What is the length of the line segment with end points (5, -7) and (5, 11)
1 answer:
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Answer: The base is 17 meters long, and the height is 12 meters.
Step-by-step explanation:
The area of a triangle is:
A = b*h/2
where:
b = base
h = height.
For this particular triangle, we have that:
b = 5m + h
A = 102 m^2 = b*h/2
We can replace the first equation into the second one:
b*h/2 = 102 m^2
(5m + h)*h/2 = 102 m^2
(5m + h)*h = 2*102 m^2 = 204m^2
5m*h + h^2 = 204m^2
We can rewrite this as a quadratic equation:
h^2 + 5m*h - 204m^2 = 0
The solutions are given by the Bhaskara formula, the solutions are:
Then the two solutions are:
h = (-5m - 29m)/2 = -17 m
This is a negative height, that is not really defined, so this will be discarded.
The other solution is:
h = (-5m + 29m)/2 = 12m
This is positive, so this is the option we will use.
Knowing h, we can find the value of b.
b = 5m + h = 5m + 12m = 17m
Then the area is:
A = 12m*17m/2 = 102m^2
See the picture in the attached figure to better understand the problem
we know that
in the right triangle ABC
tan 80°=opposite side angle 80°/adjacent side angle 80°
tan 80°=BC/AC----->BC=AC*tan 80°-----> BC=75*tan 80°---> BC=425.35 m
the approximate height of the building=BC+<span>eye level
</span>the approximate height of the building=425.35+1.5----> 426.85 m
the answer is
the approximate height of the building is 426.85 m
we know that
in the right triangle ABC
tan 80°=opposite side angle 80°/adjacent side angle 80°
tan 80°=BC/AC----->BC=AC*tan 80°-----> BC=75*tan 80°---> BC=425.35 m
the approximate height of the building=BC+<span>eye level
</span>the approximate height of the building=425.35+1.5----> 426.85 m
the answer is
the approximate height of the building is 426.85 m