Junior high school math, geometry problems about circles, hurry up, live etc.

16 questions, solution:

Connect OA, OB

acb=30°

AOB = 60° [The central angle of the circle opposite by the same arc is equal to twice the circumferential angle] OA=OB AOB is an equilateral triangle.

ab=oa=

Circle o diameter = 2 * oa =

5. Solution: ab is the diameter, c is the point on the circumference.

acb=90°

bac=30°

bc=1 2*ab=1 2*8=4cm [in a right-angled triangle, the right-angled side opposite by the 30° angle is equal to half of the hypotenuse].

ac= (ab 2-bc 2) = (8 2-4 2) = 48 = 4 3cm [Pythagorean theorem].

Question 5 In RT ABC bc bc = 4 (the right-angled edge of 30° is equal to half of the hypotenuse) ab = bc +ac (Pythagorean theorem).

Question 6 Connecting oa ob then the angle aob=60° (according to the circumferential angle, the opposite angle is equal to half of the central angle of the circle).

Because oa=ob so aob is an equilateral triangle so 0a=0b=ab= diameter d=2r= oh the whole process is written typing trouble.

16 Connect OA, OB.

oa cm.

The diameter is centimeters.

5 BC 4 cm, AC 4 3 cm.

In the first question, we can get that the radius of the sector is 3 times the radius of the circle.

The second question can be obtained as an angle of 45 degrees, which is obtained by comparing the arc length with the circumference.

The width is the root number 2, and the rest is counted by itself.

Question 3: The first question can get a pair of right angles and 2 sides corresponding to 1:2 according to the parallel and midpoint, and the second question can get a triangle, and the triangle AOD is all equal to the triangle ABC, this question should make full use of the right triangle similarity.

The center of the circle is o

If od,oe is connected, then od=oe, d= e;

and d and e are the midpoints of arc ab and arc ac respectively, so od ab; oe⊥ac.

dfb=∠egc.(The co-angles of equal angles are equal).

Therefore: afg= agf (equal to the vertex angle), af=ag

First of all, EF must be on both sides of the diameter, and it is impossible to do it on one side, right?

Because ECA= FCA, then ACB- ACB= ACB- ACF i.e. ECB= FCB, because ECB= FCB, OE=OF, OC=OC, so the triangle ECO and FCO are congruent and proved.

Answer: Known ECA= FCA, ACB- ACE=ACB- ACFECB= FCB

Then ecf is an isosceles triangle.

ec=fc

ab is the diameter of the bar:

Extend cp and intersect o at point f

ab cf arc af = arc ac

Arc AC = Arc CE

Arc AF = Arc CE

ACP= CAD (Equal Circumferential Angles Paired) AD=CD

The ab that seems to be missing in the condition is the diameter.

pca＋∠bac＝90°＝∠abc＋∠bac∴∠pca＝∠abc

Arc AC = Arc CE

ABC CAE (Equal Circumferential Angles Paired) PCA CAE

ad＝cd

Extend the intersection of cp with circle o m because c is the midpoint of arc ae so arc ac = arc ce so angle eac = angle cea because cp is perpendicular ab so cm perpendicular ab according to the diameter perpendicular to the chord arc ac = arc am so angle acp = angle aec so angle cae = angle acp so ad=cd

On the first floor, when CQ op, Qo is hypotenuse, and QP is a right-angled edge, and there can be no Qo QP

In the second floor, when the point P coincides with the point B, the point Q also coincides with the point P, and the QP degenerates into a point, and the Qo is the radius, which cannot be equal.

There are contemptible people who have no talent, and think that there is only one. CQ op, angle OCP = 60 degrees.