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In a circle, chord AB and chord CD intersect at E such that AE : EB = 2: 3 and CE: ED = 5: 2. If AB =X and CD = y, which of the following is true?

Ax2y2=1516\frac{x^2}{y^2}=\frac{15}{16}

Bxy=615\frac{x}{y}=\frac{6}{15}

Cxy=109\frac{x}{y}=\frac{10}{9}

Dx2y2=125147\frac{x^2}{y^2}=\frac{125}{147}

Answer:

x2y2=125147\frac{x^2}{y^2}=\frac{125}{147}

Read Explanation:

Using the intersecting chords theorem:

AE×EB=CE×EDAE\times EB=CE\times ED

Given:

AE:EB=2:3,CE:ED=5:2AE:EB=2:3,\qquad CE:ED=5:2

Let:


AE=2a, EB=3a, CE=5b, ED=2bAE=2a,\ EB=3a,\ CE=5b,\ ED=2b

Then:

(2a)(3a)=(5b)(2b)(2a)(3a)=(5b)(2b)
6a2=10b26a^2=10b^2

a2b2=53\frac{a^2}{b^2}=\frac53

Now,


X=AB=2a+3a=5aX=AB=2a+3a=5a

and

Y=CD=5b+2b=7bY=CD=5b+2b=7b

Therefore,


X2Y2\frac{X^2}{Y^2}


=25a249b2=\frac{25a^2}{49b^2}


=2549×53=\frac{25}{49}\times\frac53
=125147=\boxed{\frac{125}{147}}


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