2007-05-25 03:37:01 · 12 answers · asked by A++ 2 in Science & Mathematics ➔ Mathematics
sounds like a good joke
try to draw it in different units :-)
you can't make a 10,6,3 triangle (area = 0)
2007-05-25 03:45:26 · answer #1 · answered by yerffej89 3 · 2⤊ 0⤋
We can solve this by graphical construction. On a graph paper, first draw a horizontal line 10 cms long, and then draw two arcs with radius of 6 cm and 3 cm respectively from either end of the 10 cm line. The point where the two arcs intersect is the apex of the triangle. However, you will observe that the two arcs don't intersect with each other. Hence there is no triangle. The dimensions are wrong.
If the arcs intersect, the point of intersection will be the apex and from there drop a perpendicular to the base and the area is given by half the base multiplied by the height (the perpendicular is the height).
2007-05-25 03:45:06 · answer #2 · answered by Swamy 7 · 2⤊ 0⤋
Use Pythagoras theorem to locate the top of the triangle [use 0.5 a triangle]: ?(8^2 - 4^2) = top of triangle = 4?three times this via the backside [4] to get sixteen?3 because of the fact the section for 0.5 of the triangle. Double to get 32?3 as your very final answer.
2016-12-18 04:06:55 · answer #3 · answered by ? 4 · 0⤊ 0⤋
Is your leg sore from the prof pulling it.
3 6 10 does not add up.
10 the base would have to have the other two sides add up to be more than 10. 3 plus 6 will not do that.
2007-05-25 04:31:11 · answer #4 · answered by Anonymous · 2⤊ 0⤋
Ummm... are you sure those are the correct lengths? Because you cannot have a triangle with one side being longer than the sum of the length of the two other sides.
2007-05-25 03:42:14 · answer #5 · answered by chewie007 3 · 6⤊ 0⤋
In the future when you have only the length of the three sides of a triangle given, use Herron's Formula to find the area.
A = √(S(S - a)(S - b)(S - c))
where S = (a + b+ c)/2
a, b, and c are lengths of the sides of the triangle
.
2007-05-25 03:53:30 · answer #6 · answered by Robert L 7 · 0⤊ 1⤋
Cannot be done, good try by previous poster but when you drop the perpindicular you are no longer dealing with a length of 10 down the bottom, right?
2007-05-25 03:52:11 · answer #7 · answered by waltonblue 3 · 2⤊ 0⤋
halfP = (side1 + side2 + side3) / 2
TriangleArea = Sqr(halfP * (halfP - side1) * (halfP - side2) * (halfP - side3))
That's the formula but your triangle is not real because in a triangle the lenght of a side can never be greather than the sum of the other two.
10>6+3, so no triangle
2007-05-25 03:48:12 · answer #8 · answered by gartfield72 2 · 2⤊ 0⤋
I agree with the first posting.
If the lengths had been accurate, I would have used Heron's Formula: Let the lengths be a, b, c. Let the semiperimeter be (a+b+c)/2. Then the area = SQRT(s*(s-a)*(s-b)*(s-c))
2007-05-25 03:45:42 · answer #9 · answered by fcas80 7 · 2⤊ 0⤋
For triangle's the area ,A= sqrt(s(s-a)(s-b)(s-c))
s=(a+b+c)/3
2007-05-25 03:50:56 · answer #10 · answered by Anonymous · 1⤊ 1⤋