Thanks again. Drawing another square UBTO, where U is the point where the circle touches AB , T is the point where circle touches BC , and O is the centre of the circle. Extending UO to meet DC at V , UV=6 OV = 6- r where r is the radius of the circle. The area of the circle is pi times r squared. OB = 2^½ × r DQ= 2×2^½ QO= r These add to equal BD which is 6 × 2^½ BD = BO + OQ + QD = 2^½ × r + r + 2. 2^½ Multiplying all these lengths by 2^½ 12 = 2r +( 2^½ )r +4 r (2 +2^½) = 12 - 4 Multiplying this by (2 - 2^½) gives this: r × (2+2^½)(2-2^½) = 8(2-2^½) and gives r = 4(2 - 2^½) squaring this gives 16(4+2- 4×2^½) = 32 (3-2.2^½) So the area of the circle is pi × 32 (3 -2^½) which is approximately 17.25 square units This is a bit different from the video method. I used the video to correct a small error whch I had made (in the length of DQ) I guessed that my first answer was too large , the circle's area is about half of 6×6 as we can observe.
Draw DB. As DPQS and ABCD are both squares, and thus ∠PDQ = ∠ADB = 45°, then DQ is collinear with DB. As DQ is the diagonal of square DPQS, with side length 2, then DQ = 2√2. As DB is the diagonal of square ABCD, with side length 6, then DB = 6√2. QB = DB-DQ = 6√2-2√2 = 4√2. Let the center of the circle be O. As circle O is inscribed into the top right corner of ABCD, then by symmetry, diagonal DB, and thus QB, passes through O. OQ = r. Let M and N be the points of tangency between circle O and AB and BC respectively. As AB is tangent to circle O at M and BC is tangent to circle O at N, then ∠OMB = ∠BNO = 90°. As ∠MBN = 90°, then ∠NOM = 360°-3(90°) = 90°. As all internal angles are 90°, and adjacent sides OM and ON both equal r, then OMBN is a square of side length r. As OB is the diagonal of that square, OB = √2r. QB = OQ + OB 4√2 = r + √2r 4√2 = r(√2+1) r = 4√2/(√2+1) r = 4√2(√2-1)/(√2+1)(√2-1) r = (8-4√2)/(2-1) = 8 - 4√2 Circle O: A = πr² = π(8-4√2)² A = π(64-64√2+32) A = π(96-64√2) A = 32π(3-2√2) ≈ 17.248 sq units
I just set up the circle into a new right triangle. Line QB turns out to be 5.665; this is the height of a new triangle. New base of the triangle would be parallel to line AC and would be twice the height mentioned of the new triangle or 11.30. Lines AB and BC would thus be 8 in length. Total area of new triangle would be 32. Semi perimeter would be 13.65. Then just use the formula for inradius. Inradius = area/semi-perimeter. Inradius is 2.34. Area of circle is thus: 17.20. Very close to narrator's solution.
R + R/√2 = 6-2 R (1+1/√2)= 4 --> R= 2,34315cm A = πR² = 17,25 cm² ( Solved √ ) More than 12 minutes of video, and there is not a final value. Once again. Next time when you go to the market to buy some fabric or similar, say "I need 32π(3-2√2) square units" of fabric, instead of 17,25 m², and then look at the face of your salesman, and guess what he's thinking about you.
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By applying the Pythagorean theorem, we find (4-r)²+(4-r)²=r², and from it r=8-4√2, so the area of the circle is equal to π*(8-4√2)².
Thanks again.
Drawing another square UBTO, where U is the point where the circle touches AB , T is the point where circle touches BC , and O is the centre of the circle.
Extending UO to meet DC at V , UV=6 OV = 6- r where r is the radius of the circle. The area of the circle is pi times r squared.
OB = 2^½ × r DQ= 2×2^½ QO= r
These add to equal BD which is 6 × 2^½ BD = BO + OQ + QD = 2^½ × r + r + 2. 2^½
Multiplying all these lengths by 2^½ 12 = 2r +( 2^½ )r +4
r (2 +2^½) = 12 - 4 Multiplying this by (2 - 2^½) gives this: r × (2+2^½)(2-2^½) = 8(2-2^½) and gives r = 4(2 - 2^½) squaring this gives 16(4+2- 4×2^½) = 32 (3-2.2^½)
So the area of the circle is pi × 32 (3 -2^½) which is approximately 17.25 square units
This is a bit different from the video method. I used the video to correct a small error whch I had made (in the length of DQ)
I guessed that my first answer was too large , the circle's area is about half of 6×6 as we can observe.
If I'm not mistaken, it was a question identical to this one or it's a previous video on the channel, reposted.
bro, who are you, teacher or....
@@senirudilmith-n8z Do you need to be a teacher to know if a video is repeated? Your comment is very idiotic.
@@senirudilmith-n8z Your comment is idiotic. I just said that this question has already been asked or some similar question.
Draw DB. As DPQS and ABCD are both squares, and thus ∠PDQ = ∠ADB = 45°, then DQ is collinear with DB. As DQ is the diagonal of square DPQS, with side length 2, then DQ = 2√2. As DB is the diagonal of square ABCD, with side length 6, then DB = 6√2. QB = DB-DQ = 6√2-2√2 = 4√2.
Let the center of the circle be O. As circle O is inscribed into the top right corner of ABCD, then by symmetry, diagonal DB, and thus QB, passes through O. OQ = r.
Let M and N be the points of tangency between circle O and AB and BC respectively. As AB is tangent to circle O at M and BC is tangent to circle O at N, then ∠OMB = ∠BNO = 90°. As ∠MBN = 90°, then ∠NOM = 360°-3(90°) = 90°. As all internal angles are 90°, and adjacent sides OM and ON both equal r, then OMBN is a square of side length r. As OB is the diagonal of that square, OB = √2r.
QB = OQ + OB
4√2 = r + √2r
4√2 = r(√2+1)
r = 4√2/(√2+1)
r = 4√2(√2-1)/(√2+1)(√2-1)
r = (8-4√2)/(2-1) = 8 - 4√2
Circle O:
A = πr² = π(8-4√2)²
A = π(64-64√2+32)
A = π(96-64√2)
A = 32π(3-2√2) ≈ 17.248 sq units
I just set up the circle into a new right triangle. Line QB turns out to be 5.665; this is the height of a new triangle. New base of the triangle would be parallel to line AC and would be twice the height mentioned of the new triangle or 11.30. Lines AB and BC would thus be 8 in length. Total area of new triangle would be 32. Semi perimeter would be 13.65. Then just use the formula for inradius. Inradius = area/semi-perimeter. Inradius is 2.34. Area of circle is thus: 17.20. Very close to narrator's solution.
Naa apresentação como Babá A Cristina estava com Juva de Glória do Goitá.
Excellent, I solved it in the very same manner : )
We Think Alike !
Area=32π(3-2√2).❤
2+R.cos 45 +R = 6 .... R = 2.343 ... 🙂
Sir,
PQ=QS=2,
Pythagoras,
2^2+2^2=2root2,
AB=AD=6,
Pythagoras,
6^2+6^2=6root2,
DB,
DQ+QB=6root2
QB=4root2,
In circle,
radius =r,
r^2+r^2=Rroot2,
r+Rroot2=4root2,
R=8-4root2
Circle Area=Fi.r^2
=(96-64root2).Fi,
=32.Fi.(3-2root2)/=
=
Sócrates responde.
6√2 = 2√2 + r + r√2
4√2 = r + r√2
r = (4√2)/(1 + √2)
area = 32 Π/(3 + 2√2) = 32 Π (3 - 2√2) = (96 - 64√2)Π
Solving: r+r*sqrt2=6sqrt2-2sqrt2
Diameter=6sqrt2-2sqrt2=4. sqrt2. That’s all.
You might choose to have another look at QB , it is not a diameter.
(2)^2 (6)^2={4+36}=40 {90°A+90°B+90°C+90°D}=360°ABCD/40PQS=90ABCDPQS 3^30 3^10 3^5^5 3^2^3^2^3 1^1^1^2^3 2^3 (ABCDPQS ➖ 3ABCDPQS+2).
2√2+r+r√2=6√2; r=8-4√2;
S=π(8-4√2)^2
R + R/√2 = 6-2
R (1+1/√2)= 4 --> R= 2,34315cm
A = πR² = 17,25 cm² ( Solved √ )
More than 12 minutes of video, and there is not a final value. Once again.
Next time when you go to the market to buy some fabric or similar,
say "I need 32π(3-2√2) square units" of fabric, instead of 17,25 m², and then look at the face of your salesman, and guess what he's thinking about you.
I'd be pretty disappointed if this really was a math olympiad question.
It's not really at that level, is it?
No, it is about my level. It is nice to be within capacity about 70% of the time. Better for blood preeesssure readings!!!
@@kateknowles8055 🤣
@@Grizzly01-vr4pn ☑
2R^2= (4√2-R)^2
6×1.4142135624-2×1.4142135624=5.6568542496÷2.4142135624=2.343145×2.343145×3.14159268=17.097768