A triangle with sides of 5, 5, and 6 satisfies your conditions. Answer: height of 4 and base of 6.
What about height 6 and base 4?
what were your considerations?
The question is taken as is from the Geometry text.
I'm having the same problem of possibility of choices here. If they can be narrowed....what ways. Or Am I missing something and the possibilities are slim here.....because the book gives 6 and 4 as as interchangeable base and height. Any thoughts would be helpful. Thank you.
for simplicity, let them chose a right triangle. so they know the height right away and don't have to solve for it. and also, we can use Pythagoras' theorem. hopefully they know that.
we know A = (1/2)bh. so we want 12 = (1/2)bh => 24 = bh
also, we want a + b + h = 16, where a is the hypotenuse of the triangle.
by Pythagoras, we can write a in terms of b and h. and from 24 = bh, we can write either b or h in terms of h or b respectively. thus you can solve simultaneous equations.
this is a very "sophisticated" approach for someone not good with algebra, but at least there is no guessing
i guess you can develop some other kind of trial and error approach using these formulas as your guide
My trouble with this question is that it doesn't specify the triangle.
originally I was thinking that since bh=24, we can have (8,3) (6,4) (12,2) (24,1) for bh or hb pairs. (ofcourse i'm ignoring any other possibillities of answers) Then try elimination. For example 24 can be the base b'c the perimeter is 16. ---this sort of reasoning.
What makes this problem relatively easy to work out is that the solution I found is an isosceles triangle, which makes it easy to calculate the altitude from the "different" side. The altitude intersects the base at its midpoint, so you have a 3-4-5 triangle and that gives you the height of 4. You can make problems like this very difficult or impossible to solve quite easily by choosing numbers that don't lend themselves to easy integral solutions. For instance, there is no triangle with a perimeter of 16 and an area of 13.