Let, length and breadth of rectangle is L and B respectively.
It is given that :
The length rectangle is 4 cm more than 3 times the width of the rectangle.
L = 3B + 4 ......1 )
Also, area of square = area of the rectangle + 66
L² = LB + 66
Putting value of L from
L² = ( 3B + 4 )( B ) + 66
L² = 3B² + 4B + 66
( 3B + 4 )² = 3B² + 4B + 66
9B² + 16 + 24B = 3B² + 4B + 66
6B² + 20B - 50 = 0
3B² + 10B - 25 = 0
3B² + 15B - 5B -25 = 0
3B( B + 5 ) -5( B + 5 ) = 0
B = 5/3 units
L = 3( 5/3 ) + 4
L = 5 + 4 = 9 units
Hence, this is the required solution.
Answer: The distance between the two points is 13.
