Can you find area of the circle? | (Square) |
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- Learn how to find the area of the circle. Important Geometry and Algebra skills are also explained: Pythagorean theorem; area of the circle formula; square. Step-by-step tutorial by PreMath.com
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Love❤your every video sir🙏🏼
Great fan🙋🏻♂️
Thank you so much 😀❤️
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.
Thanks for the feedback ❤️
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
Excellent!
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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 √ )
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
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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.
Aapke samjhane Ka tarika bahut achcha hai
Thank you Rukhsana ji🌹
👏🏻👏🏻👏🏻
😀
Thanks ❤️
Thanks for sharing ❤️
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😅
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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.
Thank you!
You are very welcome!
Thanks for the feedback ❤️
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
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BM²=BE*BC, so BE=√2, EC=3√2;. (2r)²=EC²+AC²=50, and r²=25/2, S=πr²=25π/2.
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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
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
Excellent!
Thanks for sharing ❤️
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
Excellent!
Thanks for sharing ❤️
La flecha de la cuerda DC es "f"→ BE=f.
DC =a=8/√2=4√2 → Potencia de B respecto a la circunferencia =(a/2)²=f*a→ (4√2/2)²=f*4√2→ f=√2 → r=(a+f)/2 =(√2+4√2)/2 =5√2/2→ r²=25/2→ Área del círculo =25π/2 =12,5π ud².
Gracias y un saludo.
Excellent!
You are very welcome!
Thanks for sharing ❤️
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
Second way of solution ▶
N ∈ [DC]
the side a of the square, a
a²+a²= 64
2a²= 64
a²= 32
a= 4√2
[ON]= a-r
⇒
[ON]= 4√2 - r
[OD]= r
[DN]= a/2
[DN]= 2√2
Let's appy the Pythagorean theorem for the ΔODN :
[ON]²+[DN]²= [OD]²
(4√2 - r)²+(2√2)²= r²
32-8√2r+r²+8= r²
40= 8√2r
√2r= 5
r= 5/√2
Acircle= πr²
Acircle= π*(5/√2)²
Acircle= π*25/2
Acircle= 12,5 π
Acircle≈ 39,27 square units
Excellent!
Thanks for sharing ❤️
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.😊
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
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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
Excellent!
Thanks for sharing ❤️
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.
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I guess if the circle was a pizza with radius z and height a ...then the volume would be p • i • z • z • a ...🙂
That's a good one!
Thanks for the feedback ❤️
You can also use Tangent Secant Theorm after finding x.
Thanks for the feedback ❤️
Fine! I found the same with chords theorem ....
Thanks for the feedback ❤️
S=25π/2≈39,29
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39.28
(4√2-r)^2+(2√2)^2=r^2...32-8√2r+8=0..r=40/8√2=5/√2..mah ho fatto i calcoli a mente
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I'll say 25π
I'm wrong. I missed a division by 2 on the last step...
39.28