Very difficult at a glance and simple after your explanation. Thanks a lot for your nice task, Professor 💯👍

@PreMath

24 күн бұрын

Glad you liked it🌹 You are very welcome! Thanks for the nice feedback ❤️

Solved using the intersecting chord theorem. (x-3)^2=(x-4)x; x^2-6x+9=x^2-4x; 2x=9. Side of a square 9. r=x-2. Yellow shaded region 81-(x-2)^2Pi=81-25Pi/4.

@davyp42

24 күн бұрын

I did the same 👍

@AmirgabYT2185

23 күн бұрын

​@@davyp42+++

@allanflippin2453

23 күн бұрын

I also solved with this method. The math is much easier and one doesn't have to deal with two variables in an equation.

@DorothyMantoothIsASaint

23 күн бұрын

Intersecting Chords FTW

@Micboss1000

23 күн бұрын

Same here, way easier, took me a minute at most.

The side lengths of the square are 2r+4 Therefore, the vertical chord length is 2r-2 Chord intersection gives (r-1)(r-1) = (r+2)(r-2) r^2-2r+1 = r^2-4 -2r+1 = -4 2r-4 = 1, so r = 5/2 ((10/2)+4) = 9 for square's sides. 9^2 = 81 Yellow area is 81 - (25/4)pi 81 - 6.25pi = 81 - 19.635 = 61.365 (rounded). We took a different route, but arrived at the same place :)

Superb👍

The difference between the side of the square and the diameter of the circle = 4. So the distance between the circle and the top side = 2 (by symmetry). From this it can be deduced that PQ = r-1. And also that r+2= r+4 - OP, from which it follows that OP = 2. Pythagoras: r^2= (r-1)^2 + 4. So r= 2.5 and the side of the square = 2 (2.5) + 4 = 9. The rest is simple.

PQ=a-3; 2r+4=2a => r=a-2 OQ=r=a-2; OP=a-r=a-(a-2)=2 !!! 😁 ΔOPQ: (a-2)²=2²+(a-3)² => => a²-4a+4=4+a²-6a+9 => => a=9/2 => r=9/2-2=5/2 [yellow]=(2a)²-πr²=81-25π/4≈61.365

Square ABCD was divided into four quadrants with origin P as stated in the beginning. So, four congruent segments extend from point P to the four sides of the square. These segments are the combined red-blue/blue ones shown in the diagram. So, FP = PT. Label EP = x. So, FP = PT = x + 4. PQ + 3 = x + 4 PQ = x + 1 Point O is also in the "x-axis". It is the center of a circle tangent to side AD. Label the point of tangency as G. Additionally, the starter information helps us deduce that origin P and the "axes" form right angles. By the Perpendicular Chord Bisector Theorem, diameter EG bisects chord QR (where R is the other endpoint of the chord). So, PR = x + 1. And GP = FP = x + 4. Use the Intersecting Chords Theorem. x(x + 4) = (x + 1)(x + 1) x² + 4x = x² + 2x + 1 4x = 2x + 1 2x = 1 x = 1/2 So, each congruent segment is 4.5 units long. The starter info suggests the "axes" intersect the midpoints of the sides of square ABCD. Plus, the four quadrants are also congruent squares themselves! So, square ABCD has sides that are 9 units long each. EG = (x + 4) + x = 2x + 4 = 2(1/2) + 4 = 1 + 4 = 5 So, the radius of ⊙O is 5/2 = 2.5 units long. Yellow Region Area = Square ABCD Area - ⊙O Area A = s² = 9² = 81 A = πr² = π(5/2)² = π(25/4) = (25π)/4 Yellow Region Area = 81 - (25π)/4 So, the area of the yellow shaded region is 81 - (25π)/4 square units, a. w. a. (324 - 25π)/4 square units (exact), or about 61.37 square units (approximation).

Let's find the area: . .. ... .... ..... Let s be the side length of the square and let r be the radius of the circle. From the horizontal line through point P we can conclude: 2*r + 4 = s ⇒ 2*r = s − 4 ⇒ r = (s − 4)/2 The vertical line through point P intersects the circle at Q and Q'. For reasons of symmetry we know that PQ=PQ'=s/2−3. By applying the intersecting chords theorem we obtain: PE*(2*r − PE) = PQ*PQ' (s/2 − 4)*[(s − 4) − (s/2 − 4)] = (s/2 − 3)*(s/2 − 3) (s/2 − 4)*(s/2) = (s/2 − 3)*(s/2 − 3) s²/4 − 2*s = s²/4 − 3*s + 9 ⇒ s = 9 ⇒ r = 5/2 Now we are able to calculate the area of the yellow region: A(yellow) = A(square) − A(circle) = s² − πr² = 9² − π*(5/2)² = 81 − (25/4)π ≈ 61.37 Best regards from Germany

I love your videos so much and I love to enjoy them every day, that much that I wish you could upload ten videos each day. Thank you for your excellent explanation as usual professor

Side length of the big square: 2.c Intersecting chord theorem: (c -3)^2 = c. c -4), so c = 9/2 The area of the big square is (2.(9/2))^2 = 81 R the radius of the circle: 2.R + 4 = 2. c, so R = 5/2 The area of the circle is (25/4).Pi The yellow area is 81 - (25/4).Pi. That was very simple.

Thanks PreMath Thanks Sir Very useful to learn math With my respects ❤❤❤❤

Another approach is to use the product of the components of the intersecting chords at point P (in terms of r and a) to get the other equation to use with the equation r + 2 = a to get the same result.

If you calculate a=r+2 and substitute that from the start, the calculations become much simpler (but not as simple as intersecting chords theorem).

That was a great test. I have got tge same result using almost same approach. You rock 👍

it is just 1 linear and 1 nonlinear equation with unknown dimensions la and r: 10 vdu5:for a=0 to 15:gcola:print a:next a 20 print "premath-can you find area of the yellow shaded region" 30 dim x(3),y(3):l1=4:l2=3:sw=l1^2/(l2+l1)/10:print sw:la=l1+sw:goto 60 40 h=la/2-l2:r=(la-l1)/2:lh=la/2:dgu1=(lh^2+h^2-4*r*r)/l1^2:dgu2=((2*r-lh)^2+h^2)/l1^2 50 dg=dgu1+dgu2:return 60 gosub 40 70 dg1=dg:la1=la:la=la+sw:if la>100*l1 then stop 80 la2=la:gosub 40:if dg1*dg>0 then 70 90 la=(la1+la2)/2:gosub 40:if dg1*dg>0 then la1=la else la2=la 100 if abs(dg)>1E-10 then 90 110 print la,r 120 x(0)=0:y(0)=0:x(1)=la:y(1)=0:x(2)=la:y(2)=la:x(3)=0:y(3)=la 130 mass=850/la:goto 150 140 xbu=x*mass:ybu=y*mass:return 150 gcol 11:x=x(0):y=y(0):gosub 140:xba=xbu:yba=ybu:for a=1 to 4:ia=a:if ia=4 then ia=0 160 x=x(ia):y=y(ia):gosub 140:xbn=xbu:ybn=ybu:goto 180 170 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return 180 gosub 170:next a:gcol 8:xm=r:ym=la/2:x=xm:y=ym:gosub 140:circle xbu,ybu,r*mass 190 print "die gesuchte flaeche=";la^2 200 print "run in bbc basic sdl and hit ctrl tab to copy from the results window" 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 die gesuchte flaeche=81 runminhbbcnbasicfsdl andahit ctrlytabotoscopydfromithe results window >.228571429 9 2.5

S=81-25π/4=(324-25π)/4≈61,36

1/ 2r= 2a-4-> r=a-2 (1) 2/ Using right triangle altitude theorem or chord theorem: sq PQ=OPxPE --> sq(a-3)=a(a-4)-> a= 9/2 the side of the square= 9 and r= 5/2 Area of yellow region= 81-pi.sq(5/2)=61.37 sq units😊

Good math question 👍I solved this easier, it seems to me. I immediately found that the length of the segment OP is 2 and so on.

Let x is side length of the square ((x-6)/2)^2=x/2(x)2-(x-4)/2) So x=9 Radius of the circle=(9-4)/2=5/2 Ares of the yellow shaded region=81-π(5/2)^2=81-25π/4=61.37 square units.❤❤❤

@6:01 Once you got equations 1 & 2, it was less work to solve for r ( = a-2 ) and sub that expression into equation 1 to get: 2a² - 6a - 2a(a-2) + 9 = 0 2a² - 6a - 2a² + 4a + 9 = 0 -2a + 9 = 0 a = 9/2 = 4.5 Then, r = a - 2 = 4.5 - 2 = 2.5. The rest is the same.

I solved similar to you teacher. 2r + 4 = 2a a = r + 2 Focus ∆OPQ OQ = r OP = a - r OP = r + 2 - r OP = 2 PQ = a - 3 PQ = r+2 - 3 PQ = r - 1 Applying Pythagoras: r^2 = 2^2 + (r - 1)^2 r^2 = 4 + r^2 - 2r + 1 2r = 4 + 1 r = 5/2 2a = 2r + 4 2a = 2*(5/2) + 4 2a = 9 Square area = 9^2 Square area = 81 Circle area = πr^2 Circle area = π(5/2)^2 Circle area = (25/4)π = 6,25π Yellow area = 81 - 6,25,π Yellow area = 61,365.. unit^2

81 - 6.25 pi 61.365 Let the side of the square = 2n then the diameter will = 2n - 4 and its radius (2n-4)/2 = n -2 Hence , the distance OQ = n-2 The distance from p to either side = 2n/2 = n ( since p is the center of the square and the square = 2n) Hence, the distance from the circle's center to p is n - (n-2 ) = 2 Hence, the distance OP = 2 The distance from Q to p= n -3 (since the p is the distance n from either side of the square) Let's construct a right triangle from the circle's center, OPQ. The radius is n -2 = the hypotenuse, and the other two legs have a distance of 2 and n-3 then (n-2)^2 = 2^2 + (n-3)^2 n^2 +4- 4n = 4 + n^2 + 9 -6n -4n +6n = 9 (since n^2 and 4, which have the same sign on both sides of the equation, cancel themselves out) 2n = 9 Hence, the length of the square =9 and area = 81 and n = 9/2 or 4.5 and its radius which = 4.5 - 2 = 2.5 since the radius = n-2) Hence, the area of the circle = 2.5^2 pi or 6.25pi Hence, the area of the shaded region = 81 - 6.25pi or 61.365

เป็นความสุดยอดที่สุดและผมได้ฝึกฝนทุกวันเพื่อสอนลูกสาวของผม ขอบคุณอย่างมากครับ

Let x be the length of AT, so that 2x is the side length of square ABCD. Let r be the radius of circle O. Let M and N be the unnamed intersection points on DA and CD respectively. As ABCD is a square and assuming that T and F are the midpoints of AB and BC respectively and that ∠ATP = ∠BFP = 90°, then MF = TN = 2x. MF = OM + OE + EF 2x = r + r+ 4 = 2r + 4 x = (2r+4)/2 = r + 2 From this, as MP = x and OM = r, OP = x-r = (r+2)-r = 2. QP = x-3 = (r+2)-3 = r-1. Triangle ∆OPQ: OP² + QP² = OQ² 2² + (r-1)² = r² 4 + r² - 2r + 1 = r² 2r = 5 r = 5/2 x = r + 2 = 5/2 + 2 = 9/2 The area of the yellow shaded region is equal to the area of square ABCD minus the area of circle O: Yellow shaded area: Aʏ = (2x)² - πr² Aʏ = 9² - π(5/2)² Aʏ = 81 - 25π/4 ≈ 61.365 sq units

Since it is a Circle, QP + 3 mirrors down. I named QP as x, so TP x+3. So PE is TP-4, or x-1. Then, in AD center, I named a G point. So: 1) GP*PE = QP ² (x+3)*(x-1) = x ² x ² = x ² + 3x - x - 3 x = ³ ⁄ ₂ TP = 3 + x = 3 + ³ ⁄ ₂ = ⁹ ⁄ ₂ AB = ⁹ ⁄ ₂ * 2 = 9 GO = ⁽ ⁹ ⁻ ⁴ ⁾ ⁄ ₂ = ⁵ ⁄ ₂ Square area = 81 Circle area = π(⁵ ⁄ ₂) ² = π * ²⁵⁄₄ Yellow area = 81 - (π * ²⁵⁄₄)

2a = 2r + 4 a = r + 2 a - r = r + 2 - r a - r = 2 a - 3 = r + 2 - 3 a - 3 = r - 1 r² = (a - r)² + (a - 3)² r² = (2)² + (r - 1)² r² = 4 + r² - 2r + 1 2r = 4 + 1 2r = 5 r = 5/2 2a = 2r + 4 2a = 2(5/2) + 4 2a = 5 + 4 2a = 9 Area = (2a)² - πr² Area = (9)² - π(5/2)² Area = 81 - 25π/4

Very nice ! In my point of view it was missing that P is the ce ter of the square…

Solve 2pq+6=2r+4=2PE+8, r=5/2

L = 2R+4 = 2Y+2*3 = 2Y+6. Now [2R-X] * X = Y^2. Now L/2 = R+2 = 4+X -> X=R-2. Now 2R-X = R+2. Now 2Y+6 = 2R+4 -> 2Y=2R-2 -> Y=R-1 --> [R+2] * [R-2] = (R-1) ^2 -> R^2 - 4 = R^2 - 2R+1 -> 2R=5 -> R=5/2. Now L = 2R+4 = 9. A(shaded) = L^2 - Pi*R^2 = 9^2 - Pi*25/4 = 81 - 25*Pi/4 = 81 - 19.63 = 61.37 sg. units.

It only works if you assume that ABCD is a square.

Why not use intersecting chords in the circle:( r-1)x(r-1)=(r+2)x(r-2)?

Nice job . In your equation 1 , there is only one r . I would substitute a-2 for r in equation 1 because it would result in less algebra to simplify . I guess it is just me --- kind of lazy .

14²-pix5²=117,42 a guess, but the method is correct

Thank you!

Seems to be difficult but simple puzzle, the answer is s=9/2, r=5/2, the area is 9^2-(5/2)^2 pi.😅

81-6.25 pi or 61.365

(a - 4)² + (a - 3)² = QE² a² - 8a + 16 + a² - 6a + 9 = QE² 2a² - 14a + 25 = QE² a² + (a - 3)² = MQ² a² + a² - 6a + 9 = MQ² 2a² - 6a + 9 = MQ² MQ² + QE² = (2a - 4)² 2a² - 6a + 9 + 2a² - 14a + 25 = 4a² - 16a + 16 4a² - 20a + 34 = 4a² - 16a + 16 4a = 18 *a = 9/2* 2r = 2a - 4 r = a - 2 r = 9/2 - 2 r = 9/2 - 4/2 r = 5/2 Square Area = 9² Square Area = 81 Circle Area = π (5/2)² Circle Area = 25π/4 Yellow Shaded Region = 81 - 25π/4 YSR = 324/4 - 25π/4 *YSR = (324 - 25π)/4 Square Units* YSR = 61,3650459151 *YSR ~= 61,37 Square Units*

4 × 49 - 49 pi = 49 × .27 ( approximately)?

After solving this problem I'm taking my skills to Papa John's and and design pizza boxes. ...seems like a viable skill for employment. I need a job. 🙂

You should set up a Patreon account so I can pay you monthly. Thanks!

@PreMath

23 күн бұрын

Patreon account has been set up! Thank you very much for supporting my youtube channel. I really appreciate your kind donation. Stay blessed ❤️ Kind regards🙏