In the diagram above, The side of square ABCD measures n cm. We know that n is a natural number and that the area of quadrilateral AECF is 66 cm². What is the perimeter of the square whose side measures 2n cm?

MathyGoo13 Mar 20, 2021

#1**+4 **

First, note the areas of triangles ADF and ABE are:

ADF = 6n/2 = 3n cm^{2}

ADE = 4n/2 = 2n cm^{2}

Also, the area of the square is n^{2}.

Since the area of the square is divided into regions ADE, ADF, and AECF(given as 66 cm^{2}), the area can be represented as the equation:

3n + 2n + 66 = n^{2}

-n^{2} + 5n + 66 = 0

-(n^{2} - 5n - 66) = 0

n^{2} - 5n - 66 = 0

Now, we can simply solve the quadratic.

n^{2} - 5n - 66 = 0

(n - 11)(n + 6) = 0

n - 11 = 0

n = 11

The other solution is negative, so it doesn't work.

Thus the perimeter of a square with side 2n is

4 * 2n = 4 * 2(11)

= 4 * 22

= **88 cm ^{2}**

Solved! :)

ArithmeticBrains1234 Mar 20, 2021

#1**+4 **

Best Answer

First, note the areas of triangles ADF and ABE are:

ADF = 6n/2 = 3n cm^{2}

ADE = 4n/2 = 2n cm^{2}

Also, the area of the square is n^{2}.

Since the area of the square is divided into regions ADE, ADF, and AECF(given as 66 cm^{2}), the area can be represented as the equation:

3n + 2n + 66 = n^{2}

-n^{2} + 5n + 66 = 0

-(n^{2} - 5n - 66) = 0

n^{2} - 5n - 66 = 0

Now, we can simply solve the quadratic.

n^{2} - 5n - 66 = 0

(n - 11)(n + 6) = 0

n - 11 = 0

n = 11

The other solution is negative, so it doesn't work.

Thus the perimeter of a square with side 2n is

4 * 2n = 4 * 2(11)

= 4 * 22

= **88 cm ^{2}**

Solved! :)

ArithmeticBrains1234 Mar 20, 2021

#2**0 **

Thanks! I got to the point of making the equation:

3n + 2n + 66 = n^2.

But, I do not hold the skills to solve quadratic yet.

MathyGoo13 Mar 21, 2021