1/2sqrt((x_1^2+y_1^2)(x_2^2+y_2^2))
The area of a triangle is equal to 1/2bh (one half base times height). Since this is a right triangle, the base and height are the two legs connected to the 90* angle. To find the values of these sides, we will use Pythagorean Theorem, root a squared plus b squared.
Short leg: <x(1),y(1)>
This leg can be seen as the hypotenuse of an invisible right triangle. The x value, x(1), is how far over the x value has gone from the origon at x=0. Imagine a leg alone the x-axis, going from (0,0) to (x(1),0). The y value of the point, y(1), works the same way. This leg will go from our previous mark at (x(1),0) to the point (x(1),y(1)). This shows that the short leg of the main triangle is the hypotenuse, with a height of y(1) and base of x(1). Pythagoreum Theorem shows that the length of this leg is equal to sqrt(x_1^2+y_1^2).
Long leg: <x(2), y(2)>
The same process works here, giving us sqrt(x_2^2+y_2^2).
Now for the area, we have the b and h values. Our equation reads 1/2sqrt(x_1^2+y_1^2)sqrt(x_2^2+y_2^2).
But we can simplify this (yay). The two square roots can be written together as sqrt((x_1^2+y_1^2)(x_2^2+y_2^2))
So the correct answer is 1/2sqrt((x_1^2+y_1^2)(x_2^2+y_2^2))
Answer:
72r2+96r
Step-by-step explanation:
im just smurt i think
I am pretty sure the answer is the second option
Mes ∠AKG = mes ∠AKD + mes ∠ DKG
133° = mes ∠ AKD + 87°
133° - 87° = mes ∠AKD
∠AKD = 46°
Two lines that meet to form congruent adjacent angles.
