Quadrilateral

In the square ABCD point, P is in the middle of the DC side, and point Q is in the middle of side AD. If the area of quadrilateral BQPC is 76 cm², what is the area of ABCD?

Correct answer:

S =  121.6 cm²

Step-by-step explanation:

$S_{1} = 76 \ \text{cm}^2 \ \\ \ \\ S\ = \ a^2 \ \\ a^2\ -\ \dfrac{ 1 }{ 2 } \ (a/2)\ (a/2)\ -\ \dfrac{ 1 }{ 2 } \ a\ (a/2)\ = \ S_{1} \ \\ a^2\ -\ \dfrac{ 1 }{ 8 } \ a^2\ -\ \dfrac{ 1 }{ 2 } \ a^2\ = \ S_{1} \ \\ \ \dfrac{ 5 }{ 8 } \ a^2\ = \ S_{1} \ \\ S\ = \ a^2\ = \ \dfrac{ 8 }{ 5 } \ S_{1} \ \\ \ \\ S = \dfrac{ 8 }{ 5 } \cdot \ S_{1} = \dfrac{ 8 }{ 5 } \cdot \ 76 = \dfrac{ 8 \cdot \ 76 }{ 5 } = \dfrac{ 608 }{ 5 } = 121.6 \ \text{cm}^2$

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