Answer: a) two; b) two. Solution. a) Let the sides of the rectangle be integers $a$ and $b$ ( $a \leq b$ ). Then $a b=2(a+b)$, which is equivalent to the equation $(a-2)(b-2)=4$. The number 4 can be factored into the product of two integers in four ways: $(1 ; 4),(2,2),(-4 ;-1),(-2,-2)$, of which the first two give rectangles $(3,6)$ and $(4,4)$. b) Let the legs of the right triangle be $a$ and $b(a \leq b)$. We have the equation $a+b+\sqrt{a^{2}+b^{2}}=\frac{a b}{2}$. Let $u=a+b, v=a b$ Then the equation can be written as $2 u+2 \sqrt{u^{2}-2 v}=v$. After isolating the square root on the left side, squaring both sides, and dividing the equation by $v$, we get: $v-4 u+8=0$, i.e., $a b-4 a-4 b+8$, or $(a-4)(b-4)=8$. The number 8 can be factored into the product of two integers in four ways: ( $1 ; 8),(2,4),(-8 ;-1),(-4,-2)$, of which the first two give right triangles $(5,12,13)$ and $(6,8,10)$.
10.5. How many a) rectangles; b) right-angled triangles with integer sides exist, for which the area is numerically equal to the perimeter? (Congruent figures are counted as one).