1. **Problem statement:** In a right-angled triangle with legs $x$ and $y$, one leg is extended by 4 cm and the other shortened by 5 cm, resulting in an isosceles right-angled triangle with the same area as the original. Find the lengths $x$ and $y$ of the original legs and the new legs. 2. **Known facts and formulas:** - Area of original triangle: $A = \frac{1}{2}xy$ - New triangle is isosceles right-angled, so its legs are equal: let new legs be $a$. - New legs: one is $x + 4$, the other is $y - 5$. - Since new triangle is isosceles right-angled, $a = x + 4 = y - 5$. - Area of new triangle: $A_{new} = \frac{1}{2}a^2$. - Areas are equal: $\frac{1}{2}xy = \frac{1}{2}a^2$. 3. **Set up equations:** From isosceles condition: $$ x + 4 = y - 5 \implies y = x + 9 $$ From equal areas: $$ \frac{1}{2}xy = \frac{1}{2}a^2 \implies xy = a^2 $$ But $a = x + 4$, so: $$ xy = (x + 4)^2 $$ Substitute $y = x + 9$: $$ x(x + 9) = (x + 4)^2 $$ 4. **Expand and simplify:** $$ x^2 + 9x = x^2 + 8x + 16 $$ Subtract $x^2$ from both sides: $$ 9x = 8x + 16 $$ Subtract $8x$: $$ \cancel{9x} - \cancel{8x} = 16 \implies x = 16 $$ 5. **Find $y$ and $a$:** $$ y = x + 9 = 16 + 9 = 25 $$ $$ a = x + 4 = 16 + 4 = 20 $$ 6. **Check areas:** Original area: $$ A = \frac{1}{2} \times 16 \times 25 = 200 $$ New area: $$ A_{new} = \frac{1}{2} \times 20^2 = \frac{1}{2} \times 400 = 200 $$ Areas match, confirming the solution. **Final answer:** - Original legs: $x = 16$ cm, $y = 25$ cm - New legs: both $20$ cm