Find the area of a triangle with vertices (5,6), (10,6) and (5,14).
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The correct answer is 20.
We can check this by a very simple method for this particular problem.
We notice that that of the 3 points, 2 have common y-coordinates ([5, 6) and [10, 6]) and two have common x-coordinates ([5, 6) and [5, 14]). this means that one of the side is parallel to y-axis and the other side is parallel to x-axis. Thus the triangle is a right angel triangle with the right angle formed by the two sides parallel to x- and y-axis. These two sides constitute the base and height of the triangle. The area of this triangle is equal to:
Area of triangle = 1/2*(Base*Height)
The length of any line parallel to x-axis is equal to difference in y-coordinates.
Therefore length of base = 10 - 5 = 5
The length of any line parallel to y-axis is equal to difference in x-coordinates.
Therefore length of height = 14 - 6 = 8
Using these values of base and height of the triangle, we calculate its area as:
Area = (1/2)*5*8 = 20
We are given the vertices of the triangle as (5,6), (10,6) and ( 5,14).
Now we first need to find the lengths of their sides.
The length of side between (5,6) and (10,6)
= sqrt[(10-5)^2 + (6-6)^2]
= sqrt(25) = 5
The length of side between (10,6) and ( 5,14)
= sqrt[(10-5)^2 + (14-6)^2]
= sqrt(25 + 64) = sqrt 89
The length of side between (5,6) and (5, 14)
= sqrt[(5-5)^2 + (14-6)^2]
= sqrt(64) = 8
We, we can see that the squares of 5 and 8 add up to give 89.
Therefore this is a right angled triangle with the base and height corresponding to 5 and 8.
Therefore the area of the triangle is (1/2)*5*8 = 20
The required area of the triangle is 20.
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