Quadrilateral function questions!! Hurry!! There are plus!! 120

1) The area formula of the trapezoid is S ladder = (upper bottom + lower bottom) height 2. Top bottom = 2 (because the abscissa is 2).

oa=s ladder 2 h-cb=40 2 8—2=8 a(0,8) solution: let the analytical formula of the function image be y=kx+b

Replace x=8, y=0; x=2, y=8.

0 = 8k + b k = three-quarters of a minus.

8=2k+b b=6

The analytic formula of the function is y=—4 3x+6

2) Solution: Follow the route taken by Q and P: Q walked 12 and P walked 20, but both have the same speed.

q should be at the point c and then p should be at the place of b to a2 units, thus forming two congruent right triangles.

qb=2t bp=2t qo=po=8t∴spbqd=2t·8t=16t²

The second question doesn't know if it's right.,It might be wrong.,Try your best.,I'm sorry to pull.。。

The two bottom edges OA=10, CB=8, perpendicular to the waist of the bottom OC=2 3 , Tan OAB=2 3 10-8 = 3, OAB=60°

2) When the point A ** is on the AB, OAB=60°, PA=PA, A PA is an equilateral triangle, and QP QA, PQ=(10-X)Sin60°= 3 2 (10-X), A Q=AQ=1 2 AP=1 2 (10-X), Y=S AQP=1 2 A Q QP= 3 8 (10-X)2, when A coincides with B, AP=AB= 3 Sin60°=4, so 6 x 10 at this time;

When the extension line of the point A**segment AB, and the point Q** segment AB (does not coincide with B), the figure of the overlapping part of the paper piece is a quadrilateral (as shown in the figure, where E is the intersection point of PA and CB), when the point Q coincides with B, AP=2AB=8, the coordinate of the point P is (2,0), and when A and B coincide, the coordinate of P is (6,0), so when the figure of the overlapping part of the paper is a quadrilateral, 2 x 6;

3) Y exists at the maximum

When 6 x 10, y = 3 8 (10-x)2, to the left of the axis of symmetry x=10, the value of s decreases with the increase of x, and when x = 6, the value of y is up to 2 3;

When 2 x 6, as shown in the figure, the area of the overlapping part y=s aqp-s a eb, the height of a eb is a b sin60°, y= 3 8 (10-x)2-1 2 (10-x-4)2 3 2 = 3 8 (-x2+4x+28)=- 3 8 (x-2)2+4 3, when x=2, the maximum value of y is 4 3;

When 0 x 2, i.e., when both point A and point Q are the extension of the ** segment AB is (as shown in the figure, where E is the intersection of PA and CB, and F is the intersection of QP and CB), EFP= FPQ= EPF, the quadrilateral EPAB is isosceles, EF=EP=AB=4, Y=1 2 EF OC=1 2 4 2 3=4 3

To sum up, the maximum value of s is 4 3, and the value of x is 0 x 2

First of all, it is easy to know that BCD and MCN are similar, so dn dc=BM BC, so DN=BM*dc BC, so S ADN=AD*DN*sin=AD*BM*dc BC*sin=ab*BM*sin=s ABM;

Well, that's it.

The height between AD and BC is H1, and the height between AB and CD is H2, because it is a parallelogram, so S trapezoidal AMCD AD H1 0 5;S trapezoidal abcn ab h2 0 5

The area of the two trapezoids is half of the area of the parallelogram, so it is equal and because the two trapezoids minus the quadrilateral AMCN is the area of the two triangles, the area of the two triangles is equal in size.

1. (1) Because of the parallelogram ABCD, AD=BC, because BE=DF, AD-DF=BC-BE, that is, AF=CE, and because AD BC is AF CE, the quadrilateral AECF is a parallelogram.

2) Because bac= bae+ eac=90° b+ acb= is 90°, and because of diamond aecf, ae=ec, so eac= acb, so bae= b, ae=be=ec, because bc=10, be=5

What is Question 2?

3. Half of the two diagonals form a triangle on the side, so the value range of the third side of the triangle is the solution method, so the answer to this question is and 2, because the two parallel lines are equal in the wrong angle and because the angle bisector can be replaced by the same amount to form an isosceles triangle.

or 16, because the parallelogram area is equal to the base product height, and the question does not state which side is higher, so there are two answers.

Because in the parallelogram ABCD.

AD is parallel and equal to BC

So AF parallel ec

Because be=bf

So af=ec

So the quadrilateral AECF is a parallelogram.

Won't have so many questions? You can't always think about the answer, practice on your own, it's beneficial, and you have to rely on yourself to learn.

Solution: By known conditions, in the quadrilateral AECF, eaf=60°, AE BC is E, and AF Cd is F

So ecf=360°- eaf- afc- aec=360°- 60°- 90°- 90°=120°

So in the parallelogram ABCD, there is adc= abc=180°- ecf=180°-120°=60°.

So in the right triangle abe, bae = 90°- abe = 90°-60°=30°

then ab=cd=2*be=2*1cm=2cm

From the Pythagorean theorem, the height of the parallelogram ABCD is AE=the root number of 3cm

In the same way, ad=bc=2*df=2*

So the circumference of the parallelogram ABCD = AB+BC+CD+DA=2+3+2+3=10cm

Area = BC * AE = 3 * root number 3 square centimeters.

As the title is known, it is known to be a parallelogram, so abe 60° is known ae bc in e so aeb 90°

So bae 30° is known to be 1cm. Combine the previous 2 known conditions to obtain ab2

In a right-angled triangle, if an acute angle is equal to 30°, then the right-angled side it is facing is equal to half of the hypotenuse. One of the Pythagorean theorems).

The same as above gives FAD 30° DF= AD 3 by combining the previous 2 known conditions

Because it is a parallelogram, ab=cd ad=bc circumference 2+3+2+3 10

It is known that ab=2 be=1 so ae root number 3 (root number will not be played.) In a right triangle, if an acute angle is equal to 30°, the sum of the squares of the two right sides of the right triangle is equal to the square of the hypotenuse. One of the Pythagorean theorem).

bc*ae yields area 3*root number 3

Because eaf=60°, the quadrangular AECF is 360°, so ECF=120° so EBA=60°, and then ab=2*be=2cm, and similarly adc=60° so ad=2*df=3cm, so the perimeter comes out.

Area: Just ask for the height of ae or af Oh children will calculate this by themselves, right?? The Pythagorean theorem problem has to be done by yourself, I just haven't seen this kind of problem for a long time The ...... I did when I was a child

Because in the parallelogram ABCD.

AD is parallel and equal to BC

So AF parallel ec

Because be=bf

So af=ec

So the quadrilateral AECF is a parallelogram.

I'm in my second year of junior high school, so we haven't taught the above one yet.

Right angle AED

ae=2dae=60°

ade=30°

According to the right triangle, the opposite side of 30° is half of the hypotenuse.

So 2ae=ad

So ad=4cm

Because the parallelograms are bisected diagonally with each other.

So bo+co=6cm

s△boc=10cm

Look at Coca-Cola, Coca-Cola.

Solution: Because it's a parallelogram:

The middle angle of the triangle ABE is BAE = 45 degrees, and the angle AEB = 90 degrees. So ab=ae sin45 degrees ab=cd

The angle of the triangle AFD is FD=45 degrees, and the angle AFD=90 degrees. So ad=af sin45 degrees ad=bc

Circumference = ab + cd + bc + ad = 2 (ae + af) sin45 degrees = 4

The triangle DFC is all equal to the triangle ABE because.

ae = cfdc = ab angle dcf = angle bea the wrong angle in the parallel line is equal.

So be=df