MIKE LIVES! 8V
Seems kind of obvious…1m, 1.2m, and 5/5/8…am I missing something/
from the facebook page:
"It's been so long I forgot how to update my website :P
Trivial puzzle for today folks, unless you are in grades 2-4, in which case I can provide hints."
Yes it seems, but the triangles are kind of tricky. They covers the same area.
Ok I'm a freaking idiot.
I've create an image with both triangles: Triangle area
If you take the longer side as the base and split the triangle through the corresponding height, you can see that both triangles are the union of two 3,4,5 triangles.
so is there a simple formula to find out whether a triangle of sides length a, b, c is larger or smaller than one of sides a, b, d ?
Don't you think that Herons formula is simple?
And in the case of a-a-b, an isosceles triangle, the area is 1/2*(b)*(sqrt(a^2-(.5b)^2) -- which sounded much simpler in my mind before I typed it.
The triangles one really had me thinking for some 15 seconds while trying to mentally plot the corresponding rectangle.
Yes, Heron's formula will give you the area of a triangle from the side lengths. A=Sqrt[s(s-a)(s-b)(s-c], where a,b&c are the side lengths and s is the semiperimeter or (a+b+c)/2. It was one I had memorized back in my high school math team days. You could almost guarantee that you'd need it somewhere in every tournament at the geometry level.
Mike is alive! +1 :D
A good way to think about the triangles question: If you take two 3-4-5 triangles and glue them together along the length 4 sides you get a 5-5-6 triangle, and if you glue them together along the length 3 sides you get a 5-5-8 triangle.
s = (a+b+c)/2
area = (s-a)*(s-b)*(s-c)
I love these games
I see what you did there.
For triangles and rectangles a good rule of thumb is: skinniness implies less volume
All appear to be the same size.
All three white boxes, that is. ;-)
Except you've got to subtract all that black text.
So this got me wandering what size base for a general isosceles triangle will maximize the area. So, with just a little algebra and basic calculus techniques, I found that for an isosceles triangle whose sides are a, a and b. The b that will maximize the area, for a fixed a, is b = sqrt(2a^2). Also, The maximum area is (a^2)/2. So in the case given above, the base that will maximize area for a triangle whose other 2 sides are each 5 cm, will be sqrt(50) cm and the maximum area is 12.5 cm^2.
you could just make a one of the bases and by drawing a bit, you'd realise the altitude has to be less than or equal to a, and indeed is a when it's a right-angled triangle.
Mike, are you well? Have you been well?
Both circles have the same area, i.e., zero.
In case you don't believe me, just consider circle vs. disk.
Assume the equal side lengths of the isosceles triangle are given by a, and the base is 2x. Then area A = 2x√(a^2-^2). To maximize the area, we set the first derivative equal to zero and solve to find the local extrema. Here, A' = (2a^2-4x^2)/√(a^2-x^2). Setting equal to zero and solving for x gives x = a/√2, implying the optimal base for maximum area is 2x = (√2)a. Note that this gives a right angle at the triangle's apex, and can be thought of as one half along the diagonal of a square with side lengths a.
Here are the graphs of A and A'.
I should see the ophthalmologist. After some number of math articles, the title of this strip seemed to me as “Which has linear algebra”, which, of course, does make sense on this site.
+1 Mike Lives.
1/2 a^2 sin(t) is the area of isosceles triangle. So equal angles to left and right of PI/2 will have same area.
In this case, the if you drop perpendicular, you end up with two right triangles of lenghts 3,4,5 in both cases. With either 4,4 in base, or 3,3 in base.
So if you know a pythagorean triple a,b,c. You can make triangles: "a, a, 2b", "a, a, 2c", to have same area provided "a" is the hypotenuse of original right triangle a,b,c.