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As shown in the figure, in ▱ ABCD, the diagonal AC = 21cm, be ⊥ AC, the perpendicular foot is e, and be = 5cm, ad = 7cm, then the distance between AD and BC is______ cm．
Let the distance between AD and BC be xcm, ∵ be ⊥ AC, ∵ s △ ABC = 12 · AC · be = 12 × 21 × 5 = 1052cm2, ∵ s ▱ ABCD = 2S △ ABC = 105, ∵ ad · x = 105, ∵ x = 15, that is, the distance between AD and BC is 15cm
A parallelogram ABCD, AC, = 21, be perpendicular to AC, be = 5, ad = 7, find the distance of the line perpendicular to AD and BC
Suppose that the line perpendicular to AD and BC is CF, then CF * ad = AC * be, CF * ad and ac * be are 1 / 2 of the area of the calculated half parallelogram ABCD, I don't know if we can understand it, so CF = 21 * 5 / 7 = 15
In the parallelogram ABCD, M is the midpoint of BC, and am = 9, BD = 12, ad = 10, then what is the distance between AD and BC?
Make a vertical line AE from point a, intersect BC at e, am, BD at o
In the parallelogram ABCD,
∠OBM=∠ODA
∠OMB=∠OAD
∠AOD=∠BOM
Therefore, △ BOM is similar to △ DOA
BM/AD =BO/DO
BM/AD=MO/AO
M is the midpoint of BC, ad = 10, so BM = 10 / 2 = 5
5/10=BO/12-BO
5/10=OM/9-0M
The solution is: Bo = 4
MO=3
In △ BOM, 5 ^ 2 = 4 ^ 2 + 3 ^ 2
25 =16+9
25=25
So delta BOM is a right triangle
In right triangle BOM and right triangle AEM,
∠OBM+∠AMB=90°
∠MAE+∠AMB=90°
Therefore, OBM = MAE
∠AMB=∠AMB
So △ BOM is similar to △ AEM
AM/BM=AE/BO
AM=9,BM=5,BO=4
9/5=AE/4
AE=7.2
So the distance between AD and BC is 7.2
In the parallelogram ABCD, ∠ B = 60 ° and CD = 4cm, the distance between AD and BC is
A as AE ⊥ BC in e
∴∠AEB=90°
∵∠B=60°
CD=AB=4㎝
∴BE=½AB=2㎝
AE=√﹙4²-2²)=2√3㎝
That is, the distance between AD and BC is 2 √ 3 cm
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