Can you find area of the Yellow shaded region? | (Square and Circle) | #math #maths | #geometry

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Can you find area of the Yellow shaded region? | (Square and Circle) | #math #maths | #geometry

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[anatoliy]

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

[AlexanderRomadin]

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.

[MrPaulc]

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 :)

[rabotaakk-nwnm]

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

[ludosmets]

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.

[prossvay]

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 square units.❤❤❤

[unknownidentity]

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

[marcgriselhubert]

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.

[ChuzzleFriends]

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).

[zdrastvutye]

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
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
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 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

[pk]

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 .

[phungpham]

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😊

[KipIngram]

Let R = circle radius. Then large square side is 2*R+4, thus small squares have side R+2. Length PE is thus R+2-4 = R-2. Thus length OP = 2. Length PT is also R+2, so length PQ is R+2-3 = R-1. By Pythagoras on OPQ,

2^2 + (R-1)^2 = R^2
4 + R^2 - 2*R + 1 = R^2
5 - 2*R = 0
R = 5/2

Now we're golden. Big square area is (2*R+4)^2 = 81, circle area is pi*(5/2)^2 = 25*pi/4. So

Yellow Area = 81 - 25*pi/4.

Q.E.D.

[lior]

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

[toninhorosa]

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

[MarieAnne.]

I solved the same way, but started by solving for a in terms of r first:
2a = 2r + 4 → a = r + 2
Then we get:
OP = a − r = (r + 2) − r = 2
PQ = a − 3 = (r + 2) - 3 = r − 1
PQ = r
and using Pythagorean Theorem on △OPQ gives us:
OQ² = OP² + PQ²
r² = 2² + (r−1)²
r² = 4 + r² − 2r + 1
2r = 5
r = 5/2
Then solve for a:
a = r + 2 = 5/2 + 2 = 9/2
This makes calculations simpler than using Pythagorean Theorem with 2 variables and then substituting for a.
We then continue as shown in video:
Area = (2a)² − πr² = (9)² −π(5/2)² = 81 − 25π/4 ≈ 61.365

[quigonkenny]

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

[devondevon]

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

[LucasBritoBJJ]

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)

[timeonly]

@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.