Question

solve for the area of the triangle of root 2 root 2 1 - root 7 by 2 1 + root 7 by 2 and minus root 2 minus root 2

Answer 1

sqrt(14)

Answer 2

To find the area of the triangle with the given vertices, we can use the determinant formula for the area of a triangle.

Let the vertices be $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$. Given vertices are: $A = (\sqrt{2}, \sqrt{2})$ $B = (1 - \frac{\sqrt{7}}{2}, 1 + \frac{\sqrt{7}}{2})$ $C = (-\sqrt{2}, -\sqrt{2})$

So, we have: $x_1 = \sqrt{2}$, $y_1 = \sqrt{2}$ $x_2 = 1 - \frac{\sqrt{7}}{2}$, $y_2 = 1 + \frac{\sqrt{7}}{2}$ $x_3 = -\sqrt{2}$, $y_3 = -\sqrt{2}$

The area of the triangle is given by the formula: $\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Let's calculate each term:

1. $y_2 - y_3 = (1 + \frac{\sqrt{7}}{2}) - (-\sqrt{2}) = 1 + \frac{\sqrt{7}}{2} + \sqrt{2}$ $x_1(y_2 - y_3) = \sqrt{2}(1 + \frac{\sqrt{7}}{2} + \sqrt{2}) = \sqrt{2} + \frac{\sqrt{14}}{2} + 2$

2. $y_3 - y_1 = -\sqrt{2} - \sqrt{2} = -2\sqrt{2}$ $x_2(y_3 - y_1) = (1 - \frac{\sqrt{7}}{2})(-2\sqrt{2}) = -2\sqrt{2} + \frac{2\sqrt{14}}{2} = -2\sqrt{2} + \sqrt{14}$

3. $y_1 - y_2 = \sqrt{2} - (1 + \frac{\sqrt{7}}{2}) = \sqrt{2} - 1 - \frac{\sqrt{7}}{2}$ $x_3(y_1 - y_2) = -\sqrt{2}(\sqrt{2} - 1 - \frac{\sqrt{7}}{2}) = -2 + \sqrt{2} + \frac{\sqrt{14}}{2}$

Now, sum these three terms: Sum $= (\sqrt{2} + \frac{\sqrt{14}}{2} + 2) + (-2\sqrt{2} + \sqrt{14}) + (-2 + \sqrt{2} + \frac{\sqrt{14}}{2})$ Group terms with common radicals and constants: Sum $= (\sqrt{2} - 2\sqrt{2} + \sqrt{2}) + (\frac{\sqrt{14}}{2} + \sqrt{14} + \frac{\sqrt{14}}{2}) + (2 - 2)$ Sum $= (0) + (\frac{\sqrt{14} + 2\sqrt{14} + \sqrt{14}}{2}) + (0)$ Sum $= \frac{4\sqrt{14}}{2} = 2\sqrt{14}$

Finally, calculate the area: $\text{Area} = \frac{1}{2} |2\sqrt{14}| = \sqrt{14}$ square units.

Alternatively, we can use the base and height method. Observe that points $A(\sqrt{2}, \sqrt{2})$ and $C(-\sqrt{2}, -\sqrt{2})$ lie on the line $y=x$. The length of the base $AC$ is: $AC = \sqrt{(\sqrt{2} - (-\sqrt{2}))^2 + (\sqrt{2} - (-\sqrt{2}))^2}$ $AC = \sqrt{(2\sqrt{2})^2 + (2\sqrt{2})^2} = \sqrt{8 + 8} = \sqrt{16} = 4$ units.

The height of the triangle is the perpendicular distance from point $B(1 - \frac{\sqrt{7}}{2}, 1 + \frac{\sqrt{7}}{2})$ to the line $y=x$ (or $x-y=0$). The distance from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is given by $h = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$. Here, $(x_0, y_0) = (1 - \frac{\sqrt{7}}{2}, 1 + \frac{\sqrt{7}}{2})$, and the line is $x - y = 0$ (so $A=1, B=-1, C=0$). $h = \frac{|(1 - \frac{\sqrt{7}}{2}) - (1 + \frac{\sqrt{7}}{2})|}{\sqrt{1^2 + (-1)^2}}$ $h = \frac{|1 - \frac{\sqrt{7}}{2} - 1 - \frac{\sqrt{7}}{2}|}{\sqrt{1 + 1}}$ $h = \frac{|-\sqrt{7}|}{\sqrt{2}} = \frac{\sqrt{7}}{\sqrt{2}}$ units.

The area of the triangle is $\frac{1}{2} \times \text{base} \times \text{height}$: $\text{Area} = \frac{1}{2} \times AC \times h = \frac{1}{2} \times 4 \times \frac{\sqrt{7}}{\sqrt{2}}$ $\text{Area} = 2 \times \frac{\sqrt{7}}{\sqrt{2}} = \frac{2\sqrt{7}\sqrt{2}}{2} = \sqrt{14}$ square units.

Both methods yield the same result.