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In trapezoid ABCD, AD / / BC, ad = 3, BC = 5, points E and F are on AB and DC respectively, EF / / BC, if AE: EB = 2:3, find EF I also calculated 3. But the answer is 2
EF=3.8
The area of trapezoid can be used to calculate:
If EF = A and the height of trapezoid is 5x, then:
(3+5)*5x/2=(3+a)*2x/2+(5+a)*3x/2
A = 3.8
That is the wrong answer, believe in yourself!
As shown in the figure, in the quadrilateral ABCD, the points E and F are on AB and DC respectively, and AD / / EF / / BC, AE: EB = 2:3, ad = 3, BC = 7, find the length of EF
Because AD / / EF / / BC, ad = 3, so in parallelogram abgd, ad = eh = BG = 3, and HF / / GC, so △ DFH ∽ DCG, AE: EB = 2:3 = DF: CF, so HF: CG = 2:5, BC = 7, BG = 3, so CG = 4, so FH = 1.6ef = eh + HF = 3 + 1.6 = 4.6
As shown in the figure, in ladder ABCD, ad is parallel to BC, AE = EB, EFP is parallel to DC, if EF = 1.2, find the length of DC
Extend Da, Fe and intersect with point M
Because AD / / BC, EF / / DC
So the quadrilateral CDPF is a parallelogram
So DC = PF
Because AE = EB, AP / / BF
So PE / EF = AE / EB = 1
PE = EF = 1.2
So DC = pf = PE + EF = 2.4
The trapezoid ABCD, e, f are on AB and DC respectively, AE: EB = 2:1, EF / / BC, ad = 5, EF = 7, find the length of BC (there should be a complete process)
Both AB and DC were crossed with G
Since ad is parallel to EF, GAD is similar to GEF
So GA: Ge = ad: EF
GA: (GA + 2) = 5:7
GA = 5
So BG = GA + AE + be = 8
Because ad is parallel to BC
So GAD is similar to GBC
There are GA: GB = ad: BC
That is, 5:8 = 5: BC
So BC = 8
It is known that D and E are points on AB and AC respectively, and BD = AE, EB intersects CD at O, EF ⊥ CO at F. it is proved that (1) be = CD; (2) OE = 2 of
1) ∵AE=BD,∠A=∠A,AB=BC
∴△ABE≌△BCD (S.A.S)
2) ∵ △ABE≌△BCD
∴∠BCD=∠ABE
∴∠EOF=∠EBC+∠BCD=∠EBC+∠ABE=∠ABC=60°
Namely: ∠ OEF = 30 °
/ / EO = 2of (in RT △ the right angle of 30 ° angle is half of the bevel side)
It is known that in the equilateral triangle ABC, D and E are points on AB and AC respectively, and BD = AE, EB and CD intersect at point O, EF ⊥ CD at point F. verification: OE = 2of
It is proved that ∵ △ ABC is an equilateral triangle,
∴∠A=∠ABC=60°，AB=BC，
In △ Abe and △ BCD,
A kind of
AB＝BC
∠A＝∠ABC
BD＝AE ，
∴△ABE≌△BCD，
∴∠1=∠2，
∵ ADO is the external angle of ∵ BCD,
∴∠ADO=∠ABC+∠2=60°+∠2，
∵ ADO is the external angle of ∵ BOD,
∴∠ADO=∠1+∠BOD，
∵∠1=∠2，
∴∠BOD=∠ABC=60°，
∴∠EOF=60°，
∵EF⊥CD，
∴∠OEF=90°-∠EOF=90°-60°=30°，
∴OE=2OF．
As shown in the figure, in the equilateral triangle ABC, De is AB.AC If BD = AE, be and CD intersect point D, EF ⊥ CD and point F, OE = 2of
prove:
∵BD=AE BC=AB ∠ABC=∠A
∴△ABE≌△BCD
∴∠DCB=∠EBA
∵ △ ABC is an equilateral triangle
∴∠OBC+∠OCB=60°
∴∠BOC=120°
∴∠EOF=60°
∵EF⊥CD
∴OE=2OF
Let ⊥ be a point of ⊥ B, where ⊥ A is a point of ⊥ B, where ⊥ B is a point (1) It is proved that: ∠ AEC = ∠ C; (2) Confirmation: BD = 2Ac
(1) It is proved that: ∵ ad ⊥ ab,
ν Δ abd is a right triangle,
And ∵ point E is the midpoint of BD,
∴AE=1
2BD，
And ∵ be = 1
2BD，
∴AE=BE，
∴∠B=∠BAE，
And ? AEC = ∠ B + ∠ BAE,
∴∠AEC=∠B+∠B=2∠B，
And ∵ C = 2 ∠ B,
∴∠AEC=∠C．
(2) Proof: ∵ AEC = ∵ C,
∴AE=AC，
And ∵ AE = 1
2BD，
∴BD=2AE，
∴BD=2AC．
Triangle AB = 8cm BC = 20cm BC the median line on the edge ad = 6cm How to find the area of triangle ADC
According to the Pythagorean theorem, it can be concluded that △ bad is a right triangle. S △ bad = 8 × 6 △ 2 = 24 cm? Is the height AE of △ ABC, which is perpendicular to point E. AE = 4.8 cm ed = 3.6 cm s △ AEC = 4.8 × 13.6 △ 2 = 32.64 cm? S △ AED = 3.6 × 4.8 △ 2 = 8.64 cm? S △ ADC = s △ aec-s △
Given that ab = 12cm, BC = 30cm in the triangle ABC, and ad = 9cm of the midline on the BC side, calculate the area of the triangle ADC
Using Helen's formula:
S = √ (P (P-A) (P-B) (P-C)), where p = 1 / 2 (a + B + C)
Area of triangle ADC = area of triangle ADB
In the triangle ADB, the three sides are 9,12,15 respectively, so p = 18
The area of triangle ADB = √ (P (P-A) (P-B) (P-C)) = √ (18 (18-9) * (18-12) * (18-15)) = 54
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