If in a triangle ABC right-angled at B,s - a | vedclass

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(B) Given $s - a = 3$ and $s - c = 2$,where $s = \frac{a+b+c}{2}$.$s - a = \frac{b+c-a}{2} = 3 \Rightarrow b+c-a = 6$ ... $(i)$$s - c = \frac{a+b-c}{2

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$3, 4$

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(B) Given $s - a = 3$ and $s - c = 2$,where $s = \frac{a+b+c}{2}$.$s - a = \frac{b+c-a}{2} = 3 \Rightarrow b+c-a = 6$ ... $(i)$$s - c = \frac{a+b-c}{2} = 2 \Rightarrow a+b-c = 4$ ... $(ii)$Adding $(i)$ and $(ii)$,we get $2b = 10 \Rightarrow b = 5$.Since the triangle is right-angled at $B$,by Pythagoras theorem,$a^2 + c^2 = b^2 = 25$ ... $(iii)$.From $(i)$,$c - a = 6 - b = 6 - 5 = 1 \Rightarrow c = a + 1$.Substituting into $(iii)$: $a^2 + (a+1)^2 = 25$ $\Rightarrow a^2 + a^2 + 2a + 1 = 25$ $\Rightarrow 2a^2 + 2a - 24 = 0$ $\Rightarrow a^2 + a - 12 = 0$.$(a+4)(a-3) = 0$. Since $a > 0$,$a = 3$.Then $c = a + 1 = 4$.Thus,the values are $a = 3$ and $c = 4$.

If in a triangle $ABC$ right-angled at $B$,$s - a = 3$ and $s - c = 2$,then the values of $a$ and $c$ are respectively:

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