Answer:
If P is a point on side CD of a square ABCD such that PC= 5 cm. If the area of triangle ADP is 42 square cm. Then the length of the side square is 12cm
x= the side of the square
A ADP = AD*(DP-5)
A ADP = x(x-5)
we know that the area = 42cm²
[x(x-5)]/2=42
x²-5x=42*2
x²-5x=84
x²-5x-84=0
b²-4ac= (-5)²-4(1*-84)=361
Δ=b²−4ac=361
Δ is strictly positive, the equation x²−5x−84=0 admits two solutions
(-b-√Δ)/2a =(5-19)/2=-7
(-b+√Δ)/2a =(5+19)/2=12
a length is never negative, x = 12cm
the side of square ABCD = 12cm