In the square ABCD point P is in the middle of the DC side and point Q in the middle pages AD. If the area of quadrilateral BQPC is 49 cm2, what is the area of ABCD?

Correct result:

S =  78.4 cm2

Solution:

$S = a^2 \ \\ \ \\ a^2 - \dfrac{1}{2}\dfrac{a}{2}\dfrac{a}{2}-\dfrac{1}{2}a\dfrac{a}{2} = 49 \ \\ a^2 - \dfrac{1}{8}a^2 - \dfrac{1}{2}a^2 = 49 \ \\ \dfrac{5}{8}a^2 = 49 \ \\ a^2 = 49 \dfrac{8}{5} = 78.4 \ \text{cm}^2 \ \\ S = a^2 = 78.4 \ \text{cm}^2$

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