There are a few ways to solve for x in this problem. I will use a way that involves the formula for the area of a triangle, `A = 1/2 bh` .

Here b is one of the sides of the triangle ("b" stands for base) and h is the...

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There are a few ways to solve for x in this problem. I will use a way that involves the formula for the area of a triangle, `A = 1/2 bh` .

Here b is one of the sides of the triangle ("b" stands for base) and h is the height, or segment perpendicular to this side, dropped from the opposite vertex to this side.

Since the triangle in the image is right triangle, the sides with lengths 5 and 12 are perpendicular to each other, so one can be considered a height and another one a base. So the area of this triangle is

`A = 1/2 * 5 * 12 = 30` .

On the other hand, the segment of length x is pependicular to the side of length 13, so segment of length x is the height and the side of length 13 is the base. So the area of the same triangle is also

`A = 1/2 *x *13` .

Setting this area equal to 30, the number obtained above, we get

`1/2*x*13 = 30`

From here `x = (30*2)/13 = 60/13`

**So** `x = 60/13` .