Refering to Suiyuan’s Question on 4 squares and a rectangle within them.
Really have to applaud those members for coming up with creative solution for this question and it is correct in a way because it will not put to test the relative of the sides of smaller square to the biggier square.
But it is not possible for the area of 4 squares to be 80 cm2 if the perimeter of the rectangle is 20 cm.
If someone happen to use strategy of Guess and Check or quadratic equation, he would not be able to find any solution.
Solution 1: (Guess and Check)
1 length and 1 breadth = 20 ÷ 2 = 10
L + B = 10 cm
The most squarish rectangle ABCD is 5cm by 5 cm
Area of 4 such squares = 4 x 25 = 100 cm2 (minimum possible total area)
The thinnest rectangle ABCD is 9.9999… cm (round off to 10cm).. x 0.000…001(round off to zero)
= the area of 2 smaller squares is almost 0 cm2
The area of the 2 bigger squares
= 2 x 10 x10 = 200 cm2 (maximum possible total area)
Solution 2: (Using Quadratic Equation to find x)
Breadth = x and Length = (10 – x)
Quadratic Equation = 2xx + 2(10 –x)(10 – x) = 80 (No possible solution)
Sorry, can’t superscript. So x multiple by x = xx and so on