An algebraic equation enables the expression of equality between variable expressions
![\underline{The \ value \ of \ [AEF] \ is \ \dfrac{4}{9}}](https://tex.z-dn.net/?f=%5Cunderline%7BThe%20%5C%20value%20%5C%20of%20%5C%20%5BAEF%5D%20%5C%20is%20%5C%20%5Cdfrac%7B4%7D%7B9%7D%7D)
The reason the above value is correct is given as follows:
The given parameters are;
The symbol for the area of a triangle ΔXYZ = [XYZ]
The side length of the given square ABCD = 1
The location of point <em>E</em> = Side
on square ABCD
The location of point <em>F</em> = Side
on square ABCD
∠EAF = 45°
The area of ΔCEF, [CEF] = 1/9 (corrected by using a similar online question)
Required:
To find the value of [AEF]
Solution:
The area of a triangle = (1/2) × Base length × Height
Let <em>x</em> = EC, represent the base length of ΔCEF, and let <em>y</em> = CF represent the height of triangle ΔCEF
We get;
The area of a triangle ΔCEF, [CEF] = (1/2)·x·y = x·y/2
The area of ΔCEF, [CEF] = 1/9 (given)
∴ x·y/2 = 1/9
ΔABE:
= BC - EC = 1 - x
The area of ΔABE, [ABE] = (1/2)×AB ×BE
AB = 1 = The length of the side of the square
The area of ΔABE, [ABE] = (1/2)× 1 × (1 - x) = (1 - x)/2
ΔADF:
= CD - CF = 1 - y
The area of ΔADF, [ADF] = (1/2)×AD ×DF
AD = 1 = The length of the side of the square
The area of ΔADF, [ADF] = (1/2)× 1 × (1 - y) = (1 - y)/2
The area of ΔAEF, [AEF] = [ABCD] - [ADF] - [ABE] - [CEF]
[ABCD] = Area of the square = 1 × 1
![[AEF] = 1 - \dfrac{1 - x}{2} - \dfrac{1 - y}{2} - \dfrac{1}{19}= \dfrac{19 \cdot x + 19 \cdot y - 2}{38}](https://tex.z-dn.net/?f=%5BAEF%5D%20%3D%201%20-%20%5Cdfrac%7B1%20-%20x%7D%7B2%7D%20-%20%5Cdfrac%7B1%20-%20y%7D%7B2%7D%20-%20%5Cdfrac%7B1%7D%7B19%7D%3D%20%5Cdfrac%7B19%20%5Ccdot%20x%20%2B%2019%20%5Ccdot%20y%20-%202%7D%7B38%7D)
From
, we have;
, which gives;
![[AEF] = \dfrac{9 \cdot x + 9 \cdot y - 2}{18}](https://tex.z-dn.net/?f=%5BAEF%5D%20%3D%20%20%5Cdfrac%7B9%20%5Ccdot%20x%20%2B%209%20%5Ccdot%20y%20-%202%7D%7B18%7D)
Area of a triangle = (1/2) × The product of the length of two sides × sin(included angle between the sides)
∴ [AEF] = (1/2) ×
×
× sin(∠EAF)
= √((1 - x)² + 1),
= √((1 - y)² + 1)
[AEF] = (1/2) × √((1 - x)² + 1) × √((1 - y)² + 1) × sin(45°)
Which by using a graphing calculator, gives;
![\dfrac{1}{2} \times \sqrt{(1 - x)^2 + 1} \times \sqrt{(1 - y)^2 + 1} \times \dfrac{\sqrt{2} }{2} = \dfrac{9 \cdot x + 9 \cdot y - 2}{18}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7D%20%5Ctimes%20%5Csqrt%7B%281%20-%20x%29%5E2%20%2B%201%7D%20%5Ctimes%20%5Csqrt%7B%281%20-%20y%29%5E2%20%2B%201%7D%20%5Ctimes%20%5Cdfrac%7B%5Csqrt%7B2%7D%20%7D%7B2%7D%20%3D%20%20%5Cdfrac%7B9%20%5Ccdot%20x%20%2B%209%20%5Ccdot%20y%20-%202%7D%7B18%7D)
Squaring both sides and plugging in
, gives;
![\dfrac{(81 \cdot y^4-180 \cdot y^3 + 200 \cdot y^2 - 40\cdot y +4)\cdot y^2}{324\cdot y^4} = \dfrac{(81\cdot y^4-36\cdot y^3 + 40\cdot y^2 - 8\cdot y +4)\cdot y^2}{324\cdot y^2}](https://tex.z-dn.net/?f=%5Cdfrac%7B%2881%20%5Ccdot%20y%5E4-180%20%5Ccdot%20y%5E3%20%2B%20200%20%5Ccdot%20y%5E2%20-%2040%5Ccdot%20y%20%2B4%29%5Ccdot%20y%5E2%7D%7B324%5Ccdot%20y%5E4%7D%20%20%3D%20%5Cdfrac%7B%2881%5Ccdot%20y%5E4-36%5Ccdot%20y%5E3%20%2B%2040%5Ccdot%20y%5E2%20-%208%5Ccdot%20y%20%2B4%29%5Ccdot%20y%5E2%7D%7B324%5Ccdot%20y%5E2%7D)
Subtracting the right hand side from the equation from the left hand side gives;
![\dfrac{40\cdot y- 36\cdot y^2 + 8}{81\cdot y} = 0](https://tex.z-dn.net/?f=%5Cdfrac%7B40%5Ccdot%20y-%2036%5Ccdot%20y%5E2%20%2B%208%7D%7B81%5Ccdot%20y%7D%20%3D%200)
36·y² - 40·y + 8 = 0
![y = \dfrac{40 \pm \sqrt{(-40)^2-4 \times 36\times 8} }{2 \times 36} = \dfrac{5 \pm \sqrt{7} }{9}](https://tex.z-dn.net/?f=y%20%3D%20%5Cdfrac%7B40%20%5Cpm%20%5Csqrt%7B%28-40%29%5E2-4%20%5Ctimes%2036%5Ctimes%208%7D%20%7D%7B2%20%5Ctimes%2036%7D%20%3D%20%5Cdfrac%7B5%20%5Cpm%20%5Csqrt%7B7%7D%20%7D%7B9%7D)
![[AEF] = \dfrac{9 \cdot x + 9 \cdot y - 2}{18} = \dfrac{9 \cdot y^2-2 \cdot y + 2}{18 \cdot y}](https://tex.z-dn.net/?f=%5BAEF%5D%20%3D%20%20%5Cdfrac%7B9%20%5Ccdot%20x%20%2B%209%20%5Ccdot%20y%20-%202%7D%7B18%7D%20%3D%20%5Cdfrac%7B9%20%5Ccdot%20y%5E2-2%20%5Ccdot%20y%20%2B%202%7D%7B18%20%5Ccdot%20y%7D)
Plugging in
and rationalizing surds gives;
![[AEF] = \dfrac{9 \cdot \left(\dfrac{5 + \sqrt{7} }{9}\right) ^2-2 \cdot \left(\dfrac{5 + \sqrt{7} }{9}\right) + 2}{18 \cdot \left(\dfrac{5 + \sqrt{7} }{9}\right) } = \dfrac{\dfrac{40+8\cdot \sqrt{7} }{9} }{10+2\cdot \sqrt{7} } = \dfrac{32}{72} = \dfrac{4}{9}](https://tex.z-dn.net/?f=%5BAEF%5D%20%3D%20%20%5Cdfrac%7B9%20%5Ccdot%20%5Cleft%28%5Cdfrac%7B5%20%2B%20%5Csqrt%7B7%7D%20%7D%7B9%7D%5Cright%29%20%5E2-2%20%5Ccdot%20%5Cleft%28%5Cdfrac%7B5%20%2B%20%5Csqrt%7B7%7D%20%7D%7B9%7D%5Cright%29%20%20%2B%202%7D%7B18%20%5Ccdot%20%5Cleft%28%5Cdfrac%7B5%20%2B%20%5Csqrt%7B7%7D%20%7D%7B9%7D%5Cright%29%20%7D%20%3D%20%5Cdfrac%7B%5Cdfrac%7B40%2B8%5Ccdot%20%5Csqrt%7B7%7D%20%7D%7B9%7D%20%7D%7B10%2B2%5Ccdot%20%5Csqrt%7B7%7D%20%7D%20%3D%20%5Cdfrac%7B32%7D%7B72%7D%20%3D%20%5Cdfrac%7B4%7D%7B9%7D)
Therefore;
![\underline{[AEF]= \dfrac{4}{9}}](https://tex.z-dn.net/?f=%5Cunderline%7B%5BAEF%5D%3D%20%5Cdfrac%7B4%7D%7B9%7D%7D)
Learn more about the use of algebraic equations here:
brainly.com/question/13345893