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
Let L denote the double stroked length. By the intersecting chord theorem, we have 2*3 = 3*L^2 --> L = sqrt(2). Given this, sin(BAO) = sqrt(2)/2, so BAO is 45 degrees. This implies B is at the peak of the circle, which leads quickly to R = 5*sqrt(2)/2, R^2 = 12.5, and so our sought after semicircle area is 6.25*pi.
CD=DE=a ---> Potencia de D respecto a la circunferencia =2*3=a*3a---> a²=2---> a=√2.---> Si DE=√2--->√2*√2=2=AD--->DE=EA---> Ángulo DAE=45°---> OA=OB= Radio =r=(2+3)/√2=5/√2. ---> Área del semicírculo =π25/2*2=25π/4. Gracias y saludos.
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" !!
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.
I used intersecting theorem twice. First time, as shown, to get x = √2 Second time, I used chords CF and AG (you've used F twice in the diagram, so I've renamed F on right of circle as G) The vertical chord CF is divided into segments CE = 2√2 and EF = 2√2 The horizontal chord AG is divided into segments AE = √2 and EG = 2r−√2 √2 (2r−√2) = (2√2) (2√2) 2√2 r − 2 = 8 2√2 r = 10 r = 10/(2√2) = 5/√2 Area = 1/2 πr² = π/2 (25/2) = 25π/4
This time your solution is definitely much easier than mine. 👍 By the way: for the final calculation of R you could have used the intersecting chords theorem again: 2√2*2√2 = √2*(2R − √2) 4√2 = 2R − √2 5√2 = 2R 5√2/2 = R
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.
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
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..
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.
Can we just join OB and find the radius r ( AO and OB = r) of isosceles Triangle AOB as angle OAB and thus ABO are 45 degrees (as AO and OB = r) so angle AOB is 90 degrees. If AB is 5 and Sine 45 = 1/sqrt 2 then r = 5/sqrt 2. Then we can find the area by (pi.r squared)/2.
Sir if we join o to b than oA and oB will be radius and AB will be 5 so in a right isoscalane triangle ABO we can use Ao square +BO square=AB square by using that we can directly find r =5root2 than we can use pi r square by 2
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 l1=2:l2=3:sw=l1^2/(l1+l2)/10:l4=1+sw:goto 50 30 l3=sqr(l1^2-l4^2):r=(4*l3^2+l4^2)/2/l4:l5=(l1+l2)/l1*l3 40 l6=sqr((l1+l2)^2-l5^2):dgu1=l5^2/r^2:dgu2=(sqr((l1+l2)^2-l5^2)-r)^2/r^2 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
Let DE=CD=a and Radius=r So a(3a)=2(3) 3a^2=6 a^2=2 So a=√2 AE=√2^2-(√2)^2=√2 Connect O to C OE=r-√2 In ∆ OCE OE^2+CE^2=OC^2 (r-√2)^2+(2√2)^2=r^2 So r=5√2/2 Semicircle area=1/2(π)(5√2/2)^2=25π/4 square units =19.63 square units.❤❤❤
Once you find the x value, √2 using intersecting chord theorem, you can use the same method again to find the radius.value. calculating semicircle area is easy afterwards. ✌️
Let's draw the complete circle (radius R) Let's name x = CD = DE The intersecting chord theorem gives: 2.3 = x.(3.x), so x = sqrt(2). Now in triangle ADE: AE^2 = 4 - 2 = 2 So AE = sqrt(2, meaning that ADE is right isosceles, angle OAB = 45° Let now H be the orthogonal projection of B on (AO). The triangle ABH is also right isosceles,so BH= AH = AB/sqrt(2) = 5/sqrt(2) and OH = (5/sqrt(2)) - R In right triangle OBH we have: OB^2 = OH^2 +BH^2, so R^2 = ((5/sqrt(2) -R))^2 + (5/sqrt(2))^2 or: R^2 = (25/2) + R^2 - (10/sqrt(2)).R +(25/2) giving R = (25/10).sqrt(2) = (5/2).sqrt(2) The area of the semi circle is(Pi/2).(R^2) = (25/4).Pi.
Nice and simple one today, no lengthy calculations, just a little thinking outside the box.
Glad it was helpful!
Thanks for the feedback ❤️
Triangle AED is isosceles 45, 90, 45. Thales law angle ABF = 45, 90, 45. AD = sqrt 5^2+5^2=sqrt50=5sqrt2=diameter=2r. r=5/2sqrt2.
Area =(pi(5/2sqrt2)^2)/2=(25/4)pi
Excellent!
Thanks for sharing ❤️
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
I did it this way. Much easier.
Thanks for the feedback ❤️
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 √ )
excellente vidéo , merci beaucoup
Let L denote the double stroked length. By the intersecting chord theorem, we have 2*3 = 3*L^2 --> L = sqrt(2). Given this, sin(BAO) = sqrt(2)/2, so BAO is 45 degrees. This implies B is at the peak of the circle, which leads quickly to R = 5*sqrt(2)/2, R^2 = 12.5, and so our sought after semicircle area is 6.25*pi.
CD=DE=a ---> Potencia de D respecto a la circunferencia =2*3=a*3a---> a²=2---> a=√2.---> Si DE=√2--->√2*√2=2=AD--->DE=EA---> Ángulo DAE=45°---> OA=OB= Radio =r=(2+3)/√2=5/√2. ---> Área del semicírculo =π25/2*2=25π/4.
Gracias y saludos.
Excellent!
You are very welcome!
Thanks for sharing ❤️
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" !!
Thanks Sir
That’s very good method and understandable for solve
With my respects
❤❤❤❤❤
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.
Excellent!
You are very welcome!
Thanks for sharing ❤️
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
Excellent!
Thanks for sharing ❤️
I used intersecting theorem twice. First time, as shown, to get x = √2
Second time, I used chords CF and AG (you've used F twice in the diagram, so I've renamed F on right of circle as G)
The vertical chord CF is divided into segments CE = 2√2 and EF = 2√2
The horizontal chord AG is divided into segments AE = √2 and EG = 2r−√2
√2 (2r−√2) = (2√2) (2√2)
2√2 r − 2 = 8
2√2 r = 10
r = 10/(2√2) = 5/√2
Area = 1/2 πr² = π/2 (25/2) = 25π/4
Point B should be located directly above point O.
Let's find the area:
.
..
...
....
.....
Let's assume that O is the center of the coordinate system and that AF is located on the x-axis. Since B and C are located on the semicircle, we know:
xB² + yB² = R²
xC² + yC² = R²
The points A, D and B are located on the same line. From the slope of this line we can conclude:
(yB − yA)/(yD − yA) = (xB − xA)/(xD − xA) = AB/AD = (AD + BD)/BD = (3 + 2)/2 = 5/2
(yB − 0)/(yD − 0) = (xB + R)/(xD + R) = 5/2
yB/yD = (xB + R)/(xD + R) = 5/2
Since xC=xD=xE and CD=DE, we know:
yC − yD = yD − yE = yD − 0 = yD ⇒ yD = yC/2 ⇒ yB/yD = 2*yB/yC = (xB + R)/(xD + R) = 5/2
Applying the Pythagorean theorem to AB leads to:
(xB − xA)² + (yB − yA)² = AB² = (AD + BD)² = (2 + 3)² = 5² = 25
(xB + R)² + (yB − 0)² = 25
xB² + 2*R*xB + R² + yB² = 25
R² + 2*R*xB + R² = 25
2*R*xB = 25 − 2R²
⇒ xB = 25/(2*R) − R
yB² = R² − xB² = R² − [25/(2*R) − R]² = R² − [625/(4*R²) − 25 + R²] = R² − 625/(4*R²) + 25 − R² = 25 − 625/(4*R²)
(xB + R)/(xD + R) = 5/2
(xB + R)/(xC + R) = 5/2
xC + R = (2/5)*(xB + R) = (2/5)*[25/(2*R) − R + R] = (2/5)*25/(2*R) = 5/R
⇒ xC = 5/R − R
2*yB/yC = 5/2
yC = (4/5)*yB
yC² = (4/5)²*yB² = (16/25)*[25 − 625/(4*R²)] = 16 − 100/R²
xC² + yC² = R²
(5/R − R)² + (16 − 100/R²) = R²
25/R² − 10 + R² + 16 − 100/R² = R²
6 − 75/R² = 0
6 = 75/R²
⇒ R² = 75/6 = 25/2
Now we are able to calculate the area of the semicircle:
A = πR²/2 = π*(25/2)/2 = 25π/4
Best regards from Germany
This time your solution is definitely much easier than mine. 👍
By the way: for the final calculation of R you could have used the intersecting chords theorem again:
2√2*2√2 = √2*(2R − √2)
4√2 = 2R − √2
5√2 = 2R
5√2/2 = R
Excellent!
Thanks for sharing ❤️
Instead of joining O and F, AE and CF are two intersecting chords.
By this we can find out diameter., then area.
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
Excellent!
Thanks for sharing ❤️
Thank you!
You are very welcome!
Thanks ❤️
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.
I don’t think so. I think that, as a professor, Premath often gives a side of the problem and he leaves the rest for viewers.
Thanks for the feedback ❤️
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.
Excellent!👍
Thanks for sharing ❤️
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
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
Excellent!
Thanks for sharing ❤️
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..
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.
Excellent!
Thanks for sharing ❤️
Can we just join OB and find the radius r ( AO and OB = r) of isosceles Triangle AOB as angle OAB and thus ABO are 45 degrees (as AO and OB = r) so angle AOB is 90 degrees. If AB is 5 and Sine 45 = 1/sqrt 2 then r = 5/sqrt 2. Then we can find the area by (pi.r squared)/2.
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
Excellent!
Thanks for sharing ❤️
Sir if we join o to b than oA and oB will be radius and AB will be 5 so in a right isoscalane triangle ABO we can use Ao square +BO square=AB square by using that we can directly find r =5root2 than we can use pi r square by 2
Thanks for the feedback ❤️
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 l1=2:l2=3:sw=l1^2/(l1+l2)/10:l4=1+sw:goto 50
30 l3=sqr(l1^2-l4^2):r=(4*l3^2+l4^2)/2/l4:l5=(l1+l2)/l1*l3
40 l6=sqr((l1+l2)^2-l5^2):dgu1=l5^2/r^2:dgu2=(sqr((l1+l2)^2-l5^2)-r)^2/r^2
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
CD=a...le equazioni sono (r-√(4-a^2))^2+(2a)^2=r^2...arccos(5/2r)=arcsin(a/2)...dalle quali risultano a=√2,r=5/√2
Excellent!
Thanks for sharing ❤️
Let DE=CD=a and Radius=r
So a(3a)=2(3)
3a^2=6
a^2=2
So a=√2
AE=√2^2-(√2)^2=√2
Connect O to C
OE=r-√2
In ∆ OCE
OE^2+CE^2=OC^2
(r-√2)^2+(2√2)^2=r^2
So r=5√2/2
Semicircle area=1/2(π)(5√2/2)^2=25π/4 square units =19.63 square units.❤❤❤
Excellent!
Thanks for sharing ❤️
Once you find the x value, √2 using intersecting chord theorem, you can use the same method again to find the radius.value. calculating semicircle area is easy afterwards. ✌️
Thanks for the feedback ❤️
May I know how ABF is an isosceles triangle? Angles ABF or AFB are neither right angles nor at 45 degrees?
CD=DE=x 2*3=x*3x 3x²=6 x=√2
AE=[2²-(√2)²]=√2 AO=OF=r AF=AO+OF=r+r=2r EF=2r-√2
CE=CD+DE=2√2 2√2*2√2=√2*(2r-√2) 8=2√2r-2 2√2r=10 r=5/√2
Semicircle area = 5/√2*5/√2*π*1/2 = 25π/4
Excellent!
Thanks for sharing ❤️
19.625 Sq.units
Excellent!
Thanks for sharing ❤️
If square root is not allowed. CD = ED = x ⇒ x^2=2 ⇒ AE^2 = 2 ⇒ (r-x)^2 + (2x)^2 = r^2 ⇒ -2rx + x^2 + 4x^2 = 0 ⇒ 2rx = 2 + 2*4 ⇒ rx = 5 ⇒ r^2x^2 = 25 ⇒ r^2=25/2.😎
Thanks for sharing ❤️
Let's draw the complete circle (radius R)
Let's name x = CD = DE
The intersecting chord theorem gives: 2.3 = x.(3.x),
so x = sqrt(2). Now in triangle ADE: AE^2 = 4 - 2 = 2
So AE = sqrt(2, meaning that ADE is right isosceles,
angle OAB = 45° Let now H be the orthogonal projection of B on (AO). The triangle ABH is also right isosceles,so BH= AH = AB/sqrt(2) = 5/sqrt(2) and OH = (5/sqrt(2)) - R
In right triangle OBH we have: OB^2 = OH^2 +BH^2, so
R^2 = ((5/sqrt(2) -R))^2 + (5/sqrt(2))^2
or: R^2 = (25/2) + R^2 - (10/sqrt(2)).R +(25/2)
giving R = (25/10).sqrt(2) = (5/2).sqrt(2)
The area of the semi circle is(Pi/2).(R^2) = (25/4).Pi.
Excellent!
Thanks for sharing ❤️
19.625 Sq.units
Excellent!
Thanks for sharing ❤️