Let the required consecutive multiplies of 7 be $7 x$ and $7(x+1)$.

According to the given condition,

$(7 x)^{2}+[7(x+1)]^{2}=1225$
$\Rightarrow 49 x^{2}+49\left(x^{2}+2 x+1\right)=1225$
$\Rightarrow 49 x^{2}+49 x^{2}+98 x+49=1225$
$\Rightarrow 98 x^{2}+98 x-1176=0$
$\Rightarrow x^{2}+x-12=0$
$\Rightarrow x^{2}+4 x-3 x-12=0$
$\Rightarrow x(x+4)-3(x+4)=0$
$\Rightarrow(x+4)(x-3)=0$
$\Rightarrow x+4=0$ or $x-3=0$
$\Rightarrow x=-4$ or $x=3$
$\therefore x=3 \quad$ (Neglecting the negative value)
When $x=3$,
$\begin{array}{l}
7 x=7 \times 3=21 \\
7(x+1)=7(3+1)=7 \times 4=28
\end{array}$

Hence, the required multiples are 21 and 28 .