8th Upper Mathematics Geometry, 8th Grade Geometry Mathematics

or 162. Let the two points a and b be symmetrical with respect to the straight line mn, then (mn) is bisected perpendicularly (ab).

After A, AE is perpendicular DC to point E, and in the triangle ADE de=cos60*20=10

ae = sin60 * 20 = 10 times root 2

Do bf perpendicular to point f

Because the angle dae = 30, the angle eab = 90

So the triangle ade is similar to the triangle baf

So de af=ae bf

Because af=ae-ef=ae-bc=10 times the root 2-10, bf can be found

Then find the triangle area and the trapezoidal area, and add it up to OKL.

Do ab at the point of b

The perpendicular line of the intersection dc at the point e, and then the perpendicular line of the ad is crossed by the point e at the point f, then the angle bfc = 60 degrees, it is known that bc = 10 meters, and the angle c = 90 degrees, so be=20 root number 3

EF = AB = 20 meters, so the area of the quadrilateral meadow ABCD = triangle DEF + triangle BCE + rectangular ABEF = 200 times the root number 3 3+

50 times the root number 3 3 + 400 times the root number 3 3 = 650 times the root number 3 3

1 All 1, over AB, CD do the vertical line intersects in O, and the vertical foot is E, F;

Because ab cd

So oe=of

According to the right triangle hl

Because oe=of;

There is also a public side op

So the triangle ope and opf are congruent.

bpo=∠opd

2. Because the triangle ope and opf are congruent.

So PE=PF

And because ae=cf

So pe-ae = pf-cf

i.e. PA=PC

Remember to give me extra points......Thank you.

Solution 1) Connect.

bo=do；ao=co；ab=cd

abo≌△cdo；

ocd=∠oab；

and oa=oc, and the same as po;

pao≌△pco；

opc=∠opa；

The po divides the bpd equally

2) Ibid., Pao PCO;

pa=pc；

Connect OA, OB, OC, OD

Because ab=cd,oa=oc,od=ob, so triangle oab=triangle ocd, so angle oba=angle odc, angle boa=angle doc

Angle OAP = Angle OBA + Angle BOA Angle OCP = Angle COD + Angle ODC So Angle OAP = Angle OCP

Because oa=oc op=op, angular oap=angular ocp so triangle oad=triangle ocd so angular apo=angular cpo

So po bisects the angle bpd pa=pc

Let af=a, then fd=3a, bc=4a

In the right-angle ABF: BF= (AB2+AF2)= 16+A2) It is not difficult to prove that ABF and BCE are similar triangles, so there is:

bf bc = ab ce, i.e.:

16+a2)/4a=4/8

The solution is a=4 3

i.e. bc = 4a = 16 3

In the same way: be af = ce ab = 8 4, that is, be = 2af = 2a = 8 3 Therefore: perimeter = 8 + (16 3) + (8 3) area = 1 2 8 (8 3).