Pythagoras' Theorem

Applying Pythagoras' theorem: $a^2+b^2=c^2$, where $a=3$ and $b=4$. We set the hypotenuse as $x$ and solve $3^2+4^2=x^2$, which gives $9+16=x^2$, so $x^2=25$. Taking the square root of both sides, we get $x=5$.

We use Pythagoras' theorem: $a^2+b^2=c^2$, where $a=6$ and $b=8$. We set the hypotenuse as $x$ and solve $6^2+8^2=x^2$, which gives $36+64=x^2$, so $x^2=100$. Taking the square root of both sides, we get $x=10$.

Applying Pythagoras' theorem: $a^2+b^2=c^2$, where $a=9$ and $b=12$. We set the hypotenuse as $x$ and solve $9^2+12^2=x^2$, which gives $81+144=x^2$, so $x^2=225$. Taking the square root of both sides, we get $x=15$.

The hypotenuse is always the longest side in a right-angled triangle. Here, the longest side is 20, so it must be the hypotenuse.

Using Pythagoras' theorem: $a^2+b^2=c^2$, where $c=13$ and $a=12$. We set the missing side as $x$ and solve $12^2+x^2=13^2$, which gives $144+x^2=169$, so $x^2=25$. Taking the square root of both sides, we get $x=5$.

We form a right triangle using the horizontal and vertical distances. The horizontal distance is $8-2=6$, and the vertical distance is $10-2=8$. We set the unknown distance as $x$ and apply Pythagoras' theorem: $6^2+8^2=x^2$, which simplifies to $36+64=x^2$. We get $x^2=100$, and $x=10$.