1. **State the problem:** We have a right triangle ABC with legs $a$ and $b$, hypotenuse $h$, and area 25 m². We want to express $h$ in terms of the perimeter $P$. 2. **Recall the formulas:** - Area $= \frac{1}{2}ab = 25$ so $ab = 50$. - Perimeter $P = a + b + h$. - By Pythagoras, $h = \sqrt{a^2 + b^2}$. 3. **Express $h$ in terms of $a$ and $b$:** $$h = \sqrt{a^2 + b^2}$$ 4. **Express $h$ in terms of $P$ and $a,b$:** $$P = a + b + h \implies h = P - (a + b)$$ 5. **Use the identity for $(a + b)^2$:** $$ (a + b)^2 = a^2 + 2ab + b^2 $$ 6. **Rewrite $h^2$ using Pythagoras:** $$ h^2 = a^2 + b^2 = (a + b)^2 - 2ab $$ 7. **Substitute $ab = 50$:** $$ h^2 = (a + b)^2 - 100 $$ 8. **Recall from step 4 that $h = P - (a + b)$, so:** $$ h = P - (a + b) \implies a + b = P - h $$ 9. **Substitute $a + b = P - h$ into the expression for $h^2$:** $$ h^2 = (P - h)^2 - 100 $$ 10. **Expand and simplify:** $$ h^2 = P^2 - 2Ph + h^2 - 100 $$ 11. **Cancel $h^2$ on both sides:** $$ 0 = P^2 - 2Ph - 100 $$ 12. **Solve for $h$:** $$ 2Ph = P^2 - 100 \implies h = \frac{P^2 - 100}{2P} $$ **Final answer:** $$\boxed{h = \frac{P^2 - 100}{2P}}$$ This expresses the hypotenuse $h$ in terms of the perimeter $P$ given the area is 25 m².