Can you find area of the Blue shaded region? | (Semicircle) | #math #maths | #geometry

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Can you find area of the Blue shaded region? | (Semicircle) | #math #maths | #geometry

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[shivansmishra]

I love your voice, it makes me smile 😊😊😊😊

[phungpham]

1/ Draw the full circle. Extend DE cutting the circle at D’ and draw the chord CC’ perpendicular to the chord DD’ intersecting at point H.
We have: HE=1 and the 2 chord are equal = 4
So the distance from the center O to DD’= 1—-> r= sqrt5 and the area of the triangle DOD’=2
2/ Calculating sector DOD’
By using the cosine theorem: sq DD’= sqr+sqr -2 sqr cos DOD’
—> cos DOD ‘ = -0.6 7:47 7:47 7:47
Angle DOD’= 126.87 degrees
Area of sector DOD’= pix5x126.87/(360)=5.535
Area of the blue region=1/2 (5.535-2)=1.768 sq units

[prossvay]

Connect O to C and to D
Let OE=x ; OF=3-x; and Radius of semicircle=R
In ∆OCF
1^2+(3-x)^2=R^2
1+9-6x+x^2=R^2
so R^2=x^2-6x+10 (1)
In ∆OED
2^2+x^2=R^2
R^2=x^2+4 (2)
(1) and (2)
x^2-6x+10=x^2+4
so x=1
(2) R^2=4+1
so R=√5
Sin(LDOE)=2/√5
so LDOE=63.43°
So area of Blue square units.❤❤❤

[marcgriselhubert]

In triangle OED: OE^2 = R^2 -4 (R:radius of the circle)
In triangleOFC: OF^2 = R^2 -1 We substract thes equations and obtain: OE^2 - OF^2 = -3
Knowing that OE + OF = 3, we get that OE - OF = -1 and that OE = 1 and OF = 2
We replace OE by 1 in the first equation and get R^2 = 5.
and R = sqrt(5).
Tan(angleAOD) = ED/OE =2, so angleAOD = Arctan(2)
The area of sectorAOD is then 5.Pi.Arctan(2)
The area of triangle OED is (1/2).OE;ED = (1/2).1.2 = 1.
Finally the area of the blue region is: 5.Pi.Arctan(2) -1.

[unknownidentity]

Let's find the area:
.
..
...
....


The triangles OCF and OED are both right triangles, so we can apply the Pythagorean theorem. With R being the the radius of the semicircle we obtain:

OF² + CF² = OC²
OE² + DE² = OD²

(EF − OE)² + CF² = R²
OE² + DE² = R²

(EF − OE)² + CF² = OE² + DE²
(3 − OE)² + 1² = OE² + 2²
9 − 6*OE + OE² + 1 = OE² + 4
6 − 6*OE = 0
⇒ OE = 1

R² = OE² + DE² = 1² + 2² = 1 + 4 = 5

Now we are able to calculate the area of the blue region:

A(blue)
= A(circle sector OAD) − A(triangle OED)
= (1/2)*R²*(∠OED) − (1/2)*OE*DE
= (1/2)*R²*arctan(DE/OE) − (1/2)*OE*DE
= (1/2)*5*arctan(2/1) − (1/2)*1*2
≈ 1.768

Best regards from Germany

[misterenter-izrz]

Easy.😂4+a^2=1+(3-a)^2, 6a=6, a=1, r^2=5, therefore the answer is (5pi)(tan^(-1)(2))/360°×)) -1=

[quigonkenny]

Draw OD and OC.

Triangle ∆DEO:
DE² + OE² = OD²
2² + OE² = r²
OE² = r² - 4
OE = √(r²-4)

Triangle ∆OFC:
FC² + OF² = OC²
1² + OF² = r²
OF² = r² - 1
OF = √(r²-1)

EF = OE + OF
3 = √(r²-4) + √(r²-1)
3 - √(r²-1) = √(r²-4)
r² - 4 = 9 + (r²-1) - 6√(r²-1)
6√(r²-1) = 12
√(r²-1) = 2
r² - 1 = 4
r² = 5
r = √5

OE² = (√5)² - 4 = 5 - 4
OE = √1 = 1

Let ∠EOD = θ. The area of the half circular segment is equal to the area of the sector of angle θ minus the area of triangle ∆DEO.

sin⁻¹(2/√5) ≈ 63.435°

Area = (63.435°/360°)π(√5)² - 1(2)/2
Area ≈ 0.881π - 1 ≈ 1.768 sq units

[cyruschang]

✓(r^2 - 4) + ✓(r^2 - 1) = 3
(r^2 - 4) = 9 + (r^2 - 1) - 6✓(r^2 - 1)
✓(r^2 - 1) = 2 => r^2 - 1 = 4 => r = ✓5
Sin^-1(2/✓5) = θ (in units of π) = τ (in degrees)
blue area = (r^2)(π)(θ/2π) - (1 x 2/2) = (5θ)/2 - 1 = (5θ - 2)/2
or
blue area = (r^2)(π)(τ/360) - (1 x 2/2) = (5τπ/360) - 1 = (τπ/72) - 1 = (τπ - 72)/72

[AndreasPfizenmaier-yw]

3^2+1^2=sqrt 10 =Radius from the very beginning

[LuisdeBritoCamacho]

STEP-BY-STEP RESOLUTION PROPOSAL :

01) Let R be the Radius of the Semicircle. So, OD = OC = R
02) OE = X
03) OF = (3 - X)
04) R^2 = X^2 + 4
05) R^2 = (3 - X)^2 + 1
06) As : R^2 = R^2 ; then :
07) X^2 + 4 = (3 - X) ^2 + 1 ; 4 = 9 - 6X + 1 ; 6X = 6 ; X = 1
08) OE = 1
09) OF = 2
10) R^2 = 5 ; R = sqrt(5)
11) Calculating Area of Sector OAD. Let Angle (DOE) = (a * Pi) Radians
12) General Formula : ((a * Pi) * R^2) / 2
13) aº ~ 63, 435º
14) Angle "a" (in Radians) ~ (0, 35242 * Pi) Radians
15) Sector Area ~ 2, 768
6) Blue Region Area ~ 2, 768 - 1 ~ 1, 768
17) ANSWER : The Blue Region Area equal to 1, 768 Square Units.

Best wishes from the International Center for Islamic Studies - Department of Ancient Mathematical Thinking, Knowledge and Wisdom in Cordoba Caliphate.