Connecting O to B, and O to F, forms 2 congruent triangles OBC and OBF. Tan OBC = 1/2 = 0.5. Angle OBC = tan -1 of 0.5 = 26.565 degrees. Angle FBC = 2 x 26.565 = 53.13 degrees. Angle ABE = 90 - 53.13 = 36.87 degrees. Tan 36.87 = AE / 2. AE = 2 x tan 36.87. AE = 2 x 0.75 = 1.5.
A slightly different aproach: area of circle =π/2; area of circle = 4π = πr^2; so r = 1; so far, so good. Now considering the right side of the square, draw OB. Since OC = 1 and CB = 2, invoke Pythagoras and OB^2 = 2^2 + 2^2 = 5, so OB = √5. Applying Pythagoras to triangle OBF, we have BF^2 = OB^2 - 1^2 = 5 - 1 = 4; so BF = 2. Let x = DE = EF. They are equal from the two-tangent theorem, and/ or because they are the corresponding sides of the two congruent right triangles DEO and EFO. Coming back to triangle EAB and calling on Pythagoras again: 2^2 + (2 - x)^2 = (2 + x)^2; expanding both sides: 4 + 4 - 4x + x^2 = (2 + x)^2 = 4 + 4x + x^2; collecting terms, we have 8 - 4x = 4 + 4x; 4 = 8x; so x = 1/2; and AE = 2 -- 1/2 = 3/2. So the area of the green triangle is (1/2) (2) (3/2) = 3/2. Cheers. 🤠
Semi circle : A = ½πr² = ½π r = 1 cm Side of square: s = 2r = 2 cm Green right triangle: base = 2 height = 2-x hypotenuse = 2+x 2²+(2-x)²=(2+x)² 2²+(2²-4x+x²)=(2²+4x+x²) 8x = 4 x = 0,5 cm Area Green triangle: A = ½b.h A = ½ . 2 . (2-0,5) A = 1,5 cm² ( Solved √ )
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Once we find that the radius is 1 (and the side of the square is 2), and that by the theorem of tangents DE=BF and BF=BC, we can proceed as in the following. DEF + AEF = 180°; AEF=CBF (alternate interior angles of two parallel lines (AD ∥ BC) cut by the transversal BE); so DEF+CBF=180°. EO is the bisectrix of DEF, BO is the bisectrix of CBF so EOB is a right triangle with EOB=90°. So by Euclid's first theorem the height OF relative to the hypotenuse EB is the proportional mean between the projections EF and FB and we can write: OF²=EF·FB --> 1²=EF·2 --> EF=½ EF=DE --> AE=2-½=3/2 Area=½·AB·AE=½·2·(3/2)=3/2
Si (πr²/2)=π/2 → r²=1→r=1 → Lado del cuadrado ABCD, CD=2r=2=CB=BF → ∠OBF=∠FOE → Razón de semejanza =1/2 entre ∆EFO y ∆OFB , ambos triángulos rectángulos por ser OF⟂EB → EF=OF/2=1/2 → EF=DE=1/2 → AE=AD-ED → AE=2-(1/2)=3/2 → Área Verde =Área de ∆EAB =(1/2)2(3/2)=3/2 = 1.5 cm² Gracias y un saludo cordial.
Join the center of the semi circle to points B and E and F. Triangles OFB and OCB congruent (rhs) Triangles EDO and EFO are congruent. By labelling angles of the above mentioned triangles we can deduce OFB and EDO are similar. Since OF=1 cm and FB = 2 cm, ED =1/2 cm, EA= 1.5 cm and area of ABE=1/2x2x1.5=1.5cmsq
So beautiful design of the figure 🤩. First we note that there are two pairs of similar right angled triangles, so OF=1, FE=1/2=DE, thus AE=3/2, so the answer is 3/2x2/2=3/2, done.😊
Can we say that the quadrangle OCBF is similar to quadrangle DOFE (All four angles in both are equal) and so the relations between corresponded sides in these two quadrangles are equal so we can find DE part and then calculate the area of ABE triangle.
The easiest way I find: Getting radis = 1 => esquare is 2x2(side times side), then in the "green triangle" we get big leg(mayor hick) = 2, and cos' it's similar to 3/4/5 triangle, then 4.k=2("k" is the ratio/proportion = 1/2), so for 3 the hick is 3×(1/2) = 3/2, so taking this hick as height, and 2 as a base, we get [(3/2) × 2]/2 = 3/2 , 1.5 cm² as the area. Done! The way the professor do is good to manage/use other algebra/geometrics conceps, then is too valid also. It depende on the grade/complexity we want to teach/learn some. My greetings to all. (Sorry about my english). 😑/😅 👍😜🙃😋🤩🇦🇷⭐⭐⭐
Got me twice with the two tangent theorem. After the first one I was thinking of trying to work out the intersection by calculating the sphere intersect, but I figured you'd have something much easier, so I unpaused and you did, lol.
There is a square area 1m2. You need to do a truncated cone that has max. Volume. Find a diameter on a top & on the bottom of this cone & side surface made from this square area Draw a development of this parts on this 1m2 area.
r=1, side of square =2, side of triangle = 2, 2x, 2+x. further according to the Pythagorean theorem: (2+х) ²=2²+(2-х) ². Second method: ∆QCB=∆QFB, QF=QC=r,
Your solution was too simple. Let's see if we can make this unnecessarily complex yet still mathematically sound... Semicircle O: A = πr²/2 π/2 = πr²/2 r² = (π/2)(2/π) = 1 r = 1 By Two Tangent Theorem, BF = BC = 2r = 2. Let G be the point on CD and J be the point on AB where GJ is perpendicular to AB and CD and passes through F. Let FJ = x and GF = 2-x. Let ∠FOG = α and ∠GFO be β, where β = 90°- α. As ∠OFB = 90°, ∠BFJ = α, and as ∠FJB 90°, ∠JBF = α. Therefore ∆OGF and ∆FJB are similar. Triangle ∆OGF: OG/FO = FJ/BF OG/1 = x/2 OG = x/2 a² + b² = c² (x/2)² + (2-x)² = 1² x²/4 + 4 - 4x + x² = 1 x² + 16 - 16x + 4x² = 4 5x² - 16x + 12 = 0 x = [-(-16)±√(-16)²-4(5)12]/2(5) x = (16±√256-240)/10 x = 8/5 ± √16/10 = 8/5 ± 2/5 x = 8/5+2/5 | x = 8/5-2/5 x = 10/5 = 2 ❌ | x = 6/5 ✓ As FJ = 6/5 and BF = 2 = 10/5, ∆FJB is a 2:5 ratio 3:4:5 Pythagorean Triple triangle and JB = 4(2/5) = 8/5. As they share ∠JBF and sides JB and BF, ∆FJB and ∆EAB are similar. Triangle ∆EAB: EA/AB = FJ/JB h/2 = (6/5)/(8/5) = 3/4 h = 2(3/4) = 3/2 A = bh/2 = 2(3/2)/2 = 3/2 = 1.5 cm²
Thanks Professor. Great problem!❤🥂👍
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@@PreMath How can we solve this Professor or anybody else? 2x + (16/(2^x)) = 10 , solve for x . the answer is 1 but process?
Connecting O to B, and O to F, forms 2 congruent triangles OBC and OBF.
Tan OBC = 1/2 = 0.5.
Angle OBC = tan -1 of 0.5 = 26.565 degrees.
Angle FBC = 2 x 26.565 = 53.13 degrees.
Angle ABE = 90 - 53.13 = 36.87 degrees.
Tan 36.87 = AE / 2.
AE = 2 x tan 36.87.
AE = 2 x 0.75 = 1.5.
The tangents to the circle were the key. When I work many problems of this type the repetition helps me remember the two tangent theorem. Thank You.
A slightly different aproach:
area of circle =π/2; area of circle = 4π = πr^2; so r = 1; so far, so good.
Now considering the right side of the square, draw OB. Since OC = 1 and CB = 2, invoke Pythagoras and
OB^2 = 2^2 + 2^2 = 5, so OB = √5.
Applying Pythagoras to triangle OBF, we have BF^2 = OB^2 - 1^2 = 5 - 1 = 4; so BF = 2.
Let x = DE = EF. They are equal from the two-tangent theorem, and/ or because they are the corresponding sides of the two congruent right triangles DEO and EFO.
Coming back to triangle EAB and calling on Pythagoras again:
2^2 + (2 - x)^2 = (2 + x)^2; expanding both sides:
4 + 4 - 4x + x^2 = (2 + x)^2 = 4 + 4x + x^2; collecting terms, we have
8 - 4x = 4 + 4x; 4 = 8x; so x = 1/2; and AE = 2 -- 1/2 = 3/2.
So the area of the green triangle is (1/2) (2) (3/2) = 3/2.
Cheers. 🤠
Gracias señor.
Desde Bogota D.C.
Semi circle :
A = ½πr² = ½π
r = 1 cm
Side of square:
s = 2r = 2 cm
Green right triangle:
base = 2
height = 2-x
hypotenuse = 2+x
2²+(2-x)²=(2+x)²
2²+(2²-4x+x²)=(2²+4x+x²)
8x = 4
x = 0,5 cm
Area Green triangle:
A = ½b.h
A = ½ . 2 . (2-0,5)
A = 1,5 cm² ( Solved √ )
Perfect job! Well done! I have a question: which application you use for that drawings?
And which for the video?
You inspire us, thank you!
Hello dear, we use Camtasia Techsmith utility along with Canva + photoshop. Thanks for asking.
You are awesome. Keep smiling👍
Love and prayers from the USA! 😀
Sorry for the late response.
Once we find that the radius is 1 (and the side of the square is 2), and that by the theorem of tangents DE=BF and BF=BC, we can proceed as in the following.
DEF + AEF = 180°; AEF=CBF (alternate interior angles of two parallel lines (AD ∥ BC) cut by the transversal BE); so DEF+CBF=180°. EO is the bisectrix of DEF, BO is the bisectrix of CBF so EOB is a right triangle with EOB=90°. So by Euclid's first theorem the height OF relative to the hypotenuse EB is the proportional mean between the projections EF and FB and we can write:
OF²=EF·FB --> 1²=EF·2 --> EF=½
EF=DE --> AE=2-½=3/2
Area=½·AB·AE=½·2·(3/2)=3/2
Great explanation👍
Thanks for sharing😊😊
Si (πr²/2)=π/2 → r²=1→r=1 → Lado del cuadrado ABCD, CD=2r=2=CB=BF → ∠OBF=∠FOE → Razón de semejanza =1/2 entre ∆EFO y ∆OFB , ambos triángulos rectángulos por ser OF⟂EB → EF=OF/2=1/2 → EF=DE=1/2 → AE=AD-ED → AE=2-(1/2)=3/2 → Área Verde =Área de ∆EAB =(1/2)2(3/2)=3/2 = 1.5 cm²
Gracias y un saludo cordial.
I prefer DE=EF=x !
(2 - x)" + 2" = (2 + x)"
2" = (2 + x)" - (2 - x)" = 4•2x
x = 1/2
AE = 2 - 1/2 = 3/2
Area = (1/2)•(3/2)•2 = 1.5 cm" 😉
Excellent!
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
Love and prayers from the USA! 😀
@@PreMath How can we solve this Professor or anybody else? 2x + (16/(2^x)) = 10 , solve for x . the answer is 1 but process?
Join the center of the semi circle to points B and E and F.
Triangles OFB and OCB congruent (rhs)
Triangles EDO and EFO are congruent.
By labelling angles of the above mentioned triangles we can deduce OFB and EDO are similar. Since OF=1 cm and FB = 2 cm, ED =1/2 cm, EA= 1.5 cm and area of ABE=1/2x2x1.5=1.5cmsq
simpler and faster
S=1,5 cm²
How can we solve this Professor or anybody else? 2x + (16/(2^x)) = 10 , solve for x . the answer is 1 but process?
So beautiful design of the figure 🤩. First we note that there are two pairs of similar right angled triangles, so OF=1, FE=1/2=DE, thus AE=3/2, so the answer is 3/2x2/2=3/2, done.😊
Excellent!
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
Love and prayers from the USA! 😀
How did you get FE= 1/2
@@ashieshsharmah1326 OEF is similar to OBF.
@@misterenter-iz7rz thanks 🤩 . One more question how did you conclude that angle BOE is 90° . Please reply
@@ashieshsharmah1326 2a+2b=180=COD, so a+b=90.🙂
Can we say that the quadrangle OCBF is similar to quadrangle DOFE (All four angles in both are equal) and so the relations between corresponded sides in these two quadrangles are equal so we can find DE part and then calculate the area of ABE triangle.
Thanks for video.Good luck sir!!!!!!!!!!!!
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Didn't see the tangent pairs. Nice solve.
The easiest way I find: Getting radis = 1 => esquare is 2x2(side times side), then in the "green triangle" we get big leg(mayor hick) = 2, and cos' it's similar to 3/4/5 triangle, then 4.k=2("k" is the ratio/proportion = 1/2), so for 3 the hick is 3×(1/2) = 3/2, so taking this hick as height, and 2 as a base, we get [(3/2) × 2]/2 = 3/2 , 1.5 cm² as the area. Done!
The way the professor do is good to manage/use other algebra/geometrics conceps, then is too valid also. It depende on the grade/complexity we want to teach/learn some.
My greetings to all.
(Sorry about my english). 😑/😅
👍😜🙃😋🤩🇦🇷⭐⭐⭐
Got me twice with the two tangent theorem. After the first one I was thinking of trying to work out the intersection by calculating the sphere intersect, but I figured you'd have something much easier, so I unpaused and you did, lol.
Easy one.
Circle: r²·π/2 = π/2 => r = 1
Square: s = 2·r = 2
QB = sqrt(1²+2²) = sqrt(5)
BF = sqrt((sqrt(5)² - 1²) = sqrt(4) = 2
Then, with x = DE = EF, we have
BE² = AB² + AE²
(BF + x)² = AB² + (AD - x)²,
BF = AD = 2,
(2 + x)² = 2² + (2 - x)², which reduces to x = 1/2
Area = (2 - 0.5)·2 / 2 = 3/2 cm²
The sides of the right kites OCBF and ODEF are in a 2:1 ratio.
So the area of the green triangle is 4 - 2 - 1/2 = 3/2 = 1,5 cm².
did the same😀
BCOF and DEFO are similar deltoids => DE/DO=CO/BC => DE=1/2
Thanks professor 😊 . Very good problem and excellent explanation .
There is a square area 1m2. You need to do a truncated cone that has max. Volume. Find a diameter on a top & on the bottom of this cone & side surface made from this square area
Draw a development of this parts on this 1m2 area.
3/2 sq. units
Is the rectangle a square?
Sorry, it is clarified a time later. Happens that i try to solve the problem without waching further.
@@zzzaphod8507 You are wrught. Sorry, i did not hear it clearly the first time
I love it!
Thanks dear
👍
3/2
Yay! I solved it.
r=1, side of square =2, side of triangle = 2, 2x, 2+x. further according to the Pythagorean theorem: (2+х) ²=2²+(2-х) ².
Second method: ∆QCB=∆QFB, QF=QC=r,
Area=1.5
I made myself laugh with this one I worked out x but failed to see the significance, oh boy a definite Homer Simpson moment...Doh!🤓
I keep looking for him to make a mistake so I double check all of his work.
Just mention:
FB/OF=OF/EF or in Numbers 2/1=1/0,5 or (x)
Sem saber o teorema do cara eu percebi que bf era igual a bc,😅
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Güzel soru
Your solution was too simple. Let's see if we can make this unnecessarily complex yet still mathematically sound...
Semicircle O:
A = πr²/2
π/2 = πr²/2
r² = (π/2)(2/π) = 1
r = 1
By Two Tangent Theorem, BF = BC = 2r = 2. Let G be the point on CD and J be the point on AB where GJ is perpendicular to AB and CD and passes through F. Let FJ = x and GF = 2-x.
Let ∠FOG = α and ∠GFO be β, where β = 90°- α. As ∠OFB = 90°, ∠BFJ = α, and as ∠FJB 90°, ∠JBF = α. Therefore ∆OGF and ∆FJB are similar.
Triangle ∆OGF:
OG/FO = FJ/BF
OG/1 = x/2
OG = x/2
a² + b² = c²
(x/2)² + (2-x)² = 1²
x²/4 + 4 - 4x + x² = 1
x² + 16 - 16x + 4x² = 4
5x² - 16x + 12 = 0
x = [-(-16)±√(-16)²-4(5)12]/2(5)
x = (16±√256-240)/10
x = 8/5 ± √16/10 = 8/5 ± 2/5
x = 8/5+2/5 | x = 8/5-2/5
x = 10/5 = 2 ❌ | x = 6/5 ✓
As FJ = 6/5 and BF = 2 = 10/5, ∆FJB is a 2:5 ratio 3:4:5 Pythagorean Triple triangle and JB = 4(2/5) = 8/5.
As they share ∠JBF and sides JB and BF, ∆FJB and ∆EAB are similar.
Triangle ∆EAB:
EA/AB = FJ/JB
h/2 = (6/5)/(8/5) = 3/4
h = 2(3/4) = 3/2
A = bh/2 = 2(3/2)/2 = 3/2 = 1.5 cm²
Since the triangles DEO and BOF are similar DE is 0.5.