A dam is 20 meters long with trapezoidal cross section and 36 square meters in area. What is the volume of the dam?

A dam is 20 meters long with trapezoidal cross section and 36 square meters in area. What is the volume of the dam?

36 × 20 = 720 cubic meters. A: the volume of the dam is 720 cubic meters

How many cubic meters of soil is needed to build a 2km long dam with a trapezoidal cross section, 20m long at the top and 23m long at the bottom?

Volume = cross sectional area × length
therefore
(20 + 23) × 14 / 2 × 2000 = 602000 M3

As shown in the figure, the cross section of the reservoir dam is trapezoid, ∠ B = 30 ° and ∠ C = 45 °

The cross section of the reservoir dam is trapezoid, ∠ B = 30 degree, ∠ C = 45 degree, the crest ad = 6m, CD = 20m, then what are the length and cross section area of the bottom BC? Is that the problem?
The dam height is h,
h=10√2
BC=h√3+h+AD=6+10√6+10√2
=44.6m
Cross sectional area = (6 + 44.6) H / 2
=356.73m^3

As shown in the figure, the cross-sectional area of the reservoir dam is trapezoidal, ∠ B = 30 °∠ C = 50 ° and the crest ad = 6m and CD = 20m. Then, what are the length of BC at the bottom of the dam (the result retains one decimal place) and the cross-sectional area (the result retains an integer)?
Answer requirements: the more detailed and clear the proof process, the better
Looking forward to your reply

Dam height = H = CD * sincbc = CD * COSC + AB * CoSb + ad = CD * COSC + H / tanb + ad = 20 * cos50 ° + CD * sinc / tanb + 6 = 20 * cos50 ° + 20 * sin50 ° / tan30 ° + 6 ≈ 45.39 M. calculate the cross-sectional area s: S = 1 / 2 (AD + BC) * H = 1 / 2 (AD + 20 * cos50 ° + 20 * sin50 ° / tan30 ° + 6) * CD * sinc

As shown in the figure, the cross section of the reservoir dam is trapezoid, angle B = 30, angle c = 45, dam crest ad = 6m, dam height de = 10m, and the length and cross-sectional area of dam bottom BC are calculated
The root sign 3 is about 1.732

Khan, you can't even do that. What else can you learn
It's not good for you to tell you the answer directly. According to the answer, see if you can draw the auxiliary line and answer it,
3 + 6 + 10 area under BC = 10 times root

As shown in the figure, the cross section of a reservoir dam is an isosceles trapezoid with a crest width of 6 meters and a dam height of 10 meters. The slope of slope AB is 1:2 (AR: BR). Now, if the crest width and slope gradient are unchanged, how much earthwork is needed to reinforce a 50 meter long dam

∵ in RT △ ABR, the slope of slope AB is 1:2, ar = 10, ∵ br = 20m, ∵ trapezoid ABCD is isosceles trapezoid, then BC = 2Br + ad = 46m, ∵ s trapezoid ABCD = (AD + BC) × ar △ 2 = 260m2, ∵ the slope of AB has no change before and after the transformation, and eh ⊥ BC is made at point h through point E. according to the title meaning, eh = 10 + 2 = 12M, ∵ pH = 2eh = 24m, the bottom length of trapezoid after the transformation is 2 × 24 + 6 = 54m The area of trapezoid is (54 + 6) × 12 △ 2 = 360 square meters. After the transformation, the extra area is s trapezoid epcf-s trapezoid ABCD = 360-260 = 100 square meters, so the required earthwork is 100 × 50 = 5000 cubic meters. A: 5000 cubic meters

As shown in the figure, the cross section of the reservoir dam is trapezoidal, the dam crest is 5m, the dam height is 20m, the slope of slope AB is 1:2.5, and the slope of slope CD is 1:2=

Slope = (elevation difference / horizontal distance)
Therefore, the dam bottom ad = 5 + 20 / (1:2.5) + 20 / (1:2) = 95m

As shown in the figure, the cross-sectional area of the reservoir dam is trapezoidal, the width of the dam crest is 8m, the height of the dam is 30m, the gradient of the slope ad is I = 3:3, and the gradient of the slope CB is I ′ = 1:2. Calculate the slope angle α of the slope ad, the width ab of the dam and the length of the slope ad

In RT △ ade, ad = de2 + AE2 = 60m; in CB, I ′ = 1:2, ｜ BF = 60m. To sum up, ad = 60m; ab = AE + EF + BF = AE + DC + BF = (303 + 68m)

As shown in the figure, the cross section of the reservoir dam is trapezoidal,

Pass a and D as high AE and DF on bottom BC, and cross BC to e and F
Let DF be x meters, and in the triangular DFC, there are DF & # 178; + CF & # 178; = DC & # 178;,
That is, 2x & # 178; = 20 & # 178; X & # 178; = 400 △ 2x & # 178; = 200X = 10 √ 2
In the triangle Abe, because ∠ B = 30 ° AE = DF = 10 √ 2,
So AB = 2ae = 20 √ 2be & # 178; = AB & # 178; - AE & # 178; = (20 √ 2) & # 178; - (10 √ 2) & # 178; = 600be = 10 √ 6
Ψ BC = be + EF + FC = 10 √ 6 + 6 + 10 √ 2 ≈ 24.49 + 6 + 14.14 ≈ 44.6 (m)
Cross sectional area: (6 + 44.6) × 14.14 △ 2 ≈ 357.7 (M2)

The cross-section of the reservoir dam is trapezoidal, with 30 degrees for ∠ B and 45 degrees for ∠ C. the dam crest ad = 6m and CD = 20m, then the length of the dam crest and the area of the cross-section
The answer is indicated by a root

Dam bottom BC = CD × tan45 ° + AD + CD × tan45 ° / tan30 ° = 10sqrt [2] + 10sqrt [6] + 6 (m)
S trapezoid = (BC + AD) × CD × Tan 45 ° / 2 = 100 + 100sqrt [3] + 60sqrt [2] (M & # 178;)
Sqrt is the square root