = α → α + β = 3φ → b = 2√29 → a = (6/5)√29 → y^2 = (4/25)29(34) → y = (2/5)(√29)(√34) → sin⁡(α) = 3√34/34 → cos⁡(α) = 5√34/34 sin⁡(δ) = 2√29/29 → cos⁡(δ) = 5√29/29 → sin⁡(α + δ) = sin⁡(α)cos⁡(δ) + sin⁡(δ)cos⁡(α) = 25(√34/34)(√29/29) = x/y → x = (2/5)(√29)(√34)25(√34/34)(√29/29) = 10

Construct triangle AQ'D, with Q'=90° and PQ'=MD. Therefore triangle AQ'D is congruent to triangle CQB, because, ∠QCB=∠ABP=∠Q'AB, ∠QBC=∠PAB= ∠Q'DA and AD=AC.Thus, AQ'=4, we conclude that x=10.

Thumb without side with 7 unitis.

Method 0 : a simple translation. Immediate solution without calculating.

10

We put PB=a and <PAB=<CBQ=<MDC=α and from it tan(α)=a/6=MQ/(x-4) so MQ=(a(x-4))/6 and according to Pythagoras' theorem we find [a(x-4)²/6]+(x-4)²=a²+36 and (x-4)²(a²+36)=36(a²+36) and from it (x-4)²=36 so x-4=6 and from it x=10

Congruence best me

Drop a perpendicular CK on DM. ∆DKC is Congruent to ∆APB, hence DK = AP = 6 X= DK+ KM = 6+4=10

Second method was brilliant!

My Method : In triangle ABC, M is the center of gravity ( meeting point of the medians PC and AQ ) AM = 2*MQ and MC = 2*MP ---> Homothety of center M and ratio k = 2 transforms MQBP into MADC then ratio of area is k^2 = 4 ---> area shaded MADC = 4* area MQBP = 4*6 = 24

PBQM∞ADCM (1² : 2² = 1 : 4) ADCM = 24

I used the second method to solve

I dont understand why BPM= BQM

I don’t understand how you just chose 4??? Obviously, the answer isn’t 1 or 2. I figured it would be a whole number/integer from 3 to 5, but why just select four?

AB II DC and NOT AB II CD.

(6)^2=36 {90°A90°B+90°C+90°D}=360°ABCD/36=10ABCD 5^5 2^3^2^3 2^1^1^3 23(ABCD ➖ 3ABCD+2).

Solution#1: It can be shown that the quadrilateral of area 6 is similar to the red quadrilateral. And since corresponding sides are in ratio 2:1 then their areas are in ratio 4:1. Desired area is 4(6) = 24. Solution#2: Labeling BQ as x and the height of triangle BMQ to side BQ as h one finds that h = (2/3)x but h = 6/x so x^2 = 9. We're asked to find (2x - h)(2x) which becomes 4x^2 - 12 so the desired area is 4(9) - 12 = 24.

We have two similar right triangles with two angles and sides AQC and APC then AQ/AC=AC/AP 2R/5=5/4 R=25/8

M pertenece a la diagonal BD y equidista de AB y de BC→ Igualdad de bases, alturas y áreas de los triángulos AMP=PMB=BMQ=QMC=6/2=3 → QBA y PBC son congruentes→ Áreas: QBA+PBC=3*3+3*3=18 =ABCD/2 =ABC → El solapo PBQM =AMC=6 → ADCM=18+6=24 u². Gracias y un saludo cordial.

Circle area = pi x r^2, where r^2 = 81- sq.root(4536). I am using chromebook, so difficult to use math symbol.

What about the area of the circle? Since the circle was colored/highlighted, I misunderstood I had to solve the area of the circle. Instead of the rectangle. Anyway, I got the answer of the circle area. Can you show your answer?

X=90°-arctan(1-2*a)-arctan(a/(1-a)) (GQ=GP=a) X=arctan((1-a)/a)-arctan(1-2*a) tan(X)=((1-a)/a-(1-2*a))/(1+(1-a)/a*(1-2*a)) tan(X)=(1-2*a+2*a^2)/(a+(1-a)*(1-2*a)) tan(X)=(1-2*a+2*a^2)/(a+2*a^2-3*a+1) tan(X)=1 X=45°

I loved the 2nd method. It's commendable. You were able to see those parallel lines there and form the alternate angles. Well done, it doesn't take long from there to find the area. One thing I need to say though is, symmetry is never a geometry argument. It can be used to say a pattern and then try to show it, but never enough without using the axioms and theorems to prove it works out.

If green área = 6 cm² then white area also is 6 cm² And right triangle area, is: A=½b.h = ½ a.2a = a² = 6+3=9 cm² a = √9 = 3 cm. ; s=2a = 6cm A = s² - 12 = 6²-12 = 24 cm²

angle alpha = angle beta. why do you use two separate notation?

AB=2PB y BC=2BQ. PMQB~ADCM. AM/PM=AD/PB=4/2= 2. Entonces: AM=2PM. Si los lados de una figura geométrica se duplican, su área se cuadruplica. Sí PMB=6 u², entonces: ADCM= 4×6= 24 u².

Tan y = tan(45+a). Tan y = 2. Tan (y)= (1 +tana)/(1-tana) =2. Tan a =1/3. Let 2x = side of square. F/Sqrt(2)*x/2=1/3 F= sqrt(2)x/6. Area of quadrilateral= sqrt(2)x/6 *sqrt(2)x/2 +sqrt(2)x/2 =6 x^2=9. X^2-6=3. 4x^2-(6+3+3)= 24. 24 is the yellow shaded region area

Long method. It can be verbally solved.

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