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
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" !!
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.
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
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.
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
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.
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.
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.
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..
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
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.❤❤❤
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.
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
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
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
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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 ❤️
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" !!
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
Thanks Sir
That’s very good method and understandable for solve
With my respects
❤❤❤❤❤
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 ❤️
Thank you!
You are very welcome!
Thanks ❤️
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.
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’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 ❤️
Let's find the area:
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..
...
....
.....
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 ❤️
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.
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 ❤️
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 ❤️
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
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Thanks for sharing ❤️
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
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Instead of joining O and F, AE and CF are two intersecting chords.
By this we can find out diameter., then area.
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
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Thanks for sharing ❤️
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.
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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..
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
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.❤❤❤
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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.
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
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 ❤️
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
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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
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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?
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.😎
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19.625 Sq.units
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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.
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Thanks for sharing ❤️
19.625 Sq.units
Excellent!
Thanks for sharing ❤️