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(B) Let the two legs of the right-angled triangle be $a$ and $b$,and the hypotenuse be $c = 39\, cm$.Given the perimeter $P = a + b + c = 90\, cm$.The

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$36\, cm, 15\, cm$

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(B) Let the two legs of the right-angled triangle be $a$ and $b$,and the hypotenuse be $c = 39\, cm$.Given the perimeter $P = a + b + c = 90\, cm$.Therefore,$a + b + 39 = 90$,which implies $a + b = 51\, cm$.From the Pythagorean theorem,$a^2 + b^2 = c^2 = 39^2 = 1521$.We know that $(a + b)^2 = a^2 + b^2 + 2ab$.Substituting the values: $51^2 = 1521 + 2ab$.$2601 = 1521 + 2ab \Rightarrow 2ab = 1080 \Rightarrow ab = 540$.Now,we have $a + b = 51$ and $ab = 540$. These are roots of the quadratic equation $t^2 - 51t + 540 = 0$.Solving for $t$: $t = \frac{51 \pm \sqrt{51^2 - 4(540)}}{2} = \frac{51 \pm \sqrt{2601 - 2160}}{2} = \frac{51 \pm \sqrt{441}}{2} = \frac{51 \pm 21}{2}$.$t_1 = \frac{72}{2} = 36$ and $t_2 = \frac{30}{2} = 15$.Thus,the other two sides are $36\, cm$ and $15\, cm$.

The perimeter of a right-angled triangle is $90\, cm$ and its hypotenuse is $39\, cm$. Find its other sides.

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