abd is an isosceles triangle of base 30 cm .the altitude to the base is 20cm.the length of the other altitude (in cm)is???
I think triangles only have 1 altitude so I guess you mean what is the length of one of the sides.
a and c is the sides and c is the base and h is height.
then
a=12×√b2+(4×h2)
12×√302+(4×202)=25
Dragonlance .....we actually have two more congruent altitudes.......
See the following picture.......
Draw altitude AE to side BD......
Note that ΔCBD ≈ ΔEBA so
BD/DC = BA/AE
25/20 = 30/AE → AE = 30*20/25 = 24
And we would have another congruent altitude to this one drawn from B to side AD
I believe that my answer is correct....there are 2 different altitudes.....the given one of 20 and the other that I found of 24 [ The other altitude, as I pointed out, is also 24]
P.S. .... Dragonlance gave the correct side length, but this isn't an altitude
Sorry. I thought maybe it do have three because there are three angles in a triangle. But I look at about 20 questions on maths sites that have altitude or height as part of the problem and it only ever have one altitude part, so I didn’t know. But I do now. :)
A good thing to know is if a mod have a answer different from someone else then the mod is usually right. They always say so if they make a mistake and they not make very many anyway.