Question 266098: in a triangle, the largest side has length 2cm and one of the other sides has length square root 2. given that the area of the triangle is 1cm^2, show that the triangle is right angled and isosceles. thank you so much if you could help!!
Answer by dabanfield(803)   (Show Source):
You can put this solution on YOUR website! in a triangle, the largest side has length 2cm and one of the other sides has length square root 2. given that the area of the triangle is 1cm^2, show that the triangle is right angled and isosceles. thank you so much if you could help!!
Let x be the length of the unknown side. Let this side x also represent the height of the triangle. Then we have:
area of the triangle = (1/2)*base*height = (1/2)*sqrt(2)*x
Since the area of the triangle is 1 we have then:
1 = (1/2)*sqrt(2)*x or
2 = sqrt(2)*x
x = 2/sqrt(2)
Multiply the numerator and denominator on the right by sqrt(2):
x = (2*sqrt(2))/(sqrt(2)*sqrt(2))
x = (2*sqrt(2))/2 = sqrt(2)
So the side x is equal to the other side making the triangle isosceles.
To prove the triangle is a right triangle we need to show, by the Pythagorean Theorem, that the square of the longest side is equal to the sum of the squares of the lengths of the other two sides. Since the longer side is 2 we need to show that:
2^2 = sqrt(2)^2 + sqrt(2)^2 or
4 = sqrt(2)^2 + sqrt(2)^2
Since the sqrt(2)^2 = 2 the right-hand side above is 2 + 2 = 4 so we have a right triangle.
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