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:
Step-by-step explanation:
You do not need to where the formula comes from but, just for fun, here’s a hint
To add up the numbers 1 to 10
Write out the numbers
1 2 3 4 5 6 7 8 9 10
Write them backwards
10 9 8 7 6 5 4 3 2 1
Add up both lists
11 11 11 11 11 11 11 11 11 11
This is 10 × 11 = 110
But this is twice the sum as two lots were added together
So the sum of the numbers 1 to 10 is 110 ÷ 2 = 55
ArSeqSum Notes fig4, downloadable IGCSE & GCSE Maths revision notes
Since all the inside angles are 60, in order to add up to 180, the outside angles need to be 120.
If solving for x:
Because both equations are equal to Y, you can set both equations equal to each other:
5x+1=4x+8
Then it's a matter of simplification (subtract the 4x to the left side and subtract the 1 over to the right):
x=7
If solving for Y:
Plug x=7 into either equation and solve:
Y=4(7)+8
Y=36
