I support Math Booster and I think the explanations are elegant. However, when you have limited time on a math exam, you need the quickest and shortest way to the answer. My motto is: keep it short and sweet.
As DPQS and ABCD are square, PQ = QS = SD = DP = 2 and BC = CD = DA = AB = 6. Let O be the center of the circle and let M and N be the points of tangency between circle O and AB and BC respectively. As AB and BC are tangent to circle O at M and N respectively, ∠OMB = ∠BNO = 90°. As ∠MBN = 90°, ∠NOM must equal 90° as well. As OM = ON = r, OMBN is a square and MB = BN = r as well. Draw DB. As D is common to ABCD and DPQS and S and P are collinear with CD and DA, and as B is common to OMBN and ABCD and M and N are collinear with AB and BC, then Q and O are collinear with DB, as DB is a diagonal of ABCD and DQ and OB are diagonals of DPQS and OMBN. Triangle ∆QSD: QS² + SD² = DQ² 2² + 2² = DQ² DQ² = 4 + 4 = 8 DQ = √8 = 2√2 Triangle ∆BNO: BN² + ON² = OB² r² + r² = OB² OB² = 2r² OB = √(2r²) = √2r Triangle ∆BCD: BC² + CD² = DB² 6² + 6² = DB² DB² = 36 + 36 = 72 DB = √72 = 6√2 DB = DQ + OQ + OB 6√2 = 2√2 + r + √2r 4√2 = (1+√2)r r = 4√2/(√2+1) r = 4√2(√2-1)/(√2+1)(√2-1) r = (8-4√2)/(2-1) = (8-4√2)/1 r = 8 - 4√2 Circle O: A = πr² A = π(8-4√2)² A = π(64-64√2+32) A = (96-64√2)π ≈ 17.25 sq units
Here's a simpler way. Just create a new isoceles triangle. The base would simply be twice line QB. Line QB has a length of 5.66. Construct base of the new triangle as 11.32 in length. This new base will be tangent at point Q and parallel to a line from A to C. Lines AB and BC become the new arms to the new triangle. Going Pythagorean, each arm has a length of 8. Area of the new triangle turns out to be 31.97. Then simply use the incircle formula. Radius = area/semiperimeter. 31.97/13.65 = 2.34 for the radius, which is pretty much the narrator's answer of 2.3369 rounded up. Area of circle is thus pi*r^2 or 17.20.
Let O be the centre of the circle. The most important step is to prove that DQOB form a straight line i.e. diagonal of square ABCD. The diagonal of a square is an angle bisector dividing the right angle into two 45 angles. Considering BD as angle bisector, points on it must be equidistant from its arms. Centre O is equidistant from AB and BC by radius of circle to tangents hence it is on BD. Point Q is also equidistant from AB and BC by 6 - 2 = 4 hence it is also on BD. With that condition proven, the rest is straight forward.
As RAG981 did, extend MO and PQ until they meet. Since N is used in the video, let the intersection be point T. ΔOQT is an isosceles right triangle. In the video, math booster computes the ratio hypotenuse to side = √2. Since the hypotenuse is r, side QT = r/√2. Extend PQ until it meets BC and call the intersection W. ONWT is a rectangle. ON = WT = r. ABWP also a rectangle, so WP = AB = 6. WP = PQ + QT + WT, 6 = 2 + r/√2 + r, 4 = r (1/√2 + 1). Reorder inside the parenthesis, multiply by √2 and we get 4√2 = r(√2 + 1), the same equation reached at 9:14 in the video. Proceed as in the video to find πr². Basically, RAG981 and I computed a length equal to the side length of the square. Math Booster computed the diagonal, which is √2 times as long as the side.
Draw the line down from O and another along PQ produced to meet it at N. Then QN is 4-r and OP is 4-r. So in OQN, r^2 = 2x(4-r)^2, from which r = 8-4rt2, and area is 32(3-2rt2)pi.
Answer 17.2484 My approach. Since the length of the square = 6, the distance of the LARGE SQUARE is not covered by the square in terms of r= 6- 2r Let's go to the small square with length 2: Extend line QS through the circle to line AB, thus cutting the circle. Let's label the line QP. Since this distance = 2 (from the end of the large square), then the distance from this line to the end of the circle is 2 - (6-2r) = -4 + 2r or 2r -4. Hence, the distance from the circle's center to QP= r -(2r-4)= -r +4 or 4-r Hence, point Q is 4-r from both the horizontal and vertical from the circle's center. Construct an isosceles right triangle with the sides 4-r, 4-r, and r (hypotenuse) and employ Pythagorean Theorem Hence, (4-r)^2 + (4-r)^2 = r^2 r^2 - 8r + 16 + r^2 - 8r + 16 -r^2 = 0 r^2 -16 r - 32 = 0 Use the quadratic formula r= 8-+ 4 sqrt 2 or 13.6959 or 2.34315 Since 13.6959 is greater than the length of the square, then it is not the Answer Hence, r = 2.34315 or 8 - sqrt 4 sqrt 2 Hence, the area of the circle = 2.34315^2 pi or 17.248386
The area is [96-4sqrt(2)]pi units square. Looks like I will have to use that as a square problem ao that I can know easily it would be to remember circle theorems and Pythagorean Theorem.
A questão é bonita e bem interessante. Parabéns pela escolha. Brasil - Outubro de 2024. The question is beautiful and very interesting. Congratulations on your choice. Brazil - October 2024.
I support Math Booster and I think the explanations are elegant. However, when you have limited time on a math exam, you need the quickest and shortest way to the answer. My motto is: keep it short and sweet.
As DPQS and ABCD are square, PQ = QS = SD = DP = 2 and BC = CD = DA = AB = 6.
Let O be the center of the circle and let M and N be the points of tangency between circle O and AB and BC respectively. As AB and BC are tangent to circle O at M and N respectively, ∠OMB = ∠BNO = 90°. As ∠MBN = 90°, ∠NOM must equal 90° as well. As OM = ON = r, OMBN is a square and MB = BN = r as well.
Draw DB. As D is common to ABCD and DPQS and S and P are collinear with CD and DA, and as B is common to OMBN and ABCD and M and N are collinear with AB and BC, then Q and O are collinear with DB, as DB is a diagonal of ABCD and DQ and OB are diagonals of DPQS and OMBN.
Triangle ∆QSD:
QS² + SD² = DQ²
2² + 2² = DQ²
DQ² = 4 + 4 = 8
DQ = √8 = 2√2
Triangle ∆BNO:
BN² + ON² = OB²
r² + r² = OB²
OB² = 2r²
OB = √(2r²) = √2r
Triangle ∆BCD:
BC² + CD² = DB²
6² + 6² = DB²
DB² = 36 + 36 = 72
DB = √72 = 6√2
DB = DQ + OQ + OB
6√2 = 2√2 + r + √2r
4√2 = (1+√2)r
r = 4√2/(√2+1)
r = 4√2(√2-1)/(√2+1)(√2-1)
r = (8-4√2)/(2-1) = (8-4√2)/1
r = 8 - 4√2
Circle O:
A = πr²
A = π(8-4√2)²
A = π(64-64√2+32)
A = (96-64√2)π ≈ 17.25 sq units
Great , same answer I got
Here's a simpler way. Just create a new isoceles triangle. The base would simply be twice line QB. Line QB has a length of 5.66. Construct base of the new triangle as 11.32 in length. This new base will be tangent at point Q and parallel to a line from A to C. Lines AB and BC become the new arms to the new triangle. Going Pythagorean, each arm has a length of 8. Area of the new triangle turns out to be 31.97. Then simply use the incircle formula. Radius = area/semiperimeter. 31.97/13.65 = 2.34 for the radius, which is pretty much the narrator's answer of 2.3369 rounded up. Area of circle is thus pi*r^2 or 17.20.
Let O be the centre of the circle. The most important step is to prove that DQOB form a straight line i.e. diagonal of square ABCD. The diagonal of a square is an angle bisector dividing the right angle into two 45 angles. Considering BD as angle bisector, points on it must be equidistant from its arms. Centre O is equidistant from AB and BC by radius of circle to tangents hence it is on BD. Point Q is also equidistant from AB and BC by 6 - 2 = 4 hence it is also on BD. With that condition proven, the rest is straight forward.
NIce, I did it the very same manner : )
As RAG981 did, extend MO and PQ until they meet. Since N is used in the video, let the intersection be point T. ΔOQT is an isosceles right triangle. In the video, math booster computes the ratio hypotenuse to side = √2. Since the hypotenuse is r, side QT = r/√2. Extend PQ until it meets BC and call the intersection W. ONWT is a rectangle. ON = WT = r. ABWP also a rectangle, so WP = AB = 6. WP = PQ + QT + WT, 6 = 2 + r/√2 + r, 4 = r (1/√2 + 1). Reorder inside the parenthesis, multiply by √2 and we get 4√2 = r(√2 + 1), the same equation reached at 9:14 in the video. Proceed as in the video to find πr².
Basically, RAG981 and I computed a length equal to the side length of the square. Math Booster computed the diagonal, which is √2 times as long as the side.
Draw the line down from O and another along PQ produced to meet it at N. Then QN is 4-r and OP is 4-r. So in OQN, r^2 = 2x(4-r)^2, from which r = 8-4rt2, and area is 32(3-2rt2)pi.
DB =6√2 =2√2+r+r√2 → r=(8-4√2) → Área del círculo =πr² =π(96-64√2) =32π(3-2√2 =17,24838...ud².
Gracias y un saludo.
Answer 17.2484
My approach.
Since the length of the square = 6, the distance of the LARGE SQUARE is not
covered by the square in terms of r= 6- 2r
Let's go to the small square with length 2:
Extend line QS through the circle to line AB, thus cutting the circle. Let's label
the line QP. Since
this distance = 2 (from the end of the large square), then
the distance from this line to the end of the circle is 2 - (6-2r) = -4 + 2r
or 2r -4.
Hence, the distance from the circle's center to QP= r -(2r-4)= -r +4
or 4-r
Hence, point Q is 4-r from both the horizontal and vertical from the circle's center.
Construct an isosceles right triangle with the sides 4-r, 4-r, and r (hypotenuse) and employ
Pythagorean Theorem
Hence, (4-r)^2 + (4-r)^2 = r^2
r^2 - 8r + 16 + r^2 - 8r + 16 -r^2 = 0
r^2 -16 r - 32 = 0
Use the quadratic formula
r= 8-+ 4 sqrt 2 or 13.6959 or 2.34315
Since 13.6959 is greater than the length of the square, then it is not the Answer
Hence, r = 2.34315 or 8 - sqrt 4 sqrt 2
Hence, the area of the circle = 2.34315^2 pi
or 17.248386
Answer 17.2483
The area is [96-4sqrt(2)]pi units square. Looks like I will have to use that as a square problem ao that I can know easily it would be to remember circle theorems and Pythagorean Theorem.
(2)^2 (6)^2={4+36}=40 2A(45°)=90°A 2B(45°)=90°B 2C(45)=90°C 2D(45°)=90°D {90°A+90°B+90°C+90°D}=360°ABCD/40=9ABCD 3^2(ABCD ➖ 3ABCD+3).
A questão é bonita e bem interessante. Parabéns pela escolha. Brasil - Outubro de 2024. The question is beautiful and very interesting. Congratulations on your choice. Brazil - October 2024.
R + R/√2 + 2 = 6
R (1+1/√2) = 6 - 2 = 4
R = 2,34315 cm
A = πR² = 17,2484 cm² ( Solved √ )
R√2+R+2√2=6√2
So R=8-4√2
Circle area=π(8-4√2)^2=32π(3-2√2).❤
It was quite fun, I htought.
(Thought)
R+R√2=6√2-2√2
12
Nice baby