Question

The ratio of the area of a circle with centre O and the area of the rectangle ABCD inscribed in it is π : root3 . E is a point on AB such that ∠ADE = ∠ ODC. Find AE : AD.

Answer 1

Answer:

1 : √3 or

√3 : 1

Step-by-step explanation:

Hi,

Let the radius of the circle be 'R'

Let the rectangle inscribed in the circle be ABCD

If AB = l and BC = b

then Area of rectangle = l*b,

Area of circle = π*R²

Given that Ratio of Area of circle to Area of Rectangle

= π : √3

⇒ πR²/lb = π/√3

lb/R² = √3

let x = l/R and y = b/R, then

xy = √3

Consider triangle ABC, which is right angled at B, since ABCD is

a rectangle, so

AC² = AB² + BC²

But AC is the chord of the circle subtending 90° at B point on

circle, so AC should be diameter and AC = 2R

4R² = l² + b²

(l/R)² + (b/R)² = 4

⇒ x² + y² = 4

But we know xy = √3

So x²y² = 3

x² + 3/x² = 4

x⁴ - 4x² + 3 = 0

So x² = 1 or x² = 3

⇒ x = 1 or √3

If x = 1, then y = √3 or vice-versa

If we assume x > y, then x : y =  √3 : 1

But x : y = l : b = √3 : 1

Consider Δ ADE and ΔBDC ,

∠DAE = ∠DCB = 90°

∠ADE = ∠BDC = ∠ODC (Given)

Hence , ΔADE ≈ ΔBDC

Since both triangles are similar, ratio of the sides corresponding

to equal angles would be equal

AE/BC = AD/DC

⇒AE/AD = BC/DC = 1/√3 (Assuming BC < DC)

or √3 (if BC > DC)

Hope, it helps !