Question

If ABCD and PQRS are two squares, such that area of square PQRS is 'A' m², then find the value of $\rm \sqrt{17 \: A}$ .​

Answer 1

$\Huge\textrm{ \sqrt{17A}=12 }$

$\huge\textrm{Solution}$

$\textrm{A right triangle contains 3 similar triangles.}$

$\textrm{Let us first consider them.}$

$\Large\textrm{Leg Rule}$

$\textrm{Refer to the attachment.}$

$\cdots\longrightarrow\boxed{a^{2}=c\cdot c_{1}}$

$\cdots\longrightarrow\boxed{b^{2}=c\cdot c_{2}}$

$\large\textrm{(By leg rule)}$

$1^{2}=\sqrt{17}x$

$\therefore x=\dfrac{\sqrt{17}}{17}$

$\textrm{The \underline{identical four pieces} are uncolored.}$

$\textrm{We know the following values.}$

• $\textrm{ \overline{AS}=\dfrac{16\sqrt{17}}{17} m}$
• $\textrm{ \overline{SD}=\dfrac{4\sqrt{17}}{17} m}$

$\textrm{Let us find the area of a single triangle.}$

$\triangle ADS$

$=\dfrac{1}{2}\times\overline{AS}\times\overline{SD}$

$=\dfrac{1}{2}\times\dfrac{16\sqrt{17}}{17}\times\dfrac{4\sqrt{17}}{17}$

$=\dfrac{1}{2}\times\dfrac{64\times17}{17^{2}}$

$\textrm{ =\dfrac{32}{17} m ^{2} }$

$\textrm{We know the total area.}$

$\textrm{ \square{ABCD}=16 m ^{2} }$

$\textrm{Hence, the total area is as follows.}$

$\square{ABCD}-4\triangle{ADS}$

$=16-4\times\dfrac{32}{17}$

$=16-\dfrac{128}{17}$

$=\dfrac{272-128}{17}$

$\textrm{ =\dfrac{144}{17} m ^{2} }$

$\huge\textrm{Final Answer}$

$\textrm{The required value is as follows.}$

$\sqrt{17A}$

$=\sqrt{12^{2}}$

$=12$

The answer is 12.

Hope this helps :)