Triangle $A B C$ has vertices at $A(1,0), B(21,0)$ and $C(21,21)$. The perimeter of the triangle is
(A) 70
(B) 42
(C) 64
(D) 72
(E) 63
The perimeter of the triangle is the sum of the lengths of the sides. We make a sketch of the triangle to help us with our calculations. Since side $A B$ is along the $x$-axis and side $B C$ is parallel to the $y$-axis, then the triangle is right-angled, and we can use Pythagoras' Theorem to calculate the length of $A C$.

The length of $A B$ is 20 and the length of $B C$ is 21 .

Calculating $A C$,

$$
\begin{aligned}
A C^{2} & =A B^{2}+B C^{2} \\
A C^{2} & =20^{2}+21^{2} \\
A C^{2} & =400+441 \\
A C^{2} & =841 \\
A C & =29
\end{aligned}
$$

Therefore, the perimeter is $20+21+29=70$.

ANswer: (A)