+218 
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?
+91 First, note the areas of triangles ADF and ABE are:
ADF = 6n/2 = 3n cm2
ADE = 4n/2 = 2n cm2
Also, the area of the square is n2.
Since the area of the square is divided into regions ADE, ADF, and AECF(given as 66 cm2), the area can be represented as the equation:
3n + 2n + 66 = n2
-n2 + 5n + 66 = 0
-(n2 - 5n - 66) = 0
n2 - 5n - 66 = 0
Now, we can simply solve the quadratic.
n2 - 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 cm2
Solved! :)
+91 First, note the areas of triangles ADF and ABE are:
ADF = 6n/2 = 3n cm2
ADE = 4n/2 = 2n cm2
Also, the area of the square is n2.
Since the area of the square is divided into regions ADE, ADF, and AECF(given as 66 cm2), the area can be represented as the equation:
3n + 2n + 66 = n2
-n2 + 5n + 66 = 0
-(n2 - 5n - 66) = 0
n2 - 5n - 66 = 0
Now, we can simply solve the quadratic.
n2 - 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 cm2
Solved! :)
+218 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.
