If line segment AB measures approximately 8.6 units and is considered

1.	If line segment AB measures approximately 8.6 units and is considered the base of parallelogram ABCD, what is the approximate corresponding height of the parallelogram? Round to the nearest tenth.
Answer» Answer: 4.8 units Step-by-step explanation: In the given figure side AB = 8.6 Coordinates of A(1, 3), B(8, 8), C(12, 5) and D(5, 0) Now in the given triangle we will find the ANGLE between Ab and AD to find the length of AE which is the length of the given parallelogram. $\rule {300}{1.5}$ tanθ = $\sf\dfrac{m(AB)-m(AD)}{1+m(AB)m(AD)}$ Slope m(AB) = $\sf\dfrac{8-3}{8-1}=\dfrac{5}{7}$ Slope m(AD) = $\sf\dfrac{3-0}{1-5}=-\dfrac{3}{4}$ So ; tanθ = $\sf\dfrac{\dfrac{5}{7}-(-\dfrac{3}{4})}{1+(\dfrac{5}{7})(-\dfrac{3}{4} )}$ tanθ = $\sf\dfrac{41}{13}=3.154$ θ = $\sf\:tan^{-1}(3.154)=72.41$ ° $\rule {300}{1.5}$ Now in right angle triangle ABE sin72.41° = $\sf\dfrac{DE}{5}$ 0.95324 = $\sf\dfrac{DE}{5}$ DE = 5×4.7662 = 4.8 units