Can you find area of the Yellow Semicircle? | (Perpendiculars) |
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- Опубліковано 10 лют 2025
- Learn how to find the value of the Yellow Semicircle. Important Geometry and Algebra skills are also explained: area of the circle formula; Pythagorean theorem. Step-by-step tutorial by PreMath.com.
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Can you find area of the Yellow Semicircle? | (Perpendiculars) | #math #maths | #geometry
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I really enjoy your math challenges. Thank you.
Glad to hear that!
You are very welcome!😀
Thanks for watching❤🙏
Great question sir
I solved it 😊😊😊
Suppose AD=a and FB=b in the right triangles ABC and ABE we have AD*BD=CD² and AF*BF=FE² and from this a*(b+3)=1 and (a+3)*b=4 by solving these two equations we get a=√5-2,b=√5-1 and from this 2R=a+b+3=2√5 so R=√5 and from this the area of the shaded region is equal to π*√5²/2=5π/2
4 + Х^2 = 1 + (3 - Х)^2
Х = 1, R = 5^.5
S = 2.5 Pi
Everytime we have this ratio: 3/(2+1) = 1, then the right triangles that are formed, are congruents, then angle COE=90°
Similarity of triangles:
2/R=x/R --> x=2 cm
1/R=y/R --> y= 1cm
R²= x²+1² = 2²+y² = 5 cm²
A = ½πR² = 2,5π cm² ( Solved √ )
Pytagorean theorem, twice
R² = (3-x)²+1²=x²+2²
(3²-6x+x²)+1²=x²+2²
6x = 3²+1²-2²= 6 --> x=1
R = x²+2² = 5
A = ½πR² = 2,5π cm² ( Solved √ )
Thank you!
a+b=3...r^2=1+a^2=4+b^2...(a+b)(a-b)=3....a-b=1...2a=4..a=2...r^2=5...Ayellow=πr^2/2=5π/2
Intersecting chords theorem, twice:
b.(3+a)=3b+ab = 2²
a.(3+b)=3a+ab = 1²
Subtracting:
3b-3a= 4-1=3
b-a=1 ; b+a=1+2a
2R= 3+a+b =3+(1+2a)= 4+2a
R = 2+a= x+a --> x=2cm
Pytagorean theorem:
R²= 1²+x² =1²+2²= 5
A = ½πR² = 2,5π cm² (Solved √)
Nice
S=2,5π≈7,854≈7,85
Попробуем в систему: 1+x²=4+y²; x+y=3. x²-y²=3=(x+y)(x-y)=3(x-y), откуда x-y=1. Если пропорции на чертеже верные, из этой упрощённой системы х=2, у=1. Т. е. имеем равные треугольники с гипотенузами √5=r. Тогда искомая площадь полукруга S=5π/2=2½π.
Very easy
Bom dia Mestre
Obrigado pelos ensinamentos
Thanks sir for such questions 😊😊
Circular ring area, method:
A₁=½(¼π2²)= ½π cm²= ½π(R²-x²)
R²= 1+x²
A₂=½(¼π4²)=2π cm²= ½π(R²-(3-x)²)
R²= 4+(3-x)²
Equalling:
1+x² = 4+(3²-6x+x²)
6x = 4-1+9= 12 --> x=2 cm
R² = 1+x² = 5
A = ½πR² = 2,5π (Solved √)
I use 1983 era Big Mack’s for “units” and it works out perfectly.
CC`||AB, and C` in line EF. EC`= 1. 1*3=x*3 -> x=1 - > In circle rectangle 2x4, find diagonal 2^2+4^2=2sqrt(5) -> R=sqrt(5)
La paralela a AB por C corta a EF en M y a la circunferencia en G ---> Potencia de M respecto a la circunferencia =1*3=3*MG---> MG=1---> (2r)²=(2*CD)²+(CG)² =2²+4²---> r²=5---> Área del semicírculo =5π/2 Ud².
Gracias y un saludo cordial.
I did it a slightly different way. Not as good as Premath's but it got the job done. 2r = x + y + 3. 1^2 = x (y+3) and 2^2 = y (x+3). If you expand those last 2 equations and subtract, you get y = x +1. If you substitute that into 1 = xy + 3x you get 1 = x(x+1) + 3x. That simplifies to x^2 + 4x -1 = 0. If you use the quadratic equation that yields x = -2 + sqrt5. From the 1st equation 2r = x + y + 3 you substitute in y = x +1 which yields r = x +2. If you substitute x = -2 + sqrt5 into r = x + 2 you get r = sqrt5. Then of course the area of a semicircle will be A = 1/2* sqrt5 * pi = 5/2*pi which is approximately 7.8540 square units.
Draw the 2 radii, pythag the 2 triangles, r=r (hypotenuses), equate. x=2, re-pythag either triangle to find r=√5.
A=2.5pi
Thank you for posting easy questions for neophyts !
OF = y
DO = x
CO = r (radius)
(3-y)² + 1² = r²
y² - 6y + 10 = r²
(3-x)² + 2² = r²
x² - 6x +13 = r²
x + y = 3
x = 3 -y
y = 3 -x
(3 -x)² - 6(3 -x) + 10 = x² - 6x +13
9 - 6x + x² - 18 + 6x + 10 = x² - 6x +13
6x = 12
x = 2
r² = 2² + 1² = 5
semicircle = (1/2)πr² = 5/2π
CO=OE=r OF=x DO=3-x
x²+2²=r² (3-x)²+1²=r²
x²+4=10-6x+x² 6x=6 x=1 r²=5
Semicircle area = r²π/2 = 5π/2
Be a = OD, then OF = 3 -a; and be R the radius of the circle. In triangle ODC we have OC^2 = OD^2+DC^2, so R^2 = a^2 + 1
In triangle OFE we have OE^2 = OF^2 + FE^2, so R^2 = (3 -a)^2 + 4. From these two previous equations we get that a^2 + 1 = (a -3)^2 + 4
We develop: a^2 + 1 = a^2 -6.a + 9 + 4, we simplify and get that 6.a = 9 +4 -1 = 12, and so a = 2
Now in triangle ODC: R^2 = a^2 + 1 = 2^2 + 1 = 5. The area of the semi-circle is ((Pi.(R^2))/2, so it is (5/2).Pi
(I see that I did it just like you did...!)
2.5pi
√(r^2 - 1^2) + √(r^2 - 2^2) = 3
√(r^2 - 4) = 3 - √(r^2 - 1)
(r^2 - 4) = 9 + (r^2 - 1) - 6√(r^2 - 1)
√(r^2 - 1) = 2
r^2 = 5
Semicircle area = 5π/2
7.854 or 5/2 pi
Let the radius = r
draw two triangles with base in n and 3-n (3-n + n =3)
then r^2 = 2^2 + n^2
and r^2 = 1^2 + (3-n)^2
Hence, 2^2 + n^2 = 1 + 9 + n^2 - 6n
4- 10 = -6n
-6 = -6n
n= 1
Hence r^2 = 4 + 1
= 5
r= sqrt 5
area of circle = pi sqrt 5* sqrt 5 = 5pi
area of semi-circle = 5pi/2 Answer or 7.854
r=abc/(4S)
r²π/2=(abc)²π/(2•16S²)
=(16•18•10)π/(2•16•6²)
=5π/2
Sir you are in which country please reply me.
Solution:
Applying Pythagorean Theorem in ∆CDO
CD = 1
DO = a
CO = r
CD² + DO² = CO²
(1)² + (a)² = r²
r² = a² + 1 ... ¹
Applying, once again, Pythagorean Theorem in ∆EFO
EF = 2
FO = 3 - a
EO = r
EF² + FO² = EO²
(2)² + (3 - a)² = r²
4 + 9 - 6a + a² = r²
r² = a² - 6a + 13 ... ²
Equaling Equation ¹ and Equation ², it will be:
a² + 1 = a² - 6a + 13
1 = - 6a + 13
6a = 12
a = 2
Substituting in Equation ¹
r² = (2)² + 1
r² = 5
r = √5
Semicircle Area = ½ π r²
Semicircle Area = ½ π (√5)²
Semicircle Area = 5π/2 square units
Thus:
Semicircle Area = 5π/2 square units ✅
Semicircle Area ≈ 7.8539 square units ✅
Area is 2.5 pi
r²=1+(3-x)²
r²=4+x²
4=1+9-6x
6x=6
x=1
r=√5
Area=(1/2)5π
S= 2,5π
5π÷2
Let's find the area:
.
..
...
....
.....
The triangles OCD and OEF are right triangles, so we can apply the Pythagorean theorem. With r being the radius of the semicircle we obtain:
r² = OC² = OD² + CD²
r² = OE² = OF² + EF²
OD² + CD² = OF² + EF²
OD² + CD² = (DF − OD)² + EF²
OD² + 1² = (3 − OD)² + 2²
OD² + 1 = 9 − 6*OD + OD² + 4
6*OD = 12
⇒ OD = 12/6 = 2
⇒ OF = DF − OD = 3 − 2 = 1
r² = OD² + CD² = 2² + 1² = 4 + 1 = 5
r² = OF² + EF² = 1² + 2² = 1 + 4 = 5 ✅
Now we are able to calculate the area of the yellow semicircle:
A = πr²/2 = 5π/2
MY RESOLUTION PROPOSAL :
01) OA = OC = OE = OB = R
02) OD + OF = DF = 3
03) OD = X
04) OF = (3 - X)
05) R^2 = X^2 + 1
06) R^2 = (3 - X)^2 + 4
07) X^2 + 1 = (3 - X)^2 + 4
08) X^2 + 1 = 9 - 6X + X^2 + 4
09) 6X = 9 + 4 - 1
10) 6X = 12
11) X = 2
12) OD = 2 and OF = 1
13) R^2 = 1 + 4
14) R^2 = 5
15) R = sqrt(5)
16) Semicircle Area = 5Pi/2
Therefore,
MY BEST ANSWER IS :
Semicircle Area equal to 5Pi/2 Square Units or approx. equal to 7,854 Square Units.