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:
6th period.
Step-by-step explanation:
Answer:
12.4968
Step-by-step explanation:
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Answer:
10+3pi
Step-by-step explanation:
The perimeter of of the shaded region is
AC+CT+marcSBT+SA
*Finding AC
The diagonals of a rectangle are equal is measurement. Since RB is a radius of the circle, then RB is 6. Since AC and RB are both diagonals of the rectangle, then AC is also 6.
*Finding CT
CT=RT-RC where RC is the width of the rectangle
Also RT is a radius so we have that
CT=6-RC
*Finding marcSBT
The circumference of a whole circle is 2pi*r.
We have a quarter of this with r=6.
1/4*2pi(6)
1/4*12pi
3pi
*Finding SA
SA=RS-AR
RS is a radius of the circle and AR is the length of the rectangle.
So we have that this can be rewritten as
SA=6-AR
Let's put these parts together:
6+6-RC+3pi+6-AR
Simplifying:
18-RC-AR+3pi
18-(RC+AR)+3pi
18-8+3pi (Remember length plus width equal 8)
10+3pi
Answer: 80
Step-by-step explanation:
