3)A triangle has all integer side lengths and two of those sides have lengths 9 and 16. Consider the altitudes to the three side
s. What is the largest possible value of the ratio of any two of those altitudes
2 answers:
Answer:
2 is the largest possible value.
Step-by-step explanation:
Given Data:
Length of one side = 9
Length of second side = 16
Let AB and BC sides of triangle be of length 9 and 16 respectively as shown in figure attached.
Now, let sides AD and CF be the respective altitudes.
Also, ∠ABC = ∅ (as shown in figure)
If AD and CF are the respective altitudes then,
we have
AD = 9Sin∅ ;
CF = 16Sin∅;
By dividing both sides, we get
AD/CF = 9/16
This equation shows that is independant of angle ∅.
Now, let ∠BAC = α
Now we have, BE = 9 sinα
and FC = AC sinα
By dividing both sides, we get
BE/FC=9/AC
Similarly we also have,
BE/AD = 16/AC
ABC is a triangle as long as length of AC is ithin range of 8 to 24 i.e, 8≤AC≤24 (because sum of any two sides of triangle should be greater than length of third side)
Using these values we get ranges of:
9/24 ≤ BF/FC ≤ 9/8 ; 2/3 ≤ BE/AD ≤ 2
So,
2 is the largest possible value of the ratio of any two of these altitudes.
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Answer:
Lenghts of the sides:
Lenghts of the diagonals: and
Step-by-step explanation:
Look at the rhombus ABCD shown attached, where AC and BD de diagonals of the rhombus.
The sides of a rhombus have equal lenght. Then, since the perimeter of this one is 104 centimeters, you can find the lenght of each side as following:
You know that the diagonals are in the ratio
Then, let the diagonal AC be:
This means that AE is:
And let the diagonal BD be:
So BE is:
Since the diagonals of a rhombus are perpendicular to each other, four right triangles are formed, so you can use the Pythagorean Theorem:
Where "a" is the hypotenuse and "b" and "c" are the legs.
In this case, you can choose the triangle ABE. Then:
Substituting values and solving for "x", you get:
Therefore, the lenghts of the diagonals are:
