Answer:
for cd it's 12 I believe fs
Lateral area of triangular prism = (a+b+c)h where a, b and c are the sides of the triangular base, h is the height.
LA= (a+b+c)h=
a=4
b=6
c=?
c=9
h=8
c=√6²+4²=√80 or approximately 9
you should get An Area= (4+6+√80)8
LA= 80+8√80
If you can use a calculator, enter (4/1/4)/(4/1/2) and you will get 1/2
Answer:
8
Step-by-step explanation:
2^2+2(4-2)
4+8-4
8
Hope this helps :D
The area bounded by the 2 parabolas is A(θ) = 1/2∫(r₂²- r₁²).dθ between limits θ = a,b...
<span>the limits are solution to 3cosθ = 1+cosθ the points of intersection of curves. </span>
<span>2cosθ = 1 => θ = ±π/3 </span>
<span>A(θ) = 1/2∫(r₂²- r₁²).dθ = 1/2∫(3cosθ)² - (1+cosθ)².dθ </span>
<span>= 1/2∫(3cosθ)².dθ - 1/2∫(1+cosθ)².dθ </span>
<span>= 9/8[2θ + sin(2θ)] - 1/8[6θ + 8sinθ +sin(2θ)] .. </span>
<span>.............where I have used ∫(cosθ)².dθ=1/4[2θ + sin(2θ)] </span>
<span>= 3θ/2 +sin(2θ) - sin(θ) </span>
<span>Area = A(π/3) - A(-π/3) </span>
<span>= 3π/6 + sin(2π/3) -sin(π/3) - (-3π/6) - sin(-2π/3) + sin(-π/3) </span>
<span>= π.</span>
