Questions 5 and 11, Mathematics for the third year of junior high school

Inspect a semi-circular workpiece with a right-angle ruler, which one is qualified? Why?

Understand the concept of circumferential angle, grasp the inference of the circumferential angle theorem, and apply mathematical knowledge to practical life.

Take a look at the answers. Your second picture is correct.

It is known that A, B are two points on the circle O, the angle AOB = 120°, and C is the midpoint of the arc AB. Try to determine the shape of the quadrilateral OACB and explain why.

<> is the original question: the answer.

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The fifth question, the second one, is qualified, because the circumferential angle of the semicircle is a right angle.

Question 11, because aob=120° and c is the midpoint of the arc ab, aoc= cob=60°

and ao=oc=ob

Therefore, AOC and COB are equilateral triangles.

i.e. ao=ob=bc=ca

So it's diamond-shaped.

5.the 2nd; The chord of 90 degrees is the diameter of the circle.

11.Because c is the midpoint of the arc AB.

So the angle AOC = angle BOC = 60°

Because oa=bo=co=r

So ao=ac=oc=oc=bc

So the quadrilateral oacb is a diamond.

In the fifth question, the middle one is qualified, and the diameter of the semicircle is 90 degrees to the circumferential angle.

Analysis: According to the title: a and b are different signs, in which one of the chong must be negative and the other positive.

Because a b

The void reed takes a as a positive number and b as a negative number.

i.e. a 0, b 0

So the original formula = -a (root number -ab).

So choose A.

The difference between two adjacent odd squares = a multiple of 8.

That is: 15 squared - 13 squared = 8x7

2n+1) - (2n-1) squared = 8n

The formula for finding the root of a quadratic equation (-b b 2-4ac) 2a, the equation a=2, b=-5, c=1

Then (1) = x1x2-3(x1+x2)+9=[(5+ 17) 4][(5- 17) 4]-3[(5+ 17) 4+(5- 17) 4]+9=(25-17) 16-3*10 4+9=1 2-15 2+9=2

2) Original = (x1+1+x2+1) 2-2(x1+1)(x2+1)=(x1+x2+2) 2-2(x1x2+x1+x2+1)=(5 2+2) 2-2(1 2+5 2+1)=81 4-8=49 16

Take advantage of x|+x2=2a/b x|x2=a c finds x|+x2 and x1x2 will be the formula again, and the same kind of terms will be merged.

Connecting AC and EF is the mid-vertical line of AB and CD.

ad = ac

ABCD is square.

ad = cd

ACD is an equilateral triangle.

adc =60°，∠ada=30

DG deuces the ADA

adg=15°