A circle of radius 5 O has a point P in it, and OP 4 passes through the point P with the shortest ch

In a circle, the diameter is the longest chord,,, so the longest is the diameter of the over-op.

The string perpendicular to the OP is the shortest, which can be proved simply:

Let any string cd that passes p, let ab be a string that crosses p and is perpendicular to op.

From the intersecting string theorem, cp*dp=ap*bp=fixed value.

By the mean inequality, 4*cp*dp<=(cp+dp) 2, i.e., cp+dp>=2 times the root term cp*dp, if and only if cp=dp, take the equal sign.

cp+dp is the length of the chord through the p point, so chord length" = 2 times the root number term cp*dp if and only if cp=dp to obtain the minimum value, at this time cd coincides with ab, so ab is the shortest chord.

If you haven't been to high school and haven't learned about mean inequalities, you can also observe the changes in the graph.

Half of the chord length (d 2) 2=(r 2) 2- (chord centrosis) 2The radius of the circle is a fixed value, so when the chord centroid is the largest, d has a minimum value.

It is evident that the chord centroid distance is greatest when the chord is perpendicular to the op; When the chord coincides with the op, the chord centroid is the smallest (0).

So, when the chord is perpendicular to the op, it is the shortest; The chord is longest when it coincides with the op.

A circle of radius 5 o has a little p inside it, and op=4

Then the shortest chord length of the point p is (

6), the longest chord length is (10

Because a straight line perpendicular to a known line segment is the shortest at a point. The longest chord in a recircle is the diameter.

The longest chord length when the chord is over the center of the circle = 2 5=10 The shortest chord length when the chord is perpendicular to the op = 2 (r 2-op 2) =2 (5 2-4 2) =6 Hello classmates, if the problem has been solved, remember Oh spring early congratulations Yours is my affirmation I wish you a galloping horse

Crossing the point P to make a string CD perpendicular to the op

CD is the shortest string.

Qifan oa=5, op=1

According to the Pythagorean theorem, cp=2 6

cd=4√6

That is, the string next to the shortest one is the first to take 4 6cm

If there is a point p in the circle o of radius 5 and op = 4, the shortest chord length of the point p is (6) and the longest chord length is (10).

The shortest string is 6 and the longest is 10 (the diameter of the circle).

The longest chord is the diameter, the shortest is twice as many as the root number under 5 2-3 2, the answer is 8

Connect the OP and make a straight line L perpendicular to the OP through the P point. The straight line L intersects the garden O at two points, A and B. AB is the shortest chord, the extension OP intersects with the garden O at two points C, D, and CD is the longest chord.

ab=6cm,cd=10cm

OP is the longest string with a diameter of 10, and the shortest string perpendicular to OP is 6

The shortest chord is 6 and the longest is 10

The shortest chord length that can be found over point p is 4 5s, and the longest chord through point p is equal to 12

So 4 strings with 5<= over p point <=12 , so there are 6 strings with an integer length.

Half of the shortest string = root number (5 2-3 2) = 4

Shortest chord = 4 * 2 = 8 cm.

The longest chord is 10 cm in diameter.

When the string is perpendicular to the op, the shortest string is 8cm, and the longest string passing through the center of the circle is the diameter, and the longest string is 10cm

The Pythagorean theorem gives 4, i.e., the shortest is 8;The longest is the diameter!^_