I need help with this "the area of a riight angle triangle is 210 square meters if the hypotanure is 37 meters long find the lenght and altitude."
the problem i am having here is that i know xy=210 as its the area and (37)2=x2+y2 according to pythagorus theoram but i cant get the answer right.
You need to come up with the notation for the legs of triangle such that x represents one leg and y the other leg.
You need to remember that the area of a right angle triangle is half of product of legs of triangle such that:
`A = (xy)/2`
The problem provides the information that area of triangle is of `210 m^2` such that:
`210 = (xy)/2 =gt 420 = xy`
The problem provides the information that the hypotenuse of triangle is of 37 m, hence you may use Pythagorean theorem such that:
`x^2 + y^2 = 37^2`
You need to use the equation `420 = xy` to write x in terms of y such that:`x = 420/y.`
`(420/y)^2 + y^2 = 37^2`
You need to bring these terms to a common denominator such that:
`420^2 + y^4 - 37^2y^2 = 0`
You should come up with the notation `y^2 = z` such that:
`z^2 - 37^2z - 420^2 = 0`
`z_(1,2) = (37^2 +- sqrt(37^2 + 4*420^2))/2`
`z_(1,2) = (37^2 +- sqrt(1369 + 4*420^2))/2`
`z_(1,2) = (37^2 +- sqrt706969)/2`
`z_(1,2) = (37^2 +-840.814)/2`
`z_(1,2) = (1369 +-840.814)/2 =gt z_1 = 1104.907`
Since `z = y^` 2, the second negative value for z is not accepted.
`y = sqrt1104.907 =gt y = 33.24 m`
You need to keep only positive value for y.
`x = 420/33.24 =gt x = 12.63 m`
Hence, evaluating the lengths of legs of triangle yields x = 12.63m and y = 33.24 m.
This might help.
In the triangle if
x = lenght and y = altitude
then xy/2 = 210
because the area of a triangle is altitude times base divided by two.