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3 years ago

The vertices of a triangle are A(-2, 0), B(0, 3), and C(2, 2). Draw the figure

Mathematics

1 answer:

Roman55 [17]3 years ago

8 0

Answer:

Step-by-step explanation:

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For a triangle $XYZ$, we use $[XYZ]$ to denote its area. Let $ABCD$ be a square with side length $1$. Points $E$ and $F$ lie on

nata0808 [166]

An algebraic equation enables the expression of equality between variable expressions

$\underline{The \ value \ of \ [AEF] \ is \ \dfrac{4}{9}}$

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 E = Side $\overline{BC}$ on square ABCD

The location of point F = Side $\overline{CD}$ 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 x = EC, represent the base length of ΔCEF, and let y = 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:

$\overline{BE}$ = 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:

$\overline{DF}$ = 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

$[AEF] = 1 - \dfrac{1 - x}{2} - \dfrac{1 - y}{2} - \dfrac{1}{19}= \dfrac{19 \cdot x + 19 \cdot y - 2}{38}$

From $\dfrac{x \cdot y}{2} = \dfrac{1}{9}$ , we have;

$x = \dfrac{2}{9 \cdot y}$ , which gives;

$[AEF] = \dfrac{9 \cdot x + 9 \cdot y - 2}{18}$

Area of a triangle = (1/2) × The product of the length of two sides × sin(included angle between the sides)

∴ [AEF] = (1/2) × $\overline{AE}$ × $\overline{FA}$ × sin(∠EAF)

$\overline{AE}$ = √((1 - x)² + 1), $\overline{FA}$ = √((1 - y)² + 1)

[AEF] = (1/2) × √((1 - x)² + 1) × √((1 - y)² + 1) × sin(45°)

Which by using a graphing calculator, gives;

$\dfrac{1}{2} \times \sqrt{(1 - x)^2 + 1} \times \sqrt{(1 - y)^2 + 1} \times \dfrac{\sqrt{2} }{2} = \dfrac{9 \cdot x + 9 \cdot y - 2}{18}$

Squaring both sides and plugging in $x = \dfrac{2}{9 \cdot y}$ , gives;

$\dfrac{(81 \cdot y^4-180 \cdot y^3 + 200 \cdot y^2 - 40\cdot y +4)\cdot y^2}{324\cdot y^4} = \dfrac{(81\cdot y^4-36\cdot y^3 + 40\cdot y^2 - 8\cdot y +4)\cdot y^2}{324\cdot y^2}$

Subtracting the right hand side from the equation from the left hand side gives;

$\dfrac{40\cdot y- 36\cdot y^2 + 8}{81\cdot y} = 0$

36·y² - 40·y + 8 = 0

$y = \dfrac{40 \pm \sqrt{(-40)^2-4 \times 36\times 8} }{2 \times 36} = \dfrac{5 \pm \sqrt{7} }{9}$

$[AEF] = \dfrac{9 \cdot x + 9 \cdot y - 2}{18} = \dfrac{9 \cdot y^2-2 \cdot y + 2}{18 \cdot y}$

Plugging in $y = \dfrac{5 + \sqrt{7} }{9}$ and rationalizing surds gives;

$[AEF] = \dfrac{9 \cdot \left(\dfrac{5 + \sqrt{7} }{9}\right) ^2-2 \cdot \left(\dfrac{5 + \sqrt{7} }{9}\right) + 2}{18 \cdot \left(\dfrac{5 + \sqrt{7} }{9}\right) } = \dfrac{\dfrac{40+8\cdot \sqrt{7} }{9} }{10+2\cdot \sqrt{7} } = \dfrac{32}{72} = \dfrac{4}{9}$