The longest possible altitude of the third altitude (if it is a positive integer) is 83.
According to statement
Let h is the length of third altitude
Let a, b, and c be the sides corresponding to the altitudes of length 12, 14, and h.
From Area of triangle
A = 1/2*B*H
Substitute the values in it
A = 1/2*a*12
a = 2A / 12 -(1)
Then
A = 1/2*b*14
b = 2A / 14 -(2)
Then
A = 1/2*c*h
c = 2A / h -(3)
Now, we will use the triangle inequalities:
2A/12 < 2A/14 + 2A/h
Solve it and get
h<84
2A/14 < 2A/12 + 2A/h
Solve it and get
h > -84
2A/h < 2A/12 + 2A/14
Solve it and get
h > 6.46
From all the three inequalities we get:
6.46<h<84
So, the longest possible altitude of the third altitude (if it is a positive integer) is 83.
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Answer:
QSR = 68º
Step-by-step explanation:
TSQ + QSR = TSR
15x + (10x - 2) = 173
Combine like terms
25x = 175
Divide both sides by 25
x = 7
QSR = 10x - 2
QSR = 10(7) - 2
QSR = 68º
Shift the graph of the function y = x², 5 units down /look at the picture #1/.
/look at the picture #2/ - your answer
Answer:
the answer is letter D
Step-by-step explanation:
