# Find the area of a triangle with vertices (5,6), (10,6) and (5,14).

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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.**

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

Similarly,

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