Knowing that the circumference of the right triangle ABC is 2, find the maximum value of the area S

Let the two right-angled sides be a and b, and the hypotenuse is c, then.

a+b+c=2 (perimeter).

a + b = c (Pythagorean theorem).

a+b)=(2-c)

According to the mean inequality, get.

a+b)/2]²≤a²+b²)/2

i.e., (2-c)4c2

4+c²-4c≤2c²

c²+4c-4≥0

c -2 + 2 2 or c -2-2 2 (rounded) c is a positive number, c -2 + 2 2

a+b)²=(2-c)²

a²+b²+2ab=c²+4-4c

ab=2-2c

s=ab 2=1-c 1-(-2+2 2)=3-2 2 This is the maximum value that is sought.

Let the two right-angled sides be a and b, and the hypotenuse is c, then.

a+b+c=2

a²+b²=c²

a+b)=(2-c)

According to the mean inequality, get.

a+b)/2]²≤a²+b²)/2

i.e., (2-c)4c2

4+c²-4c≤2c²

c²+4c-4≥0

C-2+22 or C-2-222

c is a positive number, c -2 + 2 2

a+b)²=(2-c)²

a²+b²+2ab=c²+4-4c

ab=2-2c

s=ab 2=1-c 1-(-2+2 2)=3-2 2 This is the maximum value that is sought.

Let the three spike sides of the triangle be a, b, c, where c is the hypotenuse, and the circumference is l, so there is l a b c a b (a 2 b 2) because a b 2 (ab), guess the front (a 2 b 2) 2ab) so l 2 (ab) 2ab).

Substituting s s ab 2 can be obtained.

s≤l^2/[4(3＋2√2)]

That is, the maximum area of a right triangle with a circumference of a fixed value l is l 2 [4(3 2*2)].

When l 4 2 2.

S maximum (4 2 2) 2*l Spike resistance[4(3 2* 2)] 2 At this time, a b 2, the triangle abc is an isosceles right triangle generally, and there is the following general conclusion: "Of the right triangles with a certain circumference, the area is the largest when it is an isosceles right triangle".

Jiangsu Wu Yunchao answer for reference!

s=1 2*a*b*sinc<=1 2*((a+b) 2) 2*sinc=1 2*1 4*root3 2=root3 16

If and only when a=b=1 2 is the equal sign.

Therefore, the maximum raised hand of the triangle area s is (root number 3) 16, and the potato sail a=b=1 2

A 2=B 2+C 2-2BCCOSA 9=B 2+C 2-2BCCOS2 3 9=B 2+C 2+BC=(B+C) 2-BC (B+C) 2=BC+9 and BC (B+C) 2 4 Therefore: (B+C) 2 (B+C) 2 4+9 3 4(B+C) 2 9 (B+C) 2 12 B+C 2 Root No. 2 Therefore, the maximum circumference = 3+2 Root No. 2 Follow-up question: Do you copy the net? One of the numbers in this question has been changed, and the square of a is not nine!

Besides, we only know that the square of c is 4!Don't mess around looking for things I: Fart, the old lady sent the wrong question:

That is in the post once ,8,

1.Knowing that the circumference of the right triangle abc is a fixed value l, find the maximum value of the area of this triangle?

Let the three sides be a, b, c, and c as hypotenuses.

So there is a+b+c=a+b+under the root number (a side + b square) = l because a + b > = 2 (root number ab), under the root number (a side + b square) > = root number under (2ab).

So l> = (root number ab) + root number under (2ab).

Substituting s=ab 2 yields s=3+2 (ab)ab-2 (ab)-3>=0

ab)>=3 or √(ab)=9

a+b=ab-3

So the range of a+b is 6,4, and the side length is abc1,a+b+c=l

2,a*a+b*b=c*c

s=yes yes solve it yes it's very simple, 1,1 according to the conditions the right triangle isosceles has the largest area waist length l (2 + 2).

Then the area is l square (12+8 2).

2 It is impossible for a positive number ab to satisfy this condition, 1, 1Knowing that the circumference of the right triangle abc is a fixed value l, find the maximum value of the area of this triangle?

2.If the positive numbers a and b meet ab=a+b=3, then what is the range of a+b?

c= 3, therefore, to find the perimeter only requires a+b.

by the sinusoidal theorem; a c=sina sinc, b c=sinb sinc, substituting c= 3, c = 3, get: a=2sina, b=2sinbSo, a+b=2(sina + sinb).

c= 3, then a+b= -c=2 3

Then: b=2 3-a, a (0,2 3)then: a+b=2[sina+sin(2 3-a)]=2[sina+( 3 2)cosa+(1 2)sina]=2[(3 2)sina + ( 3 2)cosa] extracts 3=2 3[( 3 2)sina + (1 2)cosa]=2 3sin (a + 6).

Because a (0,2 3), then: a+ 6 ( 6,5 6), then: sin(a+ 6) (1 2,1].

Then a+b=2 3sin(a+ 6) (3,2 3], i.e., the maximum value of a+b is 2 3

Because c= 3

So, the maximum value of the perimeter is 3 3

Have fun! I hope it can help you, if you don't understand, please hi me, I wish you progress in your studies! o(∩_o

As d is a little fiber on ab, and ad:db=2:1, and at the same time as a'is the exonox of AB, and AA':

a'b = 2:1, a can be calculated'd=8 3, let the moving point c satisfy ac:cb=ad:

db, so cd is the inner angle bisector of the angular trap vertical qi c. In the same way, CA can be known'is the outer angle bisector of angle c. Since the sum of the inner and outer angles is 180°, the angle DCA'=90°, we can see that point C must be in the area of da'is the diameter of the circle.

Make a circle. When the area of triangle ABC is the largest, it must be at the highest point of C, and the height of point C relative to AB is Wang Ling A'd 2 = 4 3 with an area of 2 (4 3) 2 = 4 3

The area of the triangle is the base x height divided by 2

Therefore, the maximum value of the triangle area in the question should be the area at the farthest distance from c to ab, so when bc is perpendicular to ab, the area is the largest, so let bc=x, ac=2x, and the answer will be clear.