# A right triangle has a legs 12cm and 16cm. How far from the vertex must this triangle be?

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### 2 Answers

The request of the problem is vague, hence, if you need to evaluate the distance between hypotenuse and opposite vertex, you need to find the area of triangle using two formulas, such that:

`A = (l_1*l_2)/2`

`l_1,l_2` represent the lengths of the legs

`A = (12*16)/2 => A = 6*16 => A = 96`

`A = (AD*BC)/2`

BC represents hypotenuse of triangle

AD represents the height of triangle

You may evaluate hypotenuse of triangle using Pythagorean theorem, such that:

`BC^2 = l_1^2 + l_2^2`

`BC^2 = 16^2 + 12^2`

`BC^2 = 400 => BC = 20`

Replacing 20 for BC and 96 for A yields:

`96 = (20*AD)/2 => AD = 96/10 => AD = 9.6`

**Hence, evaluating the distance from vertex A to hypotenuse BC, yields **`AD = 9.6.`

The problem statement is not clear but if you want to know the distance of vertex from the hypotenuse then we may proceed as follows:

Length of hypotenuse = sqrt(12^2 + 16^2) = 20cm

If we draw a perpendicular from vertex on the hypotenuse then we get two similar right-angled triangle and the larger triangle has a hypotenuse equal to 16cm. The length of shorter arm of this triangle is the distance of the vertex from the hypotenuse of the original triangle and by ratio and proportion we get:

Length of shorter arm/16 = 12/20

Length of shorter arm = 16*12/20 = 48/5 = 9.6cm

**The distance of vertex from the hypotenuse is 9.6cm**