Quadrilateral

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:

$$\begin{gathered} S=a^2 \\ {a}^2-{\displaystyle \frac{1}{2}}{\displaystyle \frac{a}{2}}{\displaystyle \frac{a}{2}}-{\displaystyle \frac{1}{2}}a{\displaystyle \frac{a}{2}}=49 \\ {a}^2-{\displaystyle \frac{1}{8}}{a}^2-{\displaystyle \frac{1}{2}}{a}^2=49 \\ {\displaystyle \frac{5}{8}}{a}^2=49 \\ {a}^2=49{\displaystyle \frac{8}{5}}=78.4{\text{cm}}^2 \\ S=a^2=78.4{\text{cm}}^2 \end{gathered}$$

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