The circumference of the right triangle ABC is 4 2 2, find the maximum value of the area of this tri

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

The circumference is 4+2 2, right? Let the two right-angled edges be a and b (both a and b are greater than zero), then a+b+ (a 2+b 2)=4+2 2.

And because (a-b) 2 0, i.e., a 2 + b 2-2ab 0, a 2 + b 2 2 ab gives (a 2 b 2) 2ab) from the above equation a 2 + b 2 + 2ab 2 ab + 2ab, i.e. (a + b) 2 4ab, a b 2 (ab

So 4+2 2 2 (ab) 2ab).

And because of the area s=

Substituting the inequality gives us (4 2 2) 2 [4(3 2 2)] =2

Let the right-angled edges be a and b. Then: c= (a b).

Then: a b (a b) = 2.

then: 2=a b (a b) 2 (ab) 2ab).

Then: (2 2) (ab) 2.

AB (2 2) = 6 4 2.

Take the equal sign if and only if a=b, i.e. the maximum value of the triangle area is 3 2 2.

Basic definitions. A closed shape consisting of three line segments that are not on the same line is connected one after the other, called a triangle. The figure enclosed by three straight lines on the plane or three arcs on the sphere, and the figure enclosed by the three straight lines is called a plane triangle; The shape enclosed by three arcs is called a spherical triangle, also known as a trilateral.

The resulting closed geometry is called a triangle when three line segments are connected end to end. A triangle is the basic shape of a geometric pattern.

The answer is wrong.

The positive solution is as follows: let the right-angled edges be a, b, then: c= (a b), then: a b (a b) = 2, then: 2 = a b (a b) 2 (ab) 2ab), then:

2 2) (ab) 2, obtain: ab (2 2) = 6 4 2. Take the equal sign if and only if a=b, i.e. the maximum value of the triangle area is 3 2 2.

If the length of the two right-angled sides is x,y, then there is:

x+y+√(x^2+y^2)=2

Because x+y>=2 xy, x 2+y 2>=2xy, so:

2>=2√xy+2xy

i.e.: 0<= xy<=(5-1) 2

So (xy)max=(3- 5) 2

then the maximum value of the area = (1 2) xy = (3- 5) 4

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!

The case with the largest area is an isosceles right triangle.

In this way, the right-angled side length is (root number 2) 2, and the hypotenuse side length is 1

Area: 1 2 * (root number 2) 2 * (root number 2) 2 = 1 4

Let the edge of the right triangle be b c, and c is the hypotenuse, then a+b+c=2s=1 2ab 1 4(a 2+b 2).

s is the maximum, i.e., the equal sign holds, i.e., a=b

When a=b, c=2a

Bring in a+b+c=2, i.e., a+a+ 2a=2 to get a=2 (2+ 2) then a 2=2 (3+2 2).

Maximum area s = 1 (3+2 2).

Let the two right-angled edges be x and y respectively, then the hypotenuse is 4-x-y, which is 2+y 2=(4-x-y) 2 by the Pythagorean theorem, that is, x 2+y 2=16-8(x+y)+x 2+y 2+2xy, i.e., 2xy-8(x+y)+16=0

From x+y>=2 times the root number under xy, use x+y"2 times the root number under xy"Swap it out.

2xy-16 times xy+16>=0 (1) xy=t(t>0) under the root number

Then (1) the equivalent Zheng front number is 2t 2-16t+16>=0, and the equivalent is first called t 2-8t+8>=0

out t> = 4 + 2 times the root number 2 (more than 4 rounded) or 0

Let the two right-angled sides be a, b, and the hypotenuse is long as c, then c 2 = a 2 +b 2, and the state of the state a+b+ a 2 +b 2 = 2 +1, 2 +1=a+b+ a 2 +b 2 2 ab + 2ab =(2+ 2 ) ab, that is, ab 2 2, if and only if a=b take the equal sign

The area of the triangle is s= 1 2 ab 1 2 1 2 = 1 4, i.e., s max = 1 4

So the answer is: 1 4

Solution: Let the three sides of the triangle be a, b, c, where c is the hypotenuse and the perimeter is l, so there is l a b c a b (a 2 b 2) because a b 2 (ab), 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 [4(3 2* 2)] 2 At this time, a b 2, the triangle abc is an isosceles right triangle in general, and there is the following general conclusion: "Of the right triangles with a certain circumference, the area is largest when it is an isosceles right triangle".

Jiangsu Wu Yunchao answer for reference!

It is known that the circumference of the right triangle ABC is 4+2 root number 2, and the maximum value of the area of this triangle is found a+b+ (a 2+b 2)=4+2 2

Because a+b 2 (ab), a 2+b 2) 2ab) 4+2 2 (2+ 2) (ab).

(ab) 2

That is, the maximum value of the area of the triangle of ab 4 = 2