The area of a parallelogram is base times height.
We can find the height by deriving a right triangle from the 60 degree angle and the 6 cm hypotenuse
sin(60°) = opp/hyp = x/6
√3/2 = x/6 Multiply both sides by 6
3√3 = x
x = 3√3
The height is 3 √3
Area = base x height
Area = 3√3 * 8
Area = 24 √3 cm²
Answer is C) 24 √3 cm²
Answer
Find out the how high up the wall does the ladder reach .
To proof
let us assume that the height of the wall be x .
As given
A 25-foot long ladder is propped against a wall at an angle of 18° .
as shown in the diagram given below
By using the trignometric identity

now
Base = wall height = x
Hypotenuse = 25 foot
Put in the trignometric identity


x = 23.8 foot ( approx)
Therefore the height of the ladder be 23.8 foot ( approx) .
Answer:
x1 = 2, y1 = 0
x2 = 2, y2 = 6
Step-by-step explanation:
Each pair of coordinates is (x, y). The first pair is (x1, y1). The second pair is (x2, y2). This means you have ...
x1 = 2, y1 = 0
x2 = 2, y2 = 6
_____
<em>Additional comment</em>
The slope of the line through these points is ...
m = (y2 -y1)/(x2 -x1) = (6 -0)/(2 -2) = 6/0 = undefined
The line through these points is a vertical line with equation x = 2.
Answer:
-5.1, -5, -4.5, -4.45,-4 2/5
Triangle JKL has vertices J(2,5), K(1,1), and L(5,2). Triangle QNP has vertices Q(-4,4), N(-3,0), and P(-7,1). Is (triangle)JKL
Tems11 [23]
Answer:
Yes they are
Step-by-step explanation:
In the triangle JKL, the sides can be calculated as following:
=> JK = 
=> JL = 
=> KL = 
In the triangle QNP, the sides can be calculate as following:
=> QN = ![\sqrt{[-3-(-4)]^{2} + (0-4)^{2} } = \sqrt{1^{2}+(-4)^{2} } = \sqrt{1+16}=\sqrt{17}](https://tex.z-dn.net/?f=%5Csqrt%7B%5B-3-%28-4%29%5D%5E%7B2%7D%20%2B%20%280-4%29%5E%7B2%7D%20%20%7D%20%3D%20%5Csqrt%7B1%5E%7B2%7D%2B%28-4%29%5E%7B2%7D%20%20%7D%20%3D%20%5Csqrt%7B1%2B16%7D%3D%5Csqrt%7B17%7D)
=> QP = ![\sqrt{[-7-(-4)]^{2} + (1-4)^{2} } = \sqrt{(-3)^{2}+(-3)^{2} } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}](https://tex.z-dn.net/?f=%5Csqrt%7B%5B-7-%28-4%29%5D%5E%7B2%7D%20%2B%20%281-4%29%5E%7B2%7D%20%20%7D%20%3D%20%5Csqrt%7B%28-3%29%5E%7B2%7D%2B%28-3%29%5E%7B2%7D%20%20%7D%20%3D%20%5Csqrt%7B9%2B9%7D%3D%5Csqrt%7B18%7D%20%3D%203%5Csqrt%7B2%7D)
=> NP = ![\sqrt{[-7-(-3)]^{2} + (1-0)^{2} } = \sqrt{(-4)^{2}+1^{2} } = \sqrt{16+1}=\sqrt{17}](https://tex.z-dn.net/?f=%5Csqrt%7B%5B-7-%28-3%29%5D%5E%7B2%7D%20%2B%20%281-0%29%5E%7B2%7D%20%20%7D%20%3D%20%5Csqrt%7B%28-4%29%5E%7B2%7D%2B1%5E%7B2%7D%20%20%7D%20%3D%20%5Csqrt%7B16%2B1%7D%3D%5Csqrt%7B17%7D)
It can be seen that QPN and JKL have: JK = QN; JL = QP; KL = NP
=> They are congruent triangles