Can you find the area of the Yellow shaded semicircle? | (Triangle) | #math #maths | #geometry

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Can you find the area of the Yellow shaded semicircle? | (Triangle) | #math #maths | #geometry

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[yalchingedikgedik]

Very nice
Thanks Sir
If there are cases on algebra math.
Thanks
Good luck with glades
❤❤❤❤❤

[hongningsuen]

Method using Thales theorem, trigonometric ratio, similar triangles:
1. Join AE forming ∆AEB and ∆AEC.
2. ∆AEB is right-angled triangle in semicircle (Thales theorem)
Hence ∆CAE is also right-angled triangle.
3. In right-angled ∆CAE, AC = EC/sin45 = EC√2
4. ∆ACB ~ ∆ ECD (AAA) (general property for 2 secants from a point)
Hence AB/ED = AC/EC
AB/10 = (EC√2)/EC
AB = (10/√2)
5. Radius of circle = AB/2 = 5/√2
6. Area of yellow semicircle = (1/2)π(5/√2)^2 = 25π

This method works with any random acute angle ACB, while your method works only with angle ACB = 45.

[anthonycheng]

Use similar triangle (ext angle cyc quad) and join AE, angle CAB + angle CBA = 135^0, angle AEB = 90^0, angle DAE = 45^0, then AE = CE. Then use similar triangle to find AC/CE = sqrt(2), from similar triangle to get AB= 10*sqrt(2), r = 5*sqrt(2).,

[prossvay]

Semicircle area=1/2(π)(5√2)^2=25π =78.54 square units.❤❤❤

[jimlocke]

Focus on triangle ABC. Let <CAB = Xº and <CBA = Yº. Then, sum of angles of a triangle = 180º, so Xº + Yº + 45º = 180º and Xº + Yº = 135º. Inscribed <CAB = Xº intercepts arc DB, so DB = 2Xº. Inscribed <CBA = Yº intercepts arc AE, so AE = 2Yº. AE = AD + DE = 2Yº and DB = DE + EB = 2Xº, so AE + BD = AD + DE + DE + EB = 2Yº + 2Xº. However 2Yº + 2Xº = 2(Yº + Xº) = 270º and AD + DE + EB = AB = 180º. So arc DE = 270º - 180º = 90º. Construct OB and OD. Central <DOE = 90º because it intercepts arc DE. Triangle DOE, with sides OB = OD = r, is a isosceles right triangle. r = 5√2 and r² = 50. Proceed at 6:06.

[solimana-soli]

I solved using cosine rule, setting CE = x and CD = y
then being ACE isosceles right triangle as shown in the video
CA = x√ 2
and being also CDB isosceles right triangle
CB = y√ 2
applying cosine rule on CDE
x² + y² - √ 2xy = 100
applying cosine rule on ABC
2x² + 2y² - 2√ 2xy = 4r²
that means that
2r² = 100
so area semicircle = 25 pi
needless to say that PreMath solution is better...

[misterenter-izrz]

Very brilliant puzzle, it leads us to show the angle at the middle sector is 90°, 90=225-135=(360-135)-135 😮, therefore the radius r=10/sqrt(2), and the answer is 1/2 r^2 pi=25 pi.😊

[LuisdeBritoCamacho]

STEP-BY-STEP RESOLUTION PROPOSAL :

01) Angle of Intersecting Secants Theorem :
02) " If two secant (of a circle) lines intersect outside a circle, then the measure of an angle formed by those two lines is one half the positive difference of the measures of the intercepted arcs."
03) Angle ACB = 45º
04) Arc DE = Xº
04) Arc AB = 180º

05) 45º = (180º - Xº) / 2
06) 90º = 180º - Xº
07) Xº = 180º - 90º
08) Xº = 90º
09) So : Arc (DO) = 90º
10) Inner Angle DOE = 90º
11) OD = OE = R
12) R^2 + R^2 = 100
13) 2R^2 = 100
14) R^2 = 50
15) Semicircle Area (SA) = (Pi * R^2) / 2
16) SA = 50Pi / 2
17) SA = 25Pi Square Units.

OUR ANSWER : The Semicircle Area is 25Pi Square Units.

[yxkrurq]

Let's try this using properties of secants. Let radius of circle be "r" and let /_DOE = alpha. The angle formed by two secants (here it is /_DCE = 45 degrees) which intersect outside the circle is half the difference of the of the intersecting arcs (angles) so /_DCE = 1/2(180 - alpha) or 45 = 1/2(180 - alpha) and alpha = 90 degrees. Thus triangle DOE is a right angle triangle with hypotenuse 10 and sides "r" and "r" so r is 10/sqrt(2) or 5*sqrt(2) and the area of the semi-circle is 1/2*pi*r^2 = 25*pi square units.

[quigonkenny]

Let ∠ABE = θ. As ∆ABC is a triangle, then ∠CAB = 180°-θ-45° = 135°-θ = α.

As A and B are ends of a diameter and D and E are points on the circumference, by Thales' Theorem, ∠BEA = ∠BDA = 90°. As ∆BDA is a triangle, ∠ABD = 180°-90°-α = 90°-(135°-θ) = θ-45°. As ∠ABE = θ, then ∠DBE = 45°.

As ∠DBE covers the arc DE and is 45° at a point B on the circumference, then ∠DOE, which covers the same arc, will be 2(45°) = 90° at the center O.

As ∆DOE is a right triangle, it is subject to Pythagoras.

Triangle ∆DOE:
OE² + OD² = ED²
r² + r² = 10²
2r² = 100
r² = 100/2 = 50
r = √50 = 5√2

Semicircle O:
Aₒ = πr²/2 = 50π/2 = 25π ≈ 78.54 sq units

[marioalb]

R = 10 cos 45° = 10/√2 = 5√2 cm
A = ½πR² = ½π 5².2
A = 25π cm² ( Solved √ )

[MateusMuila]

Connecting D, E, O was a genius move.

[giuseppemalaguti]

arccos(5/r)=45...per gli angoli di un triangolo ciclico ..5/r=√2/2..r=10/√2=5√2

[bdchatto]

Angle DCE is not on the perimeter of the circle, whose chord is DC

[JSSTyger]

Strangely, I get angle DOE = 90° no matter if line DE is parallel to line AB or not. OK...r² = 50 and thus A = π(50)/2 = 25π.

[wackojacko]

3:57 ...Gonna use Central Angle Theorem for my next crop circles. 🙂