Finding the length of the base of the triangle...?

Question

The question is this:

The altitude to the base of an isosceles triangle measures 30 cm. If each of the equal sides of the triangle is 34 cm, find the length of the base of the triangle.

Any help would be appreciated, im having problems particularly with isosceles triangle. Thanks!

Answer 1

b = 2√(34^2 - 30^2) = 32 cm

Answer 2

draw it out and you'll see a rt angled triangle with a side (altitude) of 30 and a hypotenuse of 34.
Use pythagorean theoreom

30^2 + x^2 = 34^2
900+x^2=1156<< take 900 from both sides
x^2=256
x=16

But this was only half the base so the base = 32cm

Answer 3

Look at it as 2 right-angled triangles touching back to back. Then, 34 squared = (30 squared) + (X squared). Find out X, then double it for the base of the original triangle.

Answer 4

a squared + b squared = c squared.
isoscleles has two equal sides...

a (30) squared = 900
b (34) squared= 1156

1156-900= 256.
sqrt of 256 is 16

1/2 of the triangle is 16. so, double that to get the base of the original triangle, and you get 32cm

Answer 5

x^2+30^2=34^2
x^2=34^2-30^2
x^2=4*64
x=2*8=16
base=2*16=32

Answer 6

Use the Pythagorean theorem.