Can you find the area of the tiny Red shaded region? | (Semicircle and Triangle) |
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- เผยแพร่เมื่อ 9 ก.พ. 2025
- Learn how to find the area of the tiny Red shaded region. Important Geometry and Algebra skills are also explained: Thales' theorem; area of a triangle formula; area of a sector formula; Pythagorean theorem. Step-by-step tutorial by PreMath.com
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Thales Theorem is a special case of the Inscribed Angle Theorem and that Angle ABE is always 90⁰. Then Sector Area minus Area of Triangle EOD is just iceing on the cake. 🙂
Thanks ❤️🌹
Awesome, the sheer adroitness of your clarification combined with the quality of the video makes for less intricacies and obstacles!
Thanks ❤️🌹
1/Calculating the radius R:: connect EB, we have EBA is a right triangle of which the angle A= 45 degrees EBA is an right isosceles one. So the diameter EA =8sqrt2----> r= 4sqrt2.
2/Calculating the value of angle EOD: notice that the triangle ADC is a special right 30-90-60 triangle-----> angle DAC= 30 degrees --------> angle EAD= 45-30=15 degrees-----> the angle EOD = 30 degrees ( EOD is the angle at the center).
3/Calculating the Area of the sector OED:: = pi sq(4 sqrt2) , 30/360 = pi . 32. 1/12 = 8 pi/3 (1)
4/Calculating the area of the triangle OED: drop the height EH to OD. We have the triangle EHD is a special 30-90-60 too, so EH= EO/2= 2 sqrt2
the area of the triangle EOD= 1/2 . 2sqrt2 .4sqrt2= 8 (2)
5/ Area of the red segment= (1)- (2) = 8pi/3 -8 =
Got it! A very good exercise, many thanks.
Very Very nice video sir 👍
So nice of you
Thanks ❤️🌹
Can you please make some more videos on algebra also
From triangle AEC you can immediately see that angle AEC = 75 = 180 - 69 - 45.
Angle is 75 degrees can be obtained from initial given triangle itself. Also, diameter can be seen to be the hypotenuse of the isosceles triangle and diameter is 8.sqrt(2).
I solved the problem using various geometry and the law of sines, and subsequently solved it using integral calculus. In both cases I reached identical outcomes matching your result.
Nice. Alternatively, angle E=75 and after drawing radius OD we have Isosceles triangle OED which forces central angle EOD=30. To find r, inscribed angle A=45 forces central angle EOB=90 which makes triangle AOB a 45-45-90 right triangle. Hence, r=OA=4√2 so with angle EOD=30 the red area is found like you did.
Trigonometric magic!
Thanks ❤️
8 is equal to a side length of a Square, Hence the diagonal, the Radius turns out to be 8* sqrt 2.
Pretty much the same way I solved it, except that I calculated ∠OED from △ACE:
∠OED = ∠AEC = 180° − ∠ACE − ∠CAE = 180° − 60° − 45° = 75°
I calculated angle OED from triangle ACE.
Also, EA = sq.rt.128, so r = 1/2 x sq.rt 128 = sq.rt.32.
Then sector area = Pi x (sq.rt.32)^2 /12.= Pi x 32 /12 = Pi x 8 /3.
Nice, many thanks, Sir! (8/3)(π - 3)
Very good!
Thanks ❤️🌹
From D we can drop the perpendicular to EO, calling H the point of intersection; we have the 30°-60°-90° triangle HDO, so we can put 2DH= 4square root of 2, so DH= 2square root of 2, then finally the same method you finished the problem putting EDO= EO•DH/2
Thanks
The area of a circular sector is ½ R² θ where R is the radius of the circle and θ the central angle in radians.
The area of the isosceles triangles defined by the same three points as this circular sector is ½ R² sin θ.
Therefore, the area of a circular segment is ½ R² θ − ½ R² sin θ which simplifies to ½ R² (θ − sin θ).
yes, but you didn't find theta or R, so your answer is just incomplete
0.37758 Answer
A different approach
Draw a straight line from B to E, forming an isosceles right triangle ABE.
BE = 8 due to BA= BE ( 45 45 and 90 degrees)
AE = (diameter) 8 sqrt 2 due 45 45 90
Hence, the radius of the circle = 4 sqrt 2
Draw a straight line from O to D . This is the radius.
But O E is also the radius. Hence, triangle EDO is an isosceles with the angles.
75 75 and 30 degrees
length ED of triangle EDO can be derived from the cosine formula
c= sqrt(a^2 + b^2 - 2ab cos 30
a= 4 sqrt 2 , b = 4 sqrt 2 Gamma 30
c = 2.93
The area of triangle EDO can be derived from using Heron's formula
and the sides 2.93, 4 sqrt 2, and 4 sqrt 2
Area of of EDO = 8
Since the area of EDO + the small red-shaded region equal
1/12 of the circle (since the circle covers 360 degrees and 360/30 =12),
then the area of EDO + the red segment = 1/12 area of the circle
The area of the full circle = 4 sqrt 2 pi r^2 = 100.53 square units
Hence the are of EDO + the small red segment = 100.53096/12 or 8.37758
Since the area of EDO (as calculated earlier) = 8
then the area of the red segment = 8.37758 - 8 = 0.37758
Area of the red shaded region=π(4√2)^2(30/360)-1/2(4√2)^2sin(30)=8π/3-8=0.38 square units. Thanks ❤❤
Thanks ❤️🌹
The AEC=75°=180°-60°-45°....
It is only a half