This problem relies upon the unstated requirement that the line CD intersects the line AB at 90 degrees. At any other angle, the line CD could not be extended through to point O.

@user-tr5ex5jn5y

Ай бұрын

given isn't complete

@EyadAmmari

28 күн бұрын

Since the segments are equal on both sides of the radius then by definition the angle is 90 degrees. This is a basic property of circles and chords.

@davidbrisbane7206

25 күн бұрын

The intersecting cord theorem works whether angle ADC is a right angle or not, but of course because the two segments of the cord are equal, then this means the cord intersects CO at 90°.

You made hard work of this. OD=r-3. Use Pythagoras for triangle ODA and you get a quadratic with r squared. You can then solve for r. It should take you about 5 lines.

@michaeledwards2251

21 күн бұрын

The r squared appears on both sides, cancels out, leaving a linear equation.

🔺️CDB notable de (37/2)° 🔺️ODB notable de 37° DB = 9 = 3(3) Radio = OB = 5(3) = 15 Área 15^2 π 225 π

S=225π≈706,86≈707

Another solution using Analytic Geometry. Let Cartesian coordinate system Oxy and the circle (O, R=a) . So A(-9,a-3) , B(9,a-3) , C(0,a) and E(0,-a) Vectors (BC)=(-9,3) , (EB) =(9,2a-3) The inner product of the two vectors is zero cause they are vertical.: (BC) •(EB) = 0 => -9•9+ 3(2a-3)=0 => a=15 ……..

I don't Speak english Very well, but I learn Thank you

Greats solutions!!

Nice solutions. I liked 2nd solution . Thanks a lot .

By chord intersection theorem, 9×9=3×X; X=9×9/3=27, so the diameter is 3+27=30, area of circle=π×15×15= 225π

81/3 = 27. (27 + 3)/2 = 15. Area = pi * 15^2. Thank for a simple problem : ) I had been tripped on other problems by a lack of focus on my part, this one lifted my spirits : )

The triangle ABO is isosceles cause OA=0B=R. OD is the median of the triangle => OD is heigh , so OC⊥AB. Triangle BCE is orthogonal cause In orthogonal triangle BCE : BD² =DC⋅DE ( BD is heigh) => 9²=3⋅DΕ ⇒ DE=27. DE=R+OD⇒R+OD=27 (1) DC=R-OD⇒R-OD=3 (2) (1)+(2) => 2R=30⇒R=15 . So the aria of the circle = πR^2=π⋅15^2=225π

Muito bom! De Rio.

9*9=3(2r-3)

Ich sehe im linken Halbkreis einen Thaleskreis mit Durchmesser d und das rechtwinklige Dreieck EAC. (EA)² + (AC)² = d² (EA)²= (d-3)² + 9² (AC)² = 9² + 3² (d-3)² + 9² + 9² + 3² = d² d² - 6d +3² + 171 = d² 6d = 180 d = 30 r = 15 A = 125*Pi

sqrt(r*r- 9*9) +3 = r; r=15

Easier to do 9 x 9 = 3 x remainder of diameter

225 pi or 708.858

I'm going to cut my cakes like that in future so I get the biggest bit.

225π

900 pi..?

The 2nd theorem is called plotemy's theorem

@Grizzly01-vr4pn

Ай бұрын

Not quite. It's called the intersecting chords theorem, but it is closely related to Ptolemy's theorem.

@aryanjain6948

Ай бұрын

@@Grizzly01-vr4pn interesting. My teacher told me it was plotemy's. Well will definitely look up both the theorems

@Grizzly01-vr4pn

Ай бұрын

@@aryanjain6948 You can see Ptolemy's theorem in use in the 2nd method of this video on Math Booster: Russian Math Olympiad Problem kzread.info/dash/bejne/pK2C2surZqXYabA.htmlsi=vIIPNDgzP09POaQR&t=741

@aryanjain6948

Ай бұрын

@@Grizzly01-vr4pn yo thanks man

Answer 225p or 708.858 This is another method Draw a line from the circle's center to the chord's center. This line is 3 less than the radius or r-3. Draw another line from the circle's center to the chord's end. This line is the radius of the circle, r. Since a right triangle is formed with the sides r-3, r, and 9, use the Pythagorean theorem to find r. (r-3)^2 + 9^2 = r^2 r^2+ 9 -6r + 81 =r^2 9 -6r + 81 =0 90 = 6r 15 = r, the radius of the circle is 15 the area, hence, is 15^2 pi or 225 pi or 706.858. Answer

Rather tedious explanation!

普通、動画の内容は解いてから見るのでサムネにπを使うことを書いておく必要があります。でないとπ≒3.14を使う人もいます。毎回答えを惑わすと視聴者減少につながるので必ずお願いします。