Can you find area of the Semicircle? | (Chords) | #math #maths | #geometry

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Can you find area of the Semicircle? | (Chords) | #math #maths | #geometry

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#FindSemiCircleArea #SemiCircle #IntersectingChordsTheorem #Chords #GeometryMath #PythagoreanTheorem #Euclid'sTheorem #EuclideanTheorem
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[Micboss]

Nice and simple one today, no lengthy calculations, just a little thinking outside the box.

[billferrol]

Triangle AED is isosceles 45, 90, 45. Thales law angle ABF = 45, 90, 45. AD = sqrt r=5/2sqrt2.
Area =(pi(5/2sqrt2)^2)/2=(25/4)pi

[Fan_di_Euclide-]

Or once we find out that DE= square root of 2 we know that the angle EÂD is 45° and by Thales Theroem we know that ABF is a right isoscele triangle and so AF=2r= 5square root of 2 and so the area of semicircle is 25pi/4

[marioalb]

Intersecting chords theorem:
x. 3x = 2 . 3
x² = 2 , x=√2 cm
sin α = x/2 = √2/2 = 1/√2
α = 45°
R = (2+3)cos45° = 5/√2 cm
A = ½πR² = ½π(5/√2)²
A = 6, 25π cm² ( Solved √ )

[jimkramer]

I’m surprised that you don’t recognize the ratios of iosceles right triangle as1:1:sqrt2. As soon as I saw the right triangle with sides 2 and sqrt2 I knew the third side was sqrt2. I taught my students to recognize 3-4-5, 30-60-90 and 1-1-sqrt2 early. I’m 82 and taught math for 50 years.

[juanalfaro]

Let CD=DE=a. Now make the circle whole & use intersecting chord theorem: a*3a=2*3 -> 3a^2 = 6 -> a=sqrt (2). Now AE^2 = AD^2 - DE^2 = 2^2 - 2 = 2 -> AD=sqrt (2). Now EF*AE = [2*sqrt(2)] ^2 --> EF*sqrt (2) = 8 --> EF = 4*sqrt (2) = 2R - AE = 2R - sqrt (2) ---> 2R = 5*sqrt (2) --> R=5*sqrt (2) / 2 --> Area = Pi*R^2 /2 = Pi*5^2/2/2 = Pi*25/4 = 6.25*Pi sq. units

[rchandos]

Point B should be located directly above point O.

[santiagoarosam]

CD=DE=a ---> Potencia de D respecto a la circunferencia =2*3=a*3a---> a²=2---> a=√2.---> Si Ángulo DAE=45°---> OA=OB= Radio =r=(2+3)/√2=5/√2. ---> Área del semicírculo =π25/2*2=25π/4.
Gracias y saludos.

[MrPaulc]

Chord is 5 with 3 and 2 each side.each side.
Extend CE downward to the unshown perimeter at the bottom of the circle.
Instersecting chords for 6 = 3x^2
2 = x^2
DE = sqrt(2) and so is AE\
Right triangle OEC will have sides 2*sqrt(2), r-sqrt(2), and r
Square them for 8 + r^2 - 2*sqrt(2)r + 2 = r^2
Rearrange for 10 = 2*sqrt(2)r
5 = sqrt(2)r
r = 5/sqrt(2) = (5*sqrt(2))/2
r^2 = 50/4 = 25/2, so a full circle is (25/2)pi
Semicircle is (25pi)/4 so 19.635 un^2 (rounded).
Yes, unusually we went pretty much the same path.
Thank you.

[murdock]

Nice! φ = 30° → sin⁡(3φ/2) = √2/2; CF = 4k = k + 3k = CD + DF
AB = AD + BD = 2 + 3 → k = √2; ∆ AED → DAE = δ → sin⁡(δ) = √2/2 = sin⁡(3φ/2) →
δ = 3φ/2 → r = 5√2/2 → semicircle area = (π/2)r^2 = 25π/4

[michaelstahl]

Great video again. Thanks. I found the solution before I watched your video . I`m proud that I got the same. I remebered me about your step " thinking outside the box" !!

[yalchingedikgedik]

Thanks Sir
That’s very good method and understandable for solve
With my respects
❤❤❤❤❤

[rajendrab.shetty]

Instead of joining O and F, AE and CF are two intersecting chords.
By this we can find out diameter., then area.

[quigonkenny]

Let CD = DE = x and let the radius of semicircle O be r. Extend the circumference of the semicircle to form a full circle, and extend CE to point C' on the newly constructed portion of the circumference. As ∠CEO = 90°, OE bisects chord CC'.

By the intersecting chords theorem, if two chords of a circle intersect, then the products of their divided segments about the point of intersection are equal. Therefore CD(DC') = AD(DB).

CD(DC') = AD(DB)
x(3x) = 2(3)
3x² = 6
x² = 6/3 = 2
x = √2

Triangle ∆AED:
AE² + ED² = AD²
AE² + (√2)² = 2²
AE² = 4 - 2 = 2
AE = √2

Draw OP, where P is the point on AB where OP and AB are perpendicular. As OP is perpendicular to chord AB, then it bisects AB, and AP = PB = (2+3)/2 = 5/2. As ∠OPA = ∠AED = 90° and ∠DAE is common, ∆OPA and ∆AED are similar triangles.

Triangle ∆OPA:
OA/AP = AD/AE
r/(5/2) = 2/√2 = √2
r = (5/2)√2 = 5√2/2

Semicircle O:
Aₒ = πr²/2 = π(5√2/2)²/2 = π(50/4)/2 = 25π/4 ≈ 19.63 sq units

[DB-lgsq]

شكرا لكم على المجهودات
يمكن استعمال DE=a و OA=r
CE^2=AE×AF
AF^2=5^2+BF^2

a^2=2
r^2=25/2

[Waldlaeufer]

Intersecting Chord Theorem: (x = CD = DE)
x * 3x = 2 * 3
3x² = 6
x² = 2
x = √2

Pythagoras: (p = AE)
p² + x² = 2²
p² + 2 = 4
p² = 4 - 2 = 2
p = √2 = x

Intersecting Chord Theorem: (p = AE, d = AF = perimeter of circle)
p * (d - p) = (2x)²
√2 (d - √2) = 4 * 2
√2 d - 2 = 8
√2 d = 10
d = 10 / √2
r = 5 / √2

A(semicircle) = 1/2 * r² π = 1/2 * 25/2 * π = 25/4 π ≈ 19.63 square units

[hongningsuen]

Method using intersecting chords theorem, Thales theorem, Pythagoras theorem, similar triangles:
1. By intersecting chords theorem for full circle, (ED)(3ED) = (2)(3). Hence ED = sqrt2.
2. Triangle ABF is right-angled triangle by Thales theorem.
Let radius of circle be R.
BF^2 = AF^2 - AB^2 by Pythagoras theorem
BF = sqrt[(2R)^2 - 5^2] = sqrt(4R^2 - 25)
3. Triangles ABF and AED are similar triangles (AAA)
Hence AF/AD = BF/ED
(2R)/2 = sqrt(4R^2 - 25)/sqrt2
R^2 = (4R^2 - 25)/2
R^2 = 25/2
4. Area of semicircle = (1/2)pi(R^2) = (25/4)pi.

[zdrastvutye]

the solution is that the endpoint of the line l1+l2 (see line 30)
must have the distance r from the center point:
10 print "premath-can you find area of the semicircle?":nu=55
20 50
30
40
dg=dgu1+dgu2-1:return
50 gosub 30
60 dg1=dg:l41=l4:l4=l4+sw:if l4>100*l1 then stop
70 l42=l4:gosub 30:if dg1*dg>0 then 60
80 l4=(l41+l42)/2:gosub 30:if dg1*dg>0 then l41=l4 else l42=l4
90 if abs(dg)>1E-10 then 80
100 print r:masx=1200/2/r:masy=850/r:if masx<masy then mass=masx else mass=masy
110 goto 130
120 xbu=x*mass:ybu=y*mass:return
130 rg=r*mass:move rg, 0:move 0, 0:plot165, 2*rg, 0:xba=0:yba=0:for a=0 to nu
140 120:xbn=xbu:ybn=ybu:goto 160
150 line xba, yba, xbn, ybn:xba=xbn:yba=ybn:return
160 gosub 150:next a:x=l4:y=2*l3:gosub 120:xbn=xbu:ybn=ybu
170 gcol 5:gosub 150:gcol 7:xba=0:yba=0:x=l6:y=l5:gosub 120:xbn=xbu:ybn=ybu
180 gosub 150:print "die gesuchte flaeche=";pi*r^2/2
190 print "run in bbc basic sdl and hit ctrl tab to copy from the results window"
premath-can you find area of the semicircle?
3.53553391
die gesuchte flaeche=19.6349541
run in bbc basic sdl and hit ctrl tab to copy from the results window
>

[LuisdeBritoCamacho]

STEP-BY-STEP RESOLUTION PROPOSAL USING INTERSECTING CHORDS THEOREM:

01) CD = DE = X lin un
02) Droping a Vertical Line from C, we get a Chord CC' = 4X
03) X * 3X = 2 * 3 ; 3X^2 = 6 ; X^2 = 2 ; X = sqrt(2) lin un
04) AE^2 = 4 - 2 ; AE^2 = 2 ; AE = sqrt(2) lin un
05) AE * EF = CE * EC' ; sqrt(2) * EF = 8 ; EF = 8 / sqrt(2) ; EF = 4 * sqrt(2)
06) AF = AE + EF ; AF = sqrt(2) + 4 * sqrt(2) ; AF = 5 * sqrt(2)
07) Radius (R) = (5 * sqrt(2) / 2) lin un ; R^2 = (25 * 2) / 4 ; R^2 = 50 / 4 ; R^2 = 25 / 2
08) Semicircle Area (SCA) = (Pi * R^2) / 2 sq un
09) SCA = (Pi * 25 / 2) / 2
10) SCA = ((25/2 * Pi) / 2) sq un ;
11) SCA = ((25 / 4) * Pi) sq un
11) SCA ~ 19, 635 lin un

Thus,

OUR ANSWER : The Semicircle Area is equal to ((25 * Pi) / 4) Square Units or approx. equal to 19, 635 Square Units.

[nexen]

I have just figured that both right triangles ABF and AOE are similar with 45 deg angles. From there we can get the diameter and radius of the circle..