Altitude of a triangle

[jdruopp]

Could someone get me started on how to figure out the altitude of triangle whose 2 sides are 13 and bottom is 10.

Thanks

 

[jwpaine]

You have an Isosceles triangle, with two congruent sides of length 13 and a base of 10

Draw a line from the vertex of the two congruent sides, and draw it down so that it is perpendicular with the base.

As you will see, you would than have a smaller triangle, who's side A = (1/2)base C = 13 and B = unknown.

Now apply the Pythagorean theorem (a^2 + b^2 = c^2) to find the missing side, B: which is that of the height of the triangle.


Hope this helps!
John.

 

[jdruopp]

so then i would get 13^2 + 5^2 = x^2

194 = x^2

sqrt194 =x

x=14 (rounded off)

is this right?

 

[morson]

No. You misunderstood, or misread, what the other poster wrote.

You were led to solving the equation x^2 + 5^2 = 13^2 for x > 0 for x.

 

[jwpaine]

so then i would get 13^2 + 5^2 = x^2

194 = x^2

sqrt194 =x

x=14 (rounded off)

is this right?

No. I set c = hypotenuse: so:

\(\displaystyle \L b^2 + 5^2 = 13^2\)
\(\displaystyle \L b^2 + 25 = 169\)
\(\displaystyle \L b^2 = 144\)

\(\displaystyle \L b = +\sqrt{144}\)

 

[jdruopp]

Thank you so much, I understand