In the figure, ABCD is a trapezoid with legs AB and CD.
Join AC and BD.
For the triangle CAD, from the vertex C, draw an altitude CE and for the triangle ABD, from the vertex B, draw an altitude BF.
Clearly, CE = BF = h (say) --- (1)
Note that the base of the triangles CAD and ABD are the same and is AD.
It is given that ar(CAD) = 
.
Now, ar(ABD) = 
= 
from (1)
= ar(CAD)
= .
Hence, area of Δ ABD = ..
If this is asking for written word then
Twelve times f plus Twenty-four.
100-35 = 65
65% de 2300 = 1495 (mes 1)
2300-1495 = 805 (mes 2)
65% de 805 = 523,25 (mes 3)
65% de 523,25 = 340,1125 ~ 340,10 (mes 4)
9514 1404 393
Answer:
1. (256/3)π ft³
2. 33.49 ft³
Step-by-step explanation:
1. Put 4 ft into the formula where r is, then simplify.
V = (4/3)π(4 ft)³ = (256/3)π ft³
__
2. The radius is half the diameter, so is 2 ft. Put that in the formula, along with the value for π, and do the arithmetic.
V = (4/3)(3.14)(2 ft)³ ≈ 33.49 ft³
