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Please forgive my English. In a square with side AB, there is inscribed a rectangle, whose area is $\frac{3}{8}$ the square area, if the measure of AB is 28, how long can be $AP$, where $P$ is the vertex of the rectangle on the side $AB$ of the square, please have any hint?

Maryam
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  • I would let $AP=x$ and use the Pythagorean theorem, etc. to express various side lengths in terms of $x$. Using the info of the area of the rectangle, you should be able to solve for $x$. – Kenta S Mar 16 '21 at 16:45

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For simplicity let's suppose AP=AQ=x, we have:

$PQ^2=2x^2\Rightarrow PQ=x\sqrt2$

$PS^2=2(28-x)^2 \Rightarrow PS=(28-x)\sqrt 2$

$S_{PQRS}=\frac 38 (28)^2=194$

Therefore:

$PS\times PQ=2x(28-x)=294$

Or:

$x^2-28x+147=0$

Which gives $x=7$

sirous
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