Let (I) be the intersecting point of the diameter with the circumference on the right , and let it be ( H) on the left. Suppose: HD = x, then CI=x. DC=6. Let CG(side of square CEFG) = y and GI = z. According to the intersecting chords formula: CB²=CI(DC+DH), So 6² =x(6+x), then x = 3.7 => y (CG)+z(GI) = 3.7, so z=3.7-y. Again according to the intersecting chords formula: FG² = GI(GC+DC+DH), Hence y² = z(y+6+3.7) by substitution=> (3.7-y)(y+6+3.7) . Solving for y , y= 3 so area of square CEFG = 3² = 9 units²
Very detailed and meticulous, but you don't need to calculate the radius. AO^2=AD^2+OD^2 =OF^2=OG^2+GF^2 => 3^2+6^2=x^2+(3+x)^2 => 3^2-x^2 = (3+x)^2 - 6^2 which gives x=3 as only +ve root using a^2-b^2 formula.
Нужно просто треугольник AOP повернуть на угол 90° относительно точки O по часовой стрелке. OA=OF, угол AOF=90°, угол POG=90° отсюда следует, что прямоугольные треугольники равны. И длина катета FG=3. При этом считать длину радиуса окружности вообще не нужно.
Let O be the center of the semicircle. Let s be the side length of CEFG, S be the side length of ABCD, and r be the radius of semicircle O. Square ABCD: Aꜱ = S² 36 = S² S = √36 = 6 Let P be the midpoint of AB. As AB is a chord, OP is perpendicular to AB and bisects it. As ABCD is a square, and C and D are on the diameter, then this means O is equidistant from C and D just as P is from A and B. Thus OD = OC = S/2 = 3. Draw radius OB. Triangle ∆OCB: OC² + CB² = OB² 3² + 6² = r² r² = 9 + 36 = 45 r = √45 = 3√5 Draw radius OF. Triangle ∆OGF: OG² + GF² = OF² (s+3)² + s² = (3√5)² s² + 6s + 9 + s² = 45 2s² + 6s - 36 = 0 s² + 3s - 18 = 0 (s+6)(s-3) = 0 s = -6 ❌ | s = 3 Square CEFG: Aₛ = s² = 3² = 9 sq units
This is probably the first problem that I have understood almost intuitively that did not require subtract of different areas. I am starting to think that subtraction of areas would have been a little too unwieldy. I shall use that as further practice!!!
Let (I) be the intersecting point of the diameter with the circumference on the right , and let it be ( H) on the left.
Suppose: HD = x, then CI=x. DC=6. Let CG(side of square CEFG) = y and GI = z.
According to the intersecting chords formula:
CB²=CI(DC+DH),
So 6² =x(6+x), then x = 3.7 => y (CG)+z(GI) = 3.7, so z=3.7-y.
Again according to the intersecting chords formula: FG² = GI(GC+DC+DH),
Hence y² = z(y+6+3.7) by substitution=> (3.7-y)(y+6+3.7) .
Solving for y , y= 3 so area of square CEFG = 3² = 9 units²
Very detailed and meticulous, but you don't need to calculate the radius. AO^2=AD^2+OD^2
=OF^2=OG^2+GF^2 =>
3^2+6^2=x^2+(3+x)^2 =>
3^2-x^2 = (3+x)^2 - 6^2
which gives x=3 as only +ve root using a^2-b^2 formula.
[ABCD] = 36 -> AB=BC=CD=AD=sqrt (36) = 6. Since the AB endpoints are in the circumference, the center O is in the middle of the parallel segment CD. Then R^2 = (CD/2) ^2 + AD^2 = 3^2 + 6^2 = 45 -> R=3*sqrt (5).
To find small square sides, OF^2 = (OC+CD) ^2 + FG^2 = (3+x) ^2 + x^2 = R^2 = 45 -> 2x^2 + 6x + 9 = 36 --> 2x^2 + 6x - 36 = 0 --> x^2 + 3x - 18 = 0 --> (x+6) * (x-3) = 0 -->
x=3 --> [CEFG] = x^2 = 3^2 = 9 sq. units
Нужно просто треугольник AOP повернуть на угол 90° относительно точки O по часовой стрелке. OA=OF, угол AOF=90°, угол POG=90° отсюда следует, что прямоугольные треугольники равны. И длина катета FG=3. При этом считать длину радиуса окружности вообще не нужно.
👍. Длины сторон квадратов тоже можно не считать. Достаточно знать, что они отличаются в 2 раза. Поэтому, [CEFG]=[ABCD]/2²=36/4=9 кв.ед.
Отлично, это хорошо видно
Let O be the center of the semicircle. Let s be the side length of CEFG, S be the side length of ABCD, and r be the radius of semicircle O.
Square ABCD:
Aꜱ = S²
36 = S²
S = √36 = 6
Let P be the midpoint of AB. As AB is a chord, OP is perpendicular to AB and bisects it. As ABCD is a square, and C and D are on the diameter, then this means O is equidistant from C and D just as P is from A and B. Thus OD = OC = S/2 = 3. Draw radius OB.
Triangle ∆OCB:
OC² + CB² = OB²
3² + 6² = r²
r² = 9 + 36 = 45
r = √45 = 3√5
Draw radius OF.
Triangle ∆OGF:
OG² + GF² = OF²
(s+3)² + s² = (3√5)²
s² + 6s + 9 + s² = 45
2s² + 6s - 36 = 0
s² + 3s - 18 = 0
(s+6)(s-3) = 0
s = -6 ❌ | s = 3
Square CEFG:
Aₛ = s² = 3² = 9 sq units
The way the problem is solved is impressive.
A = ¼ A = ¼ . 36 = 9 cm² ( Solved √ )
If OH _|_ BF, OB=OF=R, BO=OF, EF=EC (square), then EF=BE end BC/2=3, area=9
Excellent explanation
Si aplicamos al cuadrado ABCD un giro de 90º en torno al centro del círculo, A se superpone a F---> FGCE = ABCD/4 =36/4=9.
Gracias y saludos.
Excellent comments also❤
This is probably the first problem that I have understood almost intuitively that did not require subtract of different areas. I am starting to think that subtraction of areas would have been a little too unwieldy. I shall use that as further practice!!!
Solved in no time 👍
36=6^2 r=√[3^2+6^2]=√45=3√5
EC=CG=GF=FE=x (3+x)^2+x^2=(3√5)^2 9+6x+x^2+x^2=45
2x^2+6x-36=0 x^2+3x-18=0 (x+6)(x-3)=0 x>0 , x=3
area of CEFG = x^2 = 3^2 = 9
180°ABCDEFG/36 =5 (ABCDEFG ➖ 5ABCDEFG+5).
r=√(6^2+3^2)=√45=√(l^2+(l+3)^2)....45=2l^2+6l+9...2l^2+6l-36=0....l=(-3+√(9+72))/2=3..Ablue=9
9 cm^2
Sometimes, beginning, you show simple relationships in detail. But later you are so quıckly, that i tınk you must urgent to wc.