The area of triangle ABC is _____.
1 answer:
I FOUND YOUR COMPLETE QUESTION IN OTHER SOURCES.
SEE ATTACHED IMAGE.
Using Heron's Formula we can find the area of the triangle.
A = root (s (s-a) * (s-b) * (s-c))
s = (a + b + c) / 2
Where,
s: semi-perimeter
a, b, c: sides of the triangle
Substituting values:
s = (15 + 16 + 20) / 2
s = 25.5
s = 26
The area will be:
A = root (26 * (26-15) * (26-16) * (26-20))
A = 130.9961832
A = 130 u ^ 2
Answer:
The area of triangle ABC is:
C.
130 u ^ 2
SEE ATTACHED IMAGE.
Using Heron's Formula we can find the area of the triangle.
A = root (s (s-a) * (s-b) * (s-c))
s = (a + b + c) / 2
Where,
s: semi-perimeter
a, b, c: sides of the triangle
Substituting values:
s = (15 + 16 + 20) / 2
s = 25.5
s = 26
The area will be:
A = root (26 * (26-15) * (26-16) * (26-20))
A = 130.9961832
A = 130 u ^ 2
Answer:
The area of triangle ABC is:
C.
130 u ^ 2
You might be interested in
Answer:
P(X>4)= 0.624
Step-by-step explanation:
Given that
n = 10
p= 0.5 ,q= 1 - p = 0.5
Two fifth of 10 = 2/5 x 10 =4
It means that we have to find probability P(X>4).
P(X>4)= 1 -P(X=0)-P(X=1)-P(X=2)-P(X=3)-P(X=4)
We know that
P(X>4)= 1 -P(X=0)-P(X=1)-P(X=2)-P(X=3)-P(X=4)
P(X>4)= 1 -0.0009 - 0.0097 - 0.043 - 0.117-0.205
P(X>4)= 0.624