Question 1205315: With the lengths of 10 and 5 with a 55-degree angle between them; how many quadrilaterals can be made with such information?
Found 3 solutions by greenestamps, math_tutor2020, ikleyn:
Answer by greenestamps(13209)   (Show Source):
You can put this solution on YOUR website!
Knowing only the lengths of 2 of the 4 sides and the angle between those two sides, it seems obvious that an infinite number of quadrilaterals can be formed....
Perhaps you have not posted the problem correctly?
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
How many triangles can we form? Just 1. This is because of the SAS theorem.
How many quadrilaterals? Infinitely many. The first three points are nailed down because of the SAS theorem, but the fourth point can go wherever it wants. There needs to be more information to narrow things down.
Answer by ikleyn(52884)  (Show Source):
You can put this solution on YOUR website! .
With the lengths of 10 and 5 with a 55-degree angle between them;
how many quadrilaterals can be made with such information?
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We are given 4 (four) sticks. Two of them are 5 units long; two others are 10 units long.
How many different quadrilateral shapes can be made using these 4 sticks, if the angle
between each pair (5,10) at vertex is 55 degrees ?
Let "a" denotes a stick of 5 units long, and let "b" denotes a stick of 10 units long.
If the sticks are going in this order "abab", then the quiadrilateral is a PARALLELOGRAM,
since if in a quadrilateral opposite sides are equal, it implies that this quadrilateral
is NECESSARY a parallelogram.
But then there is an angle of 180-55 = 125 degrees between some two adjacent sticks "ab",
so this case is not what we are seeking for.
It implies, that the only allowed order of the sides, which may work, is "aabb".
Such shape does exist, and it is called a "kite".
But a convex kite shape does not work, since then the angle between sides "a" and "b" is greater than 55 degrees.
Hence, the only possible shape, which may work, is a non-convex kite shape "aabb", which has
angles of 55 degrees between the sides "a" and "b".
I will not go forward with more detailed explanations - I just said enough in order for
a normal person with geometric imagination could restore this figure in his/(her) mind.
Just imagine an obtuse triangle with sides 5 and 10 units long and with the angle 55 degrees between them.
Its third side "c" has the unique length by the cosine law.
Now imagine TWO such triangles, attached along this common side "c" in a way, that they form
a non-convex kite - and it will give you the required quadrilateral.
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| Its shape is an "arrowhead". |
+----------------------------------+
So, considered to congruency, only ONE such quadrilateral exists and is possible. ANSWER
Solved.
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Really nice problem. It is non-trivial, has unexpected solution and
trains a logic and a geometric imagination, both at the same time.
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