Question

In the picture, the square on the hypotenuse of the top most right triangle is drawn.

Calculate the area and the length of a side of the square.

Answer 1

Let’s label the triangles and the associated square as below.

Consider the figure 1 below
Let's find the length of $AQ$
$$\begin{array}{cc}& {AQ}^2={AP}^2+{PQ}^2\\ & {AQ}^2=1^2+1^2\\ & {AQ}^2=2\\ & AQ=\sqrt{2}\end{array}$$
Now,
$$\begin{array}{cc}& {AR}^2={AQ}^2+{QR}^2\\ & {AR}^2=(\sqrt{2}{)}^2+1^2[\text{Since}\sqrt{2}\times \sqrt{2}=\sqrt{2\times 2}=\sqrt{4}=2]\\ & {AR}^2=2+1=3\end{array}$$
$$AR=\sqrt{3}$$
Similarly,
$$\begin{array}{cc}& {AS}^2={AR}^2+{SR}^2\\ & {AS}^2=(\sqrt{3}{)}^2+1^2=3+1\text{[Since}\sqrt{3}\times \sqrt{3}=\sqrt{3\times 3}=\sqrt{9}=3]\\ & {AS}^2=4\\ & AS=\sqrt{4}=2\end{array}$$
Now, let's find the $AB$, which is the common side for the triangle $ABS$ and square $ABCD$.
We know that,
$$\begin{array}{cc}& A{B}^2=A{S}^2+B{S}^2\\ & A{B}^2=2^2+1^2=4+1\\ & A{B}^2=5\\ & AB=\sqrt{5}\end{array}$$
Hence, the length of each side of the square $=a=AB=\sqrt{5}$ metre
Now, the area of the square $=a^2=(\sqrt{5}{)}^2=\sqrt{5}\times \sqrt{5}=\sqrt{25}=5$ square metre