In Delta ABC,AB + BC = 23,BC + AC = 32,and AB | vedclass

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(N/A) Given: $AB + BC = 23$ $(i)$,$BC + AC = 32$ $(ii)$,and $AB + AC = 25$ $(iii)$.Adding $(i)$,$(ii)$,and $(iii)$,we get: $2(AB + BC + AC) = 23 + 32

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(N/A) Given: $AB + BC = 23$ $(i)$,$BC + AC = 32$ $(ii)$,and $AB + AC = 25$ $(iii)$.
Adding $(i)$,$(ii)$,and $(iii)$,we get: $2(AB + BC + AC) = 23 + 32 + 25 = 80$.
Therefore,$AB + BC + AC = 40$ $(iv)$.
Subtracting $(ii)$ from $(iv)$: $AB = 40 - 32 = 8$.
Subtracting $(iii)$ from $(iv)$: $BC = 40 - 25 = 15$.
Subtracting $(i)$ from $(iv)$: $AC = 40 - 23 = 17$.
Now,check the sides: $AB^2 + BC^2 = 8^2 + 15^2 = 64 + 225 = 289$.
Also,$AC^2 = 17^2 = 289$.
Since $AB^2 + BC^2 = AC^2$,by the converse of the Pythagoras theorem,$\Delta ABC$ is a right-angled triangle with $\angle B = 90^\circ$.

In $\Delta ABC$,$AB + BC = 23$,$BC + AC = 32$,and $AB + AC = 25$. Show that $\Delta ABC$ is a right-angled triangle.

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