Hey MrCaira.
Hint: Use the results of similar triangles (i.e. triangles have same angles but different sides which means the ratio of corresponding sides is the same).
Hello there i am writing cause i was handed this special task by my teacher, and i do not quite know how to handle it..
I was basically handed a picture of a triangle, with a square inside. As shown on the picture below!
As you see the square is a 1 by 1 and the length of |BA| is 10
So here comes the question, how do i find the length of |BC| in this triangle ?
Thanks for trying to help me!
I have been taking a look at Cos, Sin and Tan and i even a/a1 = b/b2 = c/c2
Of course that is all invalid since we would need to know at least one more information.
I agree with you on that, the thing is i only know the 90° angles in the triangles beside the one side on each of the 3 triangles
So the a/d = b/e = c/f wont quite do cause then i would end up with something looking like this 1/d = b/1 and i cannot do anything with this equation
Remember that you have three triangles.
Also remember that some sides are 1+x=y for some sides (i.e. the sides that involve the square).
So for the bottom side, you have 1+x=CA (for example) and 1+y =BC (for the other side.
You also have pythagoras for all triangles which include the big one and the two smaller ones.
Try using the 1+x and 1+y terms and solve in terms of your three triangles.
AC = 1 + x, BC = 1 + y, AB = 10, BD = y, AE = x, AF = z and BF = 10 - z where F is the point on the hypotenuse of large side, E is on the bottom side and D is on the left side.
All the usual trig relationships follow along with the Pythagorean ones that include:
(1+x)^2 + (1+y)^2 = 10^2 (Big triangle)
x^2 + 1 = z^2 (Smallest triangle if you follow the diagram/picture)
1 + y^2 = (10-z)^2 (Other triangle)
From this you have three equations in three unknowns which means you should be able to find a solution (unique) that satisfies x,y,z (given that all are positive).
I think there should be one more input because just imagine that we have a fixed rod AB of 10 unit. We ca slide it at the right top vertex of the square and we would get many positions where in the square will be inside the right triangle with the hypotenuse of 10 units.
Right triangle ABC; a=BC, b=AC, c=10=ABCode:A F D C E B
D on AB, E on BC, F on AC such that CEDF is a 1 by 1 square
So BE = a-1 and AF = b-1
Let x = a-1 or b-1
Will lead to this equation:
x^4 + 2x^3 - 98x^2 + 2x + 1 = 0
4 solutions; 2 are positive: ~.11188 and ~8.93801
So a = 1.11188 and b = 9.93801
Your problem would be less ambiguous if stated this way:
find the longest leg of ABC, or find the area of ABC.
Equation looks not too complicated, BUT that's due to c=10 and square sides = 1;
for general case, let square sides = y; then not so simple(!):
x^4 + (2y)x^3 + (2y^2 - c^2)x^2 + (2y^3)x + y^4 = 0 where x = a-y or x = b-y
YES, it is EXACTLY (not just basically) the same problem.
What I said is that by rewording the original problem,
to as example "what is area of triangle ABC", then there
is no ambiguity since it does not matter which leg is the longest;
we are not asked to find the coordinates.