Question

A square is inscribed in larger square with each vertex on each side of the larger one. If the area of small square is 25/49 times the larger one. Then find the ratio of the vertex which divides the side of larger square

Answer 1

Let A = side of larger square

       a = side of smaller square

Area of a square = a²

By given info,

a² = 25/49 A²

Taking triangle PAQ, and applying Pythagoras’ Theorem,

PQ² = PA² + AQ²

⇒a² = x² + y²

⇒25/49 A² = x² + y²    - - - - - - - - - - - -  (1)

Now, A = x + y

Squaring on both sides,

A² = (x+y)²

⇒A² = x² + y² + 2xy

⇒A² - 2xy = x² + y² - - - - - - - - - -   (2)

Substituting (2) in (1),

25/49 A² = A² - 2xy

⇒2xy = A² - 25/49 A²

⇒2xy = 24/49 A²

⇒xy = 12/49 A²

Taking the identity

(x - y)² = (x + y)² - 4xy

⇒(x - y)² = (A)² - 4(12/49 A²)  (since A = x + y and xy = 12/49 A²)

⇒(x - y)² = A² - 48/49 A²

⇒(x - y)² = 1/49 A²

⇒x - y = 1/7 A²

Solving the equations x + y = A and x - y = 1/7 A²,

x = 4/7 A

y = 3/7 A

∴ x : y = 4 : 3