Hi

one leg of a right triangle is 12 cm longer than the other

Let x and (x+12cm) represent the lengths of the legs of this right triangle

A = (1/2)b*h

110cm^2 = (1/2)x(x+12cm)

220 = x^2 + 12x

0 = x^2 + 12 - 220

factoring

(x+22)(x-10)= 0 Note:SUM of the inner product(22x) and the outer product(-10x) = 12x

(x+22) = 0

x = -22 extraneous solution

(x-10)= 0

x = 10   10cm and 22cm are the lengths of the two legs

Letting c represent the length of the longer lefg & applying the Pythagorean Theorem

10^2 + 22^2 = c^2

584 = c^2  (discarding negative solution for length)

24.17cm  = c 

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