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post-1087

Hi KiasuChief,

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.

See Below:
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

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Yes, we will not be able to find the answer if we attempt to first find the values of x and y, the sides of the original rectangle, ie.
x + y = 10
x^2 + y^2 = 40
=> x^2 + (10 – x)^2 = 40
=> x^2 + (100 – 20x + x^2) = 40
=> 2x^2 – 20x + 60 = 0
=> x^2 – 10x + 30 = 0

We cannot solve for x because x is not an integer.

lizawa’s method solves this problem because x * y is an integer.

But lantian’s method is the best.

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post-1092

Hi,

I agree lantian’s method is best in this case. It illustrates the logic clearly.
(It does not have to check x^2 + y^2 = 40 cm^2 is correct or not.)

I’m only worry that should a student chooses to solve it with Guess and Check strategy, he would not be able to yield an answer because of the error in this question.

A question should yield the same answer no matter which method a student uses.

Below images illustrate why (x^2 + y^2) should not be less than 50 cm^2

The quadratic equation gives a constraint (10 cm) to the relationship of the x and y, therefore it finds x, a sq root of a negative integer (not a real number) if given the total area x^2 & y^2 is 40 cm^2. Hence an impossible case.

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post-1084

Hi, Hope this helps

Ref

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Let AD be x cm
Let AB be y cm

Perimeter of ABCD = 20cm
so, x+y = 20/2 = 10 cm

Area of big square = y2
Area of small square = x2

Area of 4 sqaures = 80 square cm
So area of 1 big + 1 small = 40 square cm

x2 + y2 = 40

x + y = 10
(x+y)(x+y) = 100
x2 + y2 + 2xy = 100
40 + 2xy = 100
2xy = 60
xy = 30

Area of shaded rectangle = xy = 30 square cm.

Ref

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*clap* *clap* That is the right answer. However, are children in P6 taught how to do an expansion of (x+y)^2 ? Without that step, it is impossible to solve the equation.

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