Can you find area of the circle? | (Square) | #math #maths | #geometry | #viral

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Can you find area of the circle? | (Square) | #math #maths | #geometry | #viral

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[BBMathTutorials]

Love❤your every video sir🙏🏼
Great fan🙋🏻‍♂️

[bakrantz]

Solve for side of square from diagonal. Then use tangent-secant theorem to solve for length EB. Determine Length EC. Do pythagorean theorem to solve for diameter of circle using EC, diameter ED, and square side CD.

[prossvay]

Let x is a side of the square
So x√2=8
x=8/√2=8√2/2=4√2
connect M to O to N (N on circle)
F is Middle CD
FN=y
y(4√2)=(2√2)(2√2)
So y=√2
So Diameter of circle=4√2+√2=5√2
So Radius of circle=5√2/2
Circle area=π(5√2/2)^2=25π/2square units=39.27 square units.❤❤❤ Thanks sir

[marioalb]

s = 8 cos45° = 8 / √2 = 4√2 cm
Tangent secant theorem:
(½s)² = s.(s-a)
(s-a) = ¼ s
a = ¾ s
Pytagorean theorem:
(2R)² = s² + a²
4 R² = s² + (¾s)² = 25/16 s²
πR² = π/4 . 25/16 . s²
A = 25/64 π s² = 25/64 π (4√2)²
A = 12, 5π cm² ( Solved √ )

[lifestylewithrukhsana]

Aapke samjhane Ka tarika bahut achcha hai

[quigonkenny]

Let s be the side length of square ABCD and r be the radius of circle O.

Triangle ∆BAD:
BA² + AD² = DB²
s² + s² = 8²
2s² = 64
s² = 64/2 = 32
s = √32 = 4√2

Draw OM and ON, where N is the point on circle O opposite point M. NM is therefore a diameter of circle O. Let P be the point of intersection between ON and CD. As CD is a chord and ON is a radius perpendicular to that chord, CP = PD = s/2 = 2√2.

Triangle ∆DPO:
DP² + OP² = OD²
(2√2)² + OP² = r³
OP² = r² - 8
OP = √(r²-8)

OM + OP = s
r + √(r²-8) = 4√2
√(r²-8) = 4√2 - r
r² - 8 = 32 - 8√2r + r²
8√2r = 32 + 8 = 40
r = 40/(8√2) = 5/√2 = (5√2)/2

Circle O:
Aₒ = πr² = π(5/√2)² = 25π/2 sq units

[AverageCommentor]

You can solve that length BC = 4√2 using Pythagoream theorem.
We can also find BC in terms of the radius, r. Let the midpoint of DC = F BC is MO + OF. MO is just r. OF we can find using Pythagoras again. There is a right angled triangle formed by FOC. OC is the hypotenuse and is r. FC is 2√2, half the sidelength of the square. Therefore OF^2 = r^2 - 8
Therefore BC = r+√(r^2-8).
r+√(r^2-8) = 4√2.
√(r^2-8) = 4√2 - r
r^2 - 8 = 32 - 8r√2 + r^2.
8r√2 = 40
r√2 = 5
r = 5/√2
Area of circle equal πr^2.
Area of circle = 25π/2.

[devondevon]

39.28
Another approach
First, find AB=AD
Let the side of the square = n, then
n^2 + n^2 = 8^2
2n^2= 64
n^2 = 32
n= sqrt 32 = sqrt 4* sqrt 8 = 2 sqrt 8
Hence, n/2 = sqrt 8
Hence, AM= n/2 (or one-half the square) = sqrt 8
and AD= 2 sqrt 2
Draw a line from D to M, to form right triangle ADM,
DM = sqrt 40 (Pythagorean using sqrt 8, and 2 sqrt 8 as the sides)
Let's calculate angle D of the triangle ADM
By the Law of Sine
D = sqrt 8/ sqrt 40 * sine 90
D =sqrt 8/sqrt 40 *1
D= 0.4472136
This equals 26.565 degrees
Hence, the other angles are 63.435 degrees and 90 degrees

Draw a line from the circle center to M to form a triangle, an isosceles triangle, and a DOM.
Since D0 =DM, the angles are 26.565, 26.565 and 126.87 degrees.
Notice that DO and DM = radius of the circle.
'Since 126.87 degrees correspond to sqrt 40 (see above),
then by the Law of Sine
sqrt 40/sine 126.87 = DO/sine 26.565

Sine 26.565/Sine 126.87 * sqrt 40 = D0
0.55902* sqrt 40 = D0

3.54. Hence, the radius of the circle = 3.54
Its area = 39.28

[himo]

AM=MB=x AB=AD=2x (2x)²+(2x)²=8² 8x²=64 x=√8=2√2
AD=BC=4√2 2√2*2√2=4√2*(2r-4√2) 8=8√2r-32 8√2r=40 r=5/√2
Circle area = 5/√2*5/√2*π = 25π/2

[unknownidentity]

Let's find the area:
.
..
...
....


First of all we calculate the side length s of the square from the length d of its diagonal:

s = d/√2 = 8/√2 = 4√2

Let N be the midpoint of CD. Then the triangle ODN is a right triangle and we can apply the Pythagorean theorem. With r being the radius of the circle we obtain:

OD² = DN² + ON²
OD² = (CD/2)² + ON²
r² = (s/2)² + ON²
⇒ ON = √[r² − (s/2)²] = √[r² − (4√2/2)²] = √(r² − 8)

Now we are able to calculate the radius and the area of the circle:

s = AD = MN = OM + ON = r + √(r² − 8)
4√2 = r + √(r² − 8)
4√2 − r = √(r² − 8)
(4√2 − r)² = r² − 8
32 − 8√2*r + r² = r² − 8
40 = 8√2*r
⇒ r = 40/(8√2) = 5/√2

A = πr² = π*(5/√2)² = (25/2)π

Best regards from Germany

[MrPaulc]

Square's sides are 8/sqrt(2) = 4*sqrt(2)
ABO and CDO are two isosceles triangles whose heights total 4*sqrt(2)
Midpoint of BD is N.
BMN is a right triangle. MN = 2*sqrt(2)
Call 2*sqrt(2), x
x^2 + (2x-r)^2 = r^2
x^2 + 4x^2 - 4xr + r^2 = r^2
5x^2 - 4xr = 0
Reapply x = 2*sqrt(2) for 40 - ((8*sqrt(2) * r ) = 0
Therefore, 8*sqrt(2) * r = 40
4x * r = 40
r = 40/4x
r = 10/x
r = 10/(2*sqrt(2))
r^2 = 100/8
Area = (100/8)pi
Area = 314.16/8 = 39.27 (rounded)
Just looked at the video. The same basic method as you, but the number crunching was a bit different.

[shalinisuryavanshi]

Here I did it
In triangle DBC
DB(square) = BC(square) + DC(square
8(square)= a^2 +a^2
8 = a√2
a = 4√2

AM = 4√2/2 = 2√2
Now make MG where
DO = OE = MO = r (radius of circle)
OG = AD-r
OG = 4√2 - r
Now in triangle DOG
DO(square) = OG(square) + DG( square)
r^2 = (4√2-r)^2 + (2√2)^2
√2 r = 5
r = 5/√2
Now area of circle
Pi r^2
22/7 * 25/2
11*25/7
275/7
39.28 square units. //Ans

[LuisdeBritoCamacho]

STEP-BY-STEP RESOLUTION PROPOSAL BY MEANS OF ANALYTIC GEOMETRY :

01) Square Diagonal = [Side * sqrt(2)]
02) Diagonal = 8 lin un
03) Square Side = 8 / sqrt(2) ; Square Side = (4 * sqrt(2)) lin un
04) AB = AD = BC = CD = (4*sqrt(2)) lin un
05) Let the Coordinates of Point O = (X ; Y)
06) Let the Coordinates of Ponts C ; D ; M be : C (0 ; 0) ; D (4sqrt(2) ; 0) and M (2sqrt(2) ; 4sqrt(2))
07) Distances between Point O and Points C ; D ; M are equal.
08) d[O ; D] = X^2 + Y^2
09) d[O ; C] = (4*sqrt(2) - X)^2 + Y^2
10) d[O ; M] = (2*sqrt(2) - X)^2 + (4*sqrt(2) - Y)^2
11) d[O ; D] = d[O ; C] = d[O ; M]
12) X^2 + Y^2 = (4*sqrt(2) - X)^2 + Y^2 = (2*sqrt(2) - X)^2 + (4*sqrt(2) - Y)^2
13) As : X = (2*sqrt(2)), we have :
14) (2*sqrt(2))^2 + Y^2 = ((4*sqrt(2)) - (2*sqrt(2)))^2 + Y^2 = ((2*sqrt(2)) - (2*sqrt(2)))^2 + ((4*sqrt(2)) - Y)^2
15) Solving for Y.
16) Solution : Y = (3 / sqrt(2))
17) Coordinates of Point O = (2*sqrt(2) ; 3/sqrt(2))
18) R = d[O ; D] = sqrt (X^2 + Y^2)
19) R = sqrt [(2*sqrt(2))^2 + (3/sqrt(2))^2] ; R = sqrt (8 + 9/2) ; R = sqrt((16 + 9) / 2) ; R = sqrt(25 / 2) ; R = sqrt(12, 5) R ~ 3, 54 lin un
20) Circle Area (CA) = Pi * R^2 sq un
21) CA = 12, 5*Pi sq un ; CA ~ 39, 3 sq un

Therefore,

OUR ANSWER : The Circle Area is equal to [25*Pi / 2] Square Units. Or Circle Area is approx equal to 39, 3 Square Units.


P.S. - Regards from The Islamic Institute of Mathematical Sciences.

[sergioaiex]

Solution:

Side Blue Square = 8/√2
SBS = 4√2

Then: ½ SBS = 2√2

In Triangle DNO
(2√2)² + (4√2 - r)² = r²
8 + 32 - 8√2r + r² = r²
40 - 8√2r = 0
r = 40/8√2
r = 5√2/2

A = πr²
A = π (5√2/2)²
A = π 50/4
A = 25π/2

A = 12, 5 π Square Units

A ~= 39, 27 Square Units

[qstzhhg]

BM²=BE*BC, so BE=√2, EC=3√2;. (2r)²=EC²+AC²=50, and r²=25/2, S=πr²=25π/2.

[phungpham]

1/ Because OM is perpendicular to AB so OM is perpendicular to chord CD at the midpoint of CD too.
Let a be the side of the square. We have: a=8/sqrt2
By using chord theorem:
a.(2r-a)= sq(a/2)
—> r= 5/sqrt2
Area =pi. 25/2 sq units😅

[Birol]

My way of solution ▶
N ∈ [DC]
we also take a point on the circle segment DC, P

let's find the length a of the square:
a²+a²= 8²
2a²= 64
a²= 32
a= 4√2

according to the intersecting chords theorem we can write:
[MN]*[NP] = [DN]*[NC]
[MN]= a

[MN]= 4√2
[NP]= 2r-4√2
[DN]= [NC]= a/2
[DN]= 2√2
⇒
4√2*(2r-4√2)= 2√2*2√2
2r-4√2= √2
r= 5√2/2

Acircle= πr²
Acircle= π*(5√2/2)²
Acircle= 25π/2
Acircle≈ 39, 27 square units

[marcgriselhubert]

In an orthonormal center D and first axis (DC), it is easy to find that the equation of the circle is x^2 + y^2 -4.sqrt(2).x -3.sqrt(2).y = 0, or (x - 2.sqrt(2))^2 + (y -(3/2))^2 = 25/2,
so the radius of the circle is sqrt(25/2) and its area is (25/2).Pi

[misterenter-izrz]

Creative and challenging puzzl🎉, 8^2=2×(2s)^2, 64=8s^2, s^2=8, then r^2=s^2+(2s-r)^2, 4sr=5s^2, r=(5/4)s, therefore the circle area is pi r^2=pi(5/4)^2×s^2=pi 25/2.😊

[davyp]

Fine! I found the same with chords theorem ....