Question

Find the smaller distance of the vertex of the shrunk square of area 17 sq. cm from the corner of the original square having a side of 5 cm.

Answer 1

Let ‘x’ cm be the smaller distance of vertex of the shrunk square from the corner of the original square. Then the larger distance of vertex of the shrunk square from the corner of the original square = (5 - x) cm
Area of shrunk square = Area of original square - Area of four right angled triangles
17 = (5 × 5) - 4 × ($\frac{1}{2}$) × ‘x’ × (5 - x)
25 - 17 = 2 × (5x - $x^2$)
$x^2$ - 5x + 4 = 0
$x^2$ - 1x - 4x + 4 = 0
x(x -1) -4(x -1) = 0
(x-1)(x-4) = 0
x = 1 or 4
In our case, x = 1 since x is defined as the smaller distance of vertex of the shrunk square from the corner of the original square.