ABCD=576→ AB=√576=24=4*6→ BE=2*216/24=18=3*6→ AE=5*6=30 → AECD=576-216=360 =(30r/2)+(6r/2)+(24r/2)+24[(24-r)/2] → Radio =r=4. Gracias y un saludo cordial.
A very easy solution - White region area(it's a trapezoid) = area of the square - Triangle ABC= 360. Draw some lines from the center of the circle( let o be the center) to the 4 corners of the trapezoid. Then there will be 4 Triangles. Then the sum of the area of the Triangles are= 1/2×r(6+24+30)+(24-r)×24×1/2= 360 ⇨30r+288-12r=360 r=4
Thank you for the problems/exercises you give, along with explanations on how to resolve. I got stuck trying to do this one. I took a look at the answers before watching your video. I saw what others did and it made a lot of sense. I then watched the end of your video to confirm the answer.
My approach was from Monty Python - a little bit different. The bottom side is divided into two segments, 18 plus 6. And the 6 segment can be further divided into r plus (6-r). Getting the angles of the triangle was easy enough; so the supplement of the bottom angle was 180-arctan(24/18), or about 126.87 degrees. Divide that by 2, or about 63.44 degrees. Set up equation tangent(63.44) = r/(6-r). r=4.
@@egillandersson1780 Another possible solution: 1/Yellow triangle is a 3-4-5 triples. 2/Extend AE and DC down intersecting at point I. --> the triangle ECI is also a 3-4-5 triples --> EC= 6->CI=8 and EI = 10. 3/ Build the inscribed circle (center O’) of the triangle ECI. Note that the circle ( center O) is the ex circle of triangle ECI. We have points I, O’ and O are collinear. And label r and R as radiuses the ins and ex circle We have r = 2 (easy) From O’ drop the perpendicular O’H to CI. By similarity r/R=6/(8+R) -> 2/R=6/(8+R) -> 1/R=3/(8+R) ->R=(8+R)/3 R= 4cm😅😅😅
@ 5:03 one truly enters the realm of mathematics and allows for observations to solve the problem. Throw a slice of bread in a toaster and whadd'ya get? ...ya get toast! 😊
Draw a vertical line up from E. The rectangle formed has an area of 432. As the height is 24, the horizontal is 432/24 = 18. Therefore, EC = 6. The diagonal, AE is the sqrt of 24^2 + 18^2 so 576 + 324 = 900, whose sqrt = 30. Now I will turn ADCE into a series of triangles: AOE has an area of 15r. EOC area of 3r. COD area of 12r. AOD is an odd one as its base is 24 and its height is 24-r. This has an area of 12*(24-r) = 288-12r. Add all those up for 30r + (288-12r) = 18r + 288 18r + 288 = 576 - 216. 18r + 288 = 360. 18r = 72. r = 4. This one took me a while as I went down a couple of dead ends trying to figure out a viable method.EDIT: Looking further, I could have simplified it to 15r + 3r = (6*24)/2
The side length of the square is sqrt(576) = 24, the area of the yellow triangle is (1/2).BA.BE = (1/2).24.BE = 216, so BE = 216/12 = 18 We now use an orthonormal center B and first axis (BC). We have A(0; 24), E(18; 0), VectorAC(18; -24) is colinear to VectorU(-3; 4) The equation of (AE) is (x - 18).(4) - (y).(-3) = 0 or 4.x + 3.y - 72 = 0. If R is the radius of the circle, we have O(24 - R; R) Distance from O to (AC): abs(4.(24 - R) + 3.(R) - 72)/sqrt(4^2 + 3^2) = (24 - R)/5, this distance is also equal to R, so we have (24 - R)/5 = R, and so R = 4.
Square side length = 24. BE = 18. EC = 6. Extending lines AE & DC downwards to intersect at point H. Then triangle ECH is similar to triangle ABE. So 24 /18 = CH / 6. CH = 8. Tan BAE = 18 / 24 = 0.75. Angle BAE = 36.870 degrees. Joining centre point O to H. This will bisect angle AHD. Thus angle OHG = 18.435 degrees. Joining OG which = radius r. In triangle OHG, tan 18.435 = r / (r + 8). 0.33333 = r / (r + 8). 1 / 3 = r / (r + 8) 3r = r + 8. 2r = 8 r = 4.
Square ABCD: A = s² 576 = s² s = √576 = 24 Triangle ∆ABE: A = bh/2 216 = BE(24)/2 = 12BE BE = 216/12 = 18 AB² + BE² = AE² 24² + 18² = AE² AE² = 576 + 324 = 900 AE = √900 = 30 As BC = 24 and BE = 18, EC = 6. As FC = r, EF = 6-r. As FE and PE are tangents to circle O that intersect at E, PE = FE = 6-r. AP = AE-PE = 30-(6-r) = 24+r. Draw AC. As AC is the diagonal of square ABCD, AC = 24√2. As OFCG is a square with side length r and shares C as a corresponding vertex, then O is collinear with AC. As OC is the diagonal of OFCG then OC = √2r. AO = AC-OC = 24√2-√2r = √2(24-r) Triangle ∆APO: AP² + OP² = AO² (24+r)² + r² = (√2(24-r))² 576 + 48r + r² + r² = 2(576-48r+r²) 576 + 48r + 2r² = 1152 - 96r + 2r² 144r = 576 [ r = 576/144 = 4 cm ]
Let's find the radius: . .. ... .... ..... First of all we calculate the side length s of the blue square: A(ABCD) = s² 576cm² = s² ⇒ s = √(576cm²) = 24cm Now we can calculate the side lengths AB and BE of the right triangle ABE: AB = s = 24cm A(ABE) = (1/2)*AB*BE ⇒ BE = 2*A(ABE)/AB = 2*216cm²/(24cm) = 18cm Now let's assume that B is the center of the coordinate system and that BC is located on the x-axis. With r being the radius of the circle we obtain the following coordinates: A: ( 0cm ; 24cm ) B: ( 0cm ; 0cm ) E: ( 18cm ; 0cm ) O: ( 24cm−r ; r ) P: ( xP ; yP ) The line AE can be represented by the following function: y = (yA − yE)*(x − xE)/(xA − xE) + yE = (24 − 0)*(x − 18cm)/(0 − 18) + 0cm = 4(x − 18cm)/(−3) = 24cm − 4x/3 Since AE is a tangent to the circle, AE is perpendicular to PO. Therefore the product of the slopes of AE and PO is −1 and we can conclude: (yO − yP)/(xO − xP) = −1/(−4/3) = 3/4 OP² = (xO − xP)² + (yO − yP)² r² = (xO − xP)² + (3/4)²(xO − xP)² r² = (xO − xP)² + (9/16)(xO − xP)² r² = (16/16)(xO − xP)² + (9/16)(xO − xP)² r² = (25/16)(xO − xP)² ⇒ r = (5/4)(xO − xP) = (5/3)(yO − yP) (5/4)(xO − xP) = (5/3)(yO − yP) 3(xO − xP) = 4(yO − yP) 3[(24cm − r) − xP] = 4[r − (24cm − 4xP/3)] 3(24cm − r − xP) = 4(r − 24cm + 4xP/3) 72cm − 3r − 3xP = 4r − 96cm + 16xP/3 168cm − 7r = 3xP + 16xP/3 = 9xP/3 + 16xP/3 = 25xP/3 ⇒ xP = 3(168cm − 7r)/25 r = (5/4)(xO − xP) r = (5/4)[(24cm − r) − 3(168cm − 7r)/25] r = 30cm − 5r/4 − 3(168cm − 7r)/20 r = 30cm − 5r/4 − 504cm/20 + 21r/20 20r = 600cm − 25r − 504cm + 21r 24r = 96cm ⇒ r = 4cm Best regards from Germany
STEP-BY-STEP RESOLUTION PROPOSAL : 01) Blue Square Side (BSS) = sqrt(576) ; BSS = 24 02) BE = (216 * 2) / 24 ; BE = 432 / 24 ; BE = 18 lin un 03) EC = (24 - 18) ; EC = 6 lin un 04) R = Radius 05) EF = PE = (6 - R) lin un 06) AE^2 = 324 + 576 ; AE^2 = 900 ; AE = 30 07) OP = R 08) AP = 30 - (6 - R) ; AP = (24 + R) lin un 09) AC = 24*sqrt(2) lin un 10) AO = AC - OC 11) OC = R*sqrt(2) 12) AO = (24*sqrt(2) - R*sqrt(2)) 13) AP^2 + OP^2 = AO^2 14) (24 + R)^2 + R^2 = (24*sqrt(2) - R*sqrt(2))^2 15) Solution : R = 4 Therefore, Beyond any Resonable Doubt, Radius equal to 4 Linear Units.
Solution: Blue Square Area = 576 576 = s² s = 24 Yellow Triangle Area = ½ b h 216 = ½ b h 216 = ½ 24 h h = 216/12 h = 18 Applying Pythagorean Theorem in Triangle ABE (18)² + (24)² = AE² AE² = 900 AE = 30 White Area = Square Area - Triangle Área White Area = 576 - 216 White Area = 360 White Area = ∆AEO Area + ∆CEO Area + ∆CDO Area + ∆ADO Area ... ¹ ∆AEO Area = ½ b h ∆AEO Area = ½ 30 r ∆AEO Area = 15r ∆CEO Area = ½ b h ∆CEO Area = ½ 6 r ∆CEO Area = 3r ∆CDO Area = ½ b h ∆CDO Area = ½ 24 r ∆CDO Area = 12r ∆ADO Area = ½ b h ∆ADO Area = ½ 24 (24 - r) ∆ADO Area = 288 - 12r Replacing in Equation ¹ 360 = 15r + 3r + 12r + 288 - 12r 360 - 288 = 18r 18r = 72 r = 4 cm Thus Radius = 4 cm ✅
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Square side length = √576 = 24
Triangle base = 432/24 = 18
24 : 18 = 4 : 3 => hypotenuse = 30
Trapezoid area = 576 - 216 = 360 = 30r/2 + 6r/2 + 24r/2 + 24(24 - r)/2 = (15 + 3 + 12 - 12)r + (24^2)/2
360 - 288 = 72 = 18r
r = 4
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@ Thank you. May the New Year bring you and your love ones happiness and success.
Triangle AEC area=72. Triangle AEC area =6r/2+30r/2. r=4. Thanks for the fun puzzle!
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ABCD=576→ AB=√576=24=4*6→ BE=2*216/24=18=3*6→ AE=5*6=30 → AECD=576-216=360 =(30r/2)+(6r/2)+(24r/2)+24[(24-r)/2] → Radio =r=4.
Gracias y un saludo cordial.
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A very easy solution -
White region area(it's a trapezoid) = area of the square - Triangle ABC= 360.
Draw some lines from the center of the circle( let o be the center) to the 4 corners of the trapezoid. Then there will be 4 Triangles. Then the sum of the area of the Triangles are= 1/2×r(6+24+30)+(24-r)×24×1/2= 360
⇨30r+288-12r=360
r=4
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Cool question!! 👍
Glad you liked it! 😊
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Thank you for the problems/exercises you give, along with explanations on how to resolve. I got stuck trying to do this one. I took a look at the answers before watching your video. I saw what others did and it made a lot of sense. I then watched the end of your video to confirm the answer.
My approach was from Monty Python - a little bit different.
The bottom side is divided into two segments, 18 plus 6. And the 6 segment can be further divided into r plus (6-r). Getting the angles of the triangle was easy enough; so the supplement of the bottom angle was 180-arctan(24/18), or about 126.87 degrees. Divide that by 2, or about 63.44 degrees. Set up equation tangent(63.44) = r/(6-r). r=4.
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Extend AE and OF down until they meet and label the intersection as point H. AB and OH are parallel.
I did exactly the same. Easier and quickier !
@@egillandersson1780
Another possible solution:
1/Yellow triangle is a 3-4-5 triples.
2/Extend AE and DC down intersecting at point I.
--> the triangle ECI is also a 3-4-5 triples
--> EC= 6->CI=8 and EI = 10.
3/ Build the inscribed circle (center O’) of the triangle ECI.
Note that the circle ( center O) is the ex circle of triangle ECI.
We have points I, O’ and O are collinear.
And label r and R as radiuses the ins and ex circle
We have r = 2 (easy)
From O’ drop the perpendicular O’H to CI.
By similarity
r/R=6/(8+R)
-> 2/R=6/(8+R)
-> 1/R=3/(8+R)
->R=(8+R)/3
R= 4cm😅😅😅
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Thank you
You are very welcome! ❤
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Hope you had a great Christmas Premath and everyone here
@johnryder1713 you too!
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@@PreMath And Please God a good one enjoying your fine content
3×3×3.14159268=28.27433
r=4 cm
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@@PreMath Thanks a lot 🥰 Wish you the same ❤️🎉 Happy New Year 🎊🎊🎊
@ 5:03 one truly enters the realm of mathematics and allows for observations to solve the problem. Throw a slice of bread in a toaster and whadd'ya get? ...ya get toast! 😊
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Mental calculation :
576=24^2 (=AB=BC)
216=24*9=24*18/2 (=AB*BE/2)
tan(a)=AB/BE=24/18=4/3
tan(Pi-a)=tan(-a)=-tan(a)=-4/3
OF/EF=tan((Pi-a)/2)
tan(Pi-a)=2*x/(1-x^2) with x=tan((Pi-a)/2)
-4/3=2*x/(1-x^2)
-4*(1-x^2)=6*x
4*x^2-6*x-4=0
x^2-3/2*x-1=0
(x-2)*(x+1/2)=0 but x>0 then x=2
OF/EF=tan((Pi-a)/2)=x=2
r/(6-r)=2
r=12-2*r
3*r=12
r=4
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Draw a vertical line up from E.
The rectangle formed has an area of 432.
As the height is 24, the horizontal is 432/24 = 18.
Therefore, EC = 6.
The diagonal, AE is the sqrt of 24^2 + 18^2 so 576 + 324 = 900, whose sqrt = 30.
Now I will turn ADCE into a series of triangles:
AOE has an area of 15r.
EOC area of 3r.
COD area of 12r.
AOD is an odd one as its base is 24 and its height is 24-r. This has an area of 12*(24-r) = 288-12r.
Add all those up for 30r + (288-12r) = 18r + 288
18r + 288 = 576 - 216.
18r + 288 = 360.
18r = 72.
r = 4.
This one took me a while as I went down a couple of dead ends trying to figure out a viable method.EDIT: Looking further, I could have simplified it to
15r + 3r = (6*24)/2
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The side length of the square is sqrt(576) = 24, the area of the yellow triangle is (1/2).BA.BE = (1/2).24.BE = 216, so BE = 216/12 = 18
We now use an orthonormal center B and first axis (BC). We have A(0; 24), E(18; 0), VectorAC(18; -24) is colinear to VectorU(-3; 4)
The equation of (AE) is (x - 18).(4) - (y).(-3) = 0 or 4.x + 3.y - 72 = 0. If R is the radius of the circle, we have O(24 - R; R)
Distance from O to (AC): abs(4.(24 - R) + 3.(R) - 72)/sqrt(4^2 + 3^2) = (24 - R)/5, this distance is also equal to R, so we have (24 - R)/5 = R, and so R = 4.
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Square side length = 24.
BE = 18.
EC = 6.
Extending lines AE & DC downwards to intersect at point H.
Then triangle ECH is similar to triangle ABE.
So 24 /18 = CH / 6.
CH = 8.
Tan BAE = 18 / 24 = 0.75.
Angle BAE = 36.870 degrees.
Joining centre point O to H.
This will bisect angle AHD.
Thus angle OHG = 18.435 degrees.
Joining OG which = radius r.
In triangle OHG, tan 18.435 = r / (r + 8).
0.33333 = r / (r + 8).
1 / 3 = r / (r + 8)
3r = r + 8.
2r = 8
r = 4.
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Square ABCD:
A = s²
576 = s²
s = √576 = 24
Triangle ∆ABE:
A = bh/2
216 = BE(24)/2 = 12BE
BE = 216/12 = 18
AB² + BE² = AE²
24² + 18² = AE²
AE² = 576 + 324 = 900
AE = √900 = 30
As BC = 24 and BE = 18, EC = 6. As FC = r, EF = 6-r. As FE and PE are tangents to circle O that intersect at E, PE = FE = 6-r. AP = AE-PE = 30-(6-r) = 24+r.
Draw AC. As AC is the diagonal of square ABCD, AC = 24√2. As OFCG is a square with side length r and shares C as a corresponding vertex, then O is collinear with AC. As OC is the diagonal of OFCG then OC = √2r. AO = AC-OC = 24√2-√2r = √2(24-r)
Triangle ∆APO:
AP² + OP² = AO²
(24+r)² + r² = (√2(24-r))²
576 + 48r + r² + r² = 2(576-48r+r²)
576 + 48r + 2r² = 1152 - 96r + 2r²
144r = 576
[ r = 576/144 = 4 cm ]
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1/The yellow triangle ABE is a 3-4-5 triples
AB= 24-> BE= 18-> AO= 30
2/ Label angle EAC= alpha and BAE = beta
We have tan (beta) = 3/4
So tan( alpha)=tan( 45 degrees-beta) = (1-3/4)/(1+3/4)=(1/4)/(7/4)=1/7
3/ AP= 24+r ->
r/(24+r)= 1/7
-> 7r=24+r
-> r=4cm😅😅😅
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l=24...AP+PE=AE...√((24-r)^2+(24-r)^2-r^2)+(6-r)=30...calcoli ..r=4
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Let's find the radius:
.
..
...
....
.....
First of all we calculate the side length s of the blue square:
A(ABCD) = s²
576cm² = s²
⇒ s = √(576cm²) = 24cm
Now we can calculate the side lengths AB and BE of the right triangle ABE:
AB = s = 24cm
A(ABE) = (1/2)*AB*BE ⇒ BE = 2*A(ABE)/AB = 2*216cm²/(24cm) = 18cm
Now let's assume that B is the center of the coordinate system and that BC is located on the x-axis. With r being the radius of the circle we obtain the following coordinates:
A: ( 0cm ; 24cm )
B: ( 0cm ; 0cm )
E: ( 18cm ; 0cm )
O: ( 24cm−r ; r )
P: ( xP ; yP )
The line AE can be represented by the following function:
y = (yA − yE)*(x − xE)/(xA − xE) + yE = (24 − 0)*(x − 18cm)/(0 − 18) + 0cm = 4(x − 18cm)/(−3) = 24cm − 4x/3
Since AE is a tangent to the circle, AE is perpendicular to PO. Therefore the product of the slopes of AE and PO is −1 and we can conclude:
(yO − yP)/(xO − xP) = −1/(−4/3) = 3/4
OP² = (xO − xP)² + (yO − yP)²
r² = (xO − xP)² + (3/4)²(xO − xP)²
r² = (xO − xP)² + (9/16)(xO − xP)²
r² = (16/16)(xO − xP)² + (9/16)(xO − xP)²
r² = (25/16)(xO − xP)²
⇒ r = (5/4)(xO − xP) = (5/3)(yO − yP)
(5/4)(xO − xP) = (5/3)(yO − yP)
3(xO − xP) = 4(yO − yP)
3[(24cm − r) − xP] = 4[r − (24cm − 4xP/3)]
3(24cm − r − xP) = 4(r − 24cm + 4xP/3)
72cm − 3r − 3xP = 4r − 96cm + 16xP/3
168cm − 7r = 3xP + 16xP/3 = 9xP/3 + 16xP/3 = 25xP/3
⇒ xP = 3(168cm − 7r)/25
r = (5/4)(xO − xP)
r = (5/4)[(24cm − r) − 3(168cm − 7r)/25]
r = 30cm − 5r/4 − 3(168cm − 7r)/20
r = 30cm − 5r/4 − 504cm/20 + 21r/20
20r = 600cm − 25r − 504cm + 21r
24r = 96cm
⇒ r = 4cm
Best regards from Germany
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STEP-BY-STEP RESOLUTION PROPOSAL :
01) Blue Square Side (BSS) = sqrt(576) ; BSS = 24
02) BE = (216 * 2) / 24 ; BE = 432 / 24 ; BE = 18 lin un
03) EC = (24 - 18) ; EC = 6 lin un
04) R = Radius
05) EF = PE = (6 - R) lin un
06) AE^2 = 324 + 576 ; AE^2 = 900 ; AE = 30
07) OP = R
08) AP = 30 - (6 - R) ; AP = (24 + R) lin un
09) AC = 24*sqrt(2) lin un
10) AO = AC - OC
11) OC = R*sqrt(2)
12) AO = (24*sqrt(2) - R*sqrt(2))
13) AP^2 + OP^2 = AO^2
14) (24 + R)^2 + R^2 = (24*sqrt(2) - R*sqrt(2))^2
15) Solution : R = 4
Therefore,
Beyond any Resonable Doubt, Radius equal to 4 Linear Units.
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Radius=4 units.❤
r=4
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ANSWER : Radius equal to 4 Linear Units.
Later I will come with my Reasoning!! Bye.
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Solution:
Blue Square Area = 576
576 = s²
s = 24
Yellow Triangle Area = ½ b h
216 = ½ b h
216 = ½ 24 h
h = 216/12
h = 18
Applying Pythagorean Theorem in Triangle ABE
(18)² + (24)² = AE²
AE² = 900
AE = 30
White Area = Square Area - Triangle Área
White Area = 576 - 216
White Area = 360
White Area = ∆AEO Area + ∆CEO Area + ∆CDO Area + ∆ADO Area ... ¹
∆AEO Area = ½ b h
∆AEO Area = ½ 30 r
∆AEO Area = 15r
∆CEO Area = ½ b h
∆CEO Area = ½ 6 r
∆CEO Area = 3r
∆CDO Area = ½ b h
∆CDO Area = ½ 24 r
∆CDO Area = 12r
∆ADO Area = ½ b h
∆ADO Area = ½ 24 (24 - r)
∆ADO Area = 288 - 12r
Replacing in Equation ¹
360 = 15r + 3r + 12r + 288 - 12r
360 - 288 = 18r
18r = 72
r = 4 cm
Thus Radius = 4 cm ✅
Good job
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