CQ has a length of 9 cm.
Case 1:
△CQN, a 45-45-90 triangle, is constructed using CQ as the hypotenuse.
Since it is a 45-45-90 triangle, the lengths of the sides are in the ratio as shown above.
Since CQ = 9 cm
[tex]\begin{gathered} x\sqrt[]{2}=9 \\ x=\frac{9}{\sqrt[]{2}} \\ x=\frac{9}{\sqrt[]{2}}\times\frac{\sqrt[]{2}}{\sqrt[]{2}} \\ x=\frac{9\sqrt[]{2}}{2} \end{gathered}[/tex]Therefore, the lengths of the sides are
CQ = 9 cm
QN = (9√2)/2 cm
CN = (9√2)/2 cm
Case 2:
△PQC, a 45-45-90 triangle, is constructed using CQ as a leg.
Since it is a 45-45-90 triangle, the lengths of the sides are in the ratio as shown above.
Since CQ = 9 then it means x = 9
Therefore, the lengths of the sides are
PQ = 9√2 cm
CQ = 9 cm
PC = 9 cm
Now let us find the area of both triangles.
Recall that the area of a triangle is given by
[tex]A=\frac{1}{2}\cdot b\cdot h[/tex]Where b is the base and h is the height of the triangle.
Area of △CQN:
Base = (9√2)/2 cm
Height = (9√2)/2 cm
[tex]\begin{gathered} A_{\text{CQN}}=\frac{1}{2}\cdot\frac{9\sqrt[]{2}}{2}\cdot\frac{9\sqrt[]{2}}{2} \\ A_{\text{CQN}}=\frac{81(2)}{8} \\ A_{\text{CQN}}=\frac{81}{4}\: cm^2 \end{gathered}[/tex]Area of △PQC:
Base = 9 cm
Height = 9 cm
[tex]\begin{gathered} A_{\text{PQC}}=\frac{1}{2}\cdot9\cdot9 \\ A_{\text{PQC}}=\frac{81}{2}\: cm^2 \end{gathered}[/tex]Finally, let us compare the area of △PQC and △CQN
[tex]\begin{gathered} \frac{A_{\text{PQC}}}{A_{\text{CQN}}}=\frac{\frac{81}{2}}{\frac{81}{4}}=\frac{81}{2}\times\frac{4}{81}=\frac{4}{2}=2 \\ \frac{A_{\text{PQC}}}{A_{\text{CQN}}}=2 \\ A_{\text{PQC}}=2\cdot A_{\text{CQN}} \end{gathered}[/tex]Therefore, △PQC will have an area that is 2 times the area of △CQN
