In the third year of junior high school, the top math students come to solve it, and if it is correc

The center of the circle is O, and the perpendicular lines from the center of the circle O to AB and CD are intersected respectively with P and Q, then OP is perpendicular to AB and OQ is perpendicular to CD, and AP=BP=6, CQ=DQ=9

Then the two triangles OPB and OQC are both right-angled triangles, OC and OB are radius, OC=ob=10

According to the Pythagorean theorem: op square = ob square - bp squared.

The solution yields op=8

Similarly: oq squared = oc squared - cq squared.

The solution is oq = root number 19

Connecting 0e, we can see that OEQ is a direct triangle, and EQ = OP = 8 Pythagorean theorem: OE square = OQ square + EQ square.

The solution is oe = root number 81

Knowing that the radius is 10 and ab is 12, the distance from the center of the circle to ab is 8, and the distance from the center of the circle to cd can be balled out as the root number 9, so the focus e and the center of the circle and the focus of ab cd form a rectangle, and the center of the circle to the focus e is diagonal, so the distance from the center of the circle to the focus e is the root number 83

The square of the chord centrometric = the square of the radius - the square of the half of the chord.

The distance from O to AB is 10 squares - (12 2) squares = 8o The distance from CD is 10 squares - (18 2) squares = root number 19, so the distance from the center of the circle to the intersection of the two strings E.

Under the root number (8 squares + 19 square roots).

Root number 83

Looking at the picture, it took a long time to draw the picture.

Ask a question, the hard work just answered, why did you close?

1) Since a is on the root number 3x of the straight line y = 3, and the abscissa of a is the root number 3, then substituting a (root number 3, 1) is obtained

Since a is on the hyperbola y = k of x, the expression of the hyperbola that is substituted for a = root number 3 is y = root number 3 x

2) The ordinate of c is 3, so substituting the hyperbola gives c (root number 3 3,3) so oa*oa = root number 3 * root number 3 + 1 * 1 to get oa = 2oc * oc = root number 3 3 * root number 3 3 + 3 * 3 * 3 to get oc = 2 root number 21 3

Solution: Through the AD x-axis, the intersection point is d, in OAB, ab=ao=5, ob=6, od=3, ad=4, and the coordinates of point A are (-3,4);

The point A is on the hyperbola, k=-12, the analytic formula of the hyperbola is: y=-12 x, the point c is on the hyperbola, the abscissa of point c is 6, y=-2, the coordinate of point d is (6, -2), and the analytic formula of the line ac is: y=-2 3 x+2

The vertex is (1,-4), and the parabola is y=(x-1) 2-4=x 2-2x-3 from y=x 2-2x-3=0 to get a(-1,0),b(3,0).

So ao=1, ab=4, there is ap*aq=ao*ab

Therefore, the triangle AOQ is similar to ABP, the angle apb=90 passes P as PM perpendicular AB to m, and let the P coordinate (X, X 2-2X-3) then m(x,0), ma=x-(-1)=x+1, MB=3-X, MP= |x^2-2x-3|

It can be proved that the right triangle apm is similar to PBM, AM MP=MP MB, that is, MP 2=AM*MB

So |x^2-2x-3|2=(x+1)(3-x),x=1 3,x 2-2x-3=2-3, so the p-coordinate is (1 3,2-3).

1, c=90°, a=60°, b=30°, ac=ab 2=5 2=, as oe bc, perpendicular foot e, oe=1= radius of o, at this time bo=2oe=2, o is tangent to bc;

So bo<2,, o intersect with bc;

bo=2, tangent;

bo>2, separated;

2, connect oe, ae bisect bac, bae= dae, oa=oe, bae= oee, oea= dae, oe ad, oed= ade=90°, oe ed, so the straight line ed is the tangent of o.

I'm a junior high school student, so it's possible to kill all my brain cells. 1、 (1 2 3 … n) -n-1)*250/7 = n(1 n)/2 - n-1)*250/7 (250/7) *2=

Is a quadrilateral cdefr a parallelogram? r in**?

If it is proved that the CFF is a parallelogram, according to the question D should be at the midpoint of BC.

CFF is a parallelogram.

then ef=dc, and ef dc

de=cf, and de cf

def=30°=∠dcf=1/2∠acb

ABC is an equilateral triangle.

then f is the ab midpoint.

AED is an equilateral triangle.

then de=ad=cf

then d is the midpoint of BC.