The value of x for the area of rectangle ABCD is 59 cm more than the area of rectangle PQRS is 3
Find the diagram attached
The area of a rectangle is expressed as:
A = lw
For the rectangle ABCD
A1 = (x + 3)(2x + 5)
A1 = 2x²+11x+15
For the rectangle PQRS:
A2 = x(x+4)
A2 = x²+4x
If the area of rectangle ABCD is 59 cm more than the area of rectangle PQRS, then;
A1 = A2 + 59
2x²+11x+15 = x²+4x + 59
Collect the like terms
2x²- x² + 11x - 4x + 15 - 59 = 0
x² + 7x - 44 = 0
This gives the required proof
Factorize the result to get the value of"x"
x² + 11x - 4x- 44 = 0
x(x+11)-4(x+11) = 0
x - 4 = 0
x = 3
Hence the value of x for the area of rectangle ABCD is 59 cm more than the area of rectangle PQRS is 3
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