Isosceles trapezoid ABCD, ab = CD = 5, ad = 4, BC = 10, e, f on AB, BC, EF bisects the circumference of trapezoid, BF = x, and uses X to express the area of △ bef AB DC is waist

Isosceles trapezoid ABCD, ab = CD = 5, ad = 4, BC = 10, e, f on AB, BC, EF bisects the circumference of trapezoid, BF = x, and uses X to express the area of △ bef AB DC is waist

When crossing point a, make am vertical BC, crossing EF to P, crossing point D, make DN vertical BC, crossing EF to Q isosceles trapezoid, the perimeter is 5 + 5 + 4 + 10 = 24, because EF bisects the perimeter of trapezoid, so EB + BC + CF = 12, because BC = 10, EB = CF, so EB = CF = (12-10) / 2 = 1, because ad = 4, BC = 10, BM = NC, so BM = (BC-AD) / 2 = 3

In the quadrilateral ABCD, ab = AC = ad, and the angle BAC = 50 °, what is the angle BDC

Solution
Draw a circle with a as the center and ab as the radius
Then B, C and D are all on the circle
The angle of arc BC is the center of the circle and the angle of arc BC is the circle
Therefore, BDC = 1 / 2 and BAC = 25 degrees

In the isosceles trapezoid ABCD, ad is parallel to BC, ab = CD, P is a point on BC, PE is parallel to AB, AC intersects with E, PF is parallel to CD, BD intersects with F

I don't know if you've learned congruent triangles. Let's assume you've learned that f is the parallel line of PC intersecting DC with G. because pf / / DC, G is on DC, so PF / / GC and FG / / PC. therefore, the quadrilateral fgcp is a parallelogram (the parallelogram with two opposite sides parallel to each other is a parallelogram), so FG = PC (1), FP = GC (2)

In the quadrilateral ABCD, ab = AC = ad, ∠ DAC = 2 ∠ BAC

Let ∠ BAC = t. ∠ CBD = y. ∠ CDB = X
∠ABD=∠ABC-∠CBD＝[（180°－T）/2]-Y＝（180°-3T）/2
＝[（180°-2T）/2]-X.
The two formulas eliminate t and get y = 2x

As shown in the figure, P is any point on the side BC of square ABCD, and PE is perpendicular to BD and PE is perpendicular to AC and F. if AC = 10, find EP + FP
KUAI

A quadrilateral ABCD is a square
Ψ ob = OC = 1 / 2Ac = 5 (o is the intersection of AC and BD)
Connect op
Then s △ OBC = s △ OBP + s △ OCP
∴1/2*OC*OB=1/2*OB*PE+1/2OC*PF
∵OB=OC
∴OB=PE+PF
∴PE+PF=OB=5

In quadrilateral ABCD, ab = 30, ad = 48, BC = 14, CD = 40 and ∠ abd + ∠ BDC = 90 degree ∠ ADB + ∠ DBC = 90 degree, the area of quadrilateral ABCD can be calculated

Transposition thinking: AB and AD have two right triangles with 30 and 40 right sides
Similarly, there are two right triangles with right sides 48 and 14
According to Pythagorean theorem, the quadrilateral area is 30 * 40 / 2 + 48 * 14 / 2 = 936

As shown in the figure, in the quadrilateral ABCD, ab = 2cm, BC = 5cm, CD = 5cm, ad = 4cm, ∠ B = 90 ° calculate the area of quadrilateral ABCD

Connecting AC, in RT △ ABC, AC = AB2 + BC2 = 22 + 52 = 3cm, in △ ACD, ∵ ac2 + ad2 = 9 + 16 = 25cd2 = 25 ∵ ac2 + ad2 = CD2, so △ ACD right triangle s quadrilateral ABCD = s △ ABC + s △ ACD = 12ab · BC + 12ac · ad = 12 × 2 × 5 + 12 × 3 × 4 = (5 + 6) cm2

In abcd-efgh, ab = 3cm, BC = 9cm, BF = 4cm?

It's very simple. I don't know why I asked this question, but I'd like to answer it. It's 64cm in total
(3 + 9) * 4 + 4 * 4 = 64, or (3 + 9 + 4) * 4 = 64

In rectangular abcd-efgh, (1) what is the relationship between edge BF and plane aegc? Why? (2) what is the relationship between AC and plane efgh? Why?

The relationship between edge BF and plane aegc is line plane parallel, because the parallel push out lines and planes of AE in BF and plane aegc are parallel; the relationship between edge AC and plane efgh is also parallel, because AC is parallel to the parallel push out lines and planes of eg in plane efgh

As shown in the figure, given that e, F, G and H are the midpoint of each side of the quadrilateral ABCD, then the values of efgh: ABCD are___ ．

Because g and F are the midpoint of CD and BC, so GF = 12dB. Because △ CGF \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\cd = 1:2