Question

The figure below is formed by overlapping a square BGFE and a rectangle ABCD. Given that C is the centre of the square and E is the mid point of DC, the fraction of the figure , that is unshaded, can be written as p/q in its simplest form. Find p + q.

Answer 1

Hey here is your answer!
Since, Figure is not to scale , use it for indicative
purposes only, but it is labelled.
Final result :p/q=6/7
i.e p+q=13.

Let the side of the square BEFG be 'a' units.

Since , C is centre of square BEFG.
=> BC = CE = x (say)

In triangle BCE, by Pythagoras Theorem
${x}^{2} + {x}^{2} = {a}^{2} \\ x = \frac{a}{ \sqrt{2} }$

Since, E is the mid -point of CD
=> DE = CE=a/√2

Length of rectangle , CD= 2*a/√2=a√2.
Breadth of rectangle, BC= x=a/√2.

Now,
Area of shaded part = Area of triangle BCE
=1/2 * BC * CE
= 1/2 * x * x
= a^2/4.

Since, shaded part is present in both rectangle
ABCD and square BEFG
=>
Area of full figure =
Area of rect . ABCD + Area of square ABCD -Area of shaded part.

= AB * BC + a * a - a^2/4
= a√2 * (a/√2) +a*a -a^2/4
= a^2+a^2-a^2/4
= 7/4a^2.

Area of unshaded part
= Area of full figure - Area of shaded part
= 7/4 a^2-a^2/4
=(7/4-1/4)a^2
= 3/2a^2

Now, finally
Fraction of unshaded part
= (Area of unshaded part )/(Area of full figure)
$= \frac{ \frac{3 {a}^{2} }{2} }{ \frac{7 {a}^{2} }{4} } \\ = \frac{6}{7}$

Hope, you understand my answer and it may helps you .