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:
The answer is 1/8
Step-by-step explanation:
So I found a half of a half and divided it by 6.
Y = kx
k = y/x
k = 2/12
k = 1/6
so when x = 12
y = kx
y = 1/6(5)
y = 5/6
answer
when x = 5 y = 5/6
Answer: g^2+2g+1
Step-by-step explanation: Since she wants each side to be 1 inch larger, each side has a length of g+1. You need to find the area by squaring (g+1).
Write this as (g+1)(g+1). Use the FOIL method to get g^2+g+g+1. Simplify this to g^2+2g+1.
Answer:
D. -11/12
Step-by-step explanation:
