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:

$$\begin{gathered} S_1=76{\text{cm}}^2 \\ S=a^2 \\ {a}^2-\frac{1}{2}(a/2)(a/2)-\frac{1}{2}a(a/2)=S_1 \\ {a}^2-\frac{1}{8}{a}^2-\frac{1}{2}{a}^2=S_1 \\ \frac{5}{8}{a}^2=S_1 \\ S=a^2=\frac{8}{5}{S}_1 \\ S=\frac{8}{5}\cdot S_1=\frac{8}{5}\cdot 76=\frac{8\cdot 76}{5}=\frac{608}{5}=\frac{608}{5}{\text{cm}}^2=121.6{\text{cm}}^2 \end{gathered}$$

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