![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
Enumerate some algorithms under radical
0
0
3 + 0 △ (root 2 - negative root 12) = 3
In the right angle △ ABC, the length of the hypotenuse is 2 and the circumference is 2+ 6, then the area of △ ABC is______ .
Let the hypotenuse of right angle △ ABC be C, and the two right angle sides be a and B,
According to the meaning of the title, a + B is obtained=
6，a2+b2=c2=4，
Then the area of △ ABC = 1
2ab=1
4[（a+b）2-（a2+b2）]=1
4（6-4）=1
2．
So the answer is 1
2．
Given that the circumference of RT △ ABC is 2 + root 2, find the maximum value of its area and the length of each side at this time
Let C be a right angle
Perimeter L = a + B + C = C * Sina + C * cosa + C = C (Sina + cosa + 1)
=c(2^(1/2)sin(A+(pi/4))+1)
c=L/(2^(1/2)sin(A+(pi/4))+1)
And: C ^ 2 = a ^ 2 + B ^ 2 = (a + b) ^ 2-2ab
2ab=(a+b)^2-c^2=(a+b+c)(a+b-c)=(a+b+c)((a+b+c)-2c)
=(a+b+c)^2-2(a+b+c)c
=L^2-2Lc
=L^2-2L^2/(2^(1/2)sin(A+(pi/4))+1)
Area = (1 / 2) AB = (1 / 4) L ^ 2 - (1 / 2) L ^ 2 / (2 ^ (1 / 2) sin (a + (PI / 4)) + 1)
When sin (a + (PI / 4)) = 1, that is, a = pi / 4, and △ ABC is an isosceles right triangle, the area is the largest
Maximum area = (1 / 4) L ^ 2 - (1 / 2) L ^ 2 / (2 ^ (1 / 2) + 1) = 1 / 2
Then: (1 / 2) AB = 1 / 2
ab=1
a=b=1
The perimeter of RT △ ABC is 4 + 2 * root 3, and the length of hypotenuse AB is 2 * radical 3. Where CE ⊥ AB, what is the area of ⊥ ABC?
AC + BC = 4; AC ^ 2 + BC ^ 2 = (2 √ 3) ^ 2 = 12
So (AC + BC) ^ 2 = 4 ^ 2, that is, AC ^ 2 + BC ^ 2-2 * ac * BC = 16, and AC ^ 2 + BC ^ 2 = 12
Ac * BC = 2
So the area: 1 / 2 * (AC * BC) = 1
In RT △ ABC, ∠ C = 90 ° and its circumference is 4 + 2 radical 3, and the median line CD on the hypotenuse AB is root 3, then what is s △ ABC?
The hypotenuse is twice the center line, so the hypotenuse is 2 √ 3
So two right angles AC + BC = 4
AC^2+BC^2=AB^2=12
So the area of the triangle is: 1 / 2Ac * BC = 1 / 2 * 1 / 2 [(AC + BC) ^ 2 - (AC ^ 2 + BC ^ 2)]
=1/4*4
=1
In the RT triangle ABC, the center line CD on the hypotenuse is the root 3, and the circumference is 4 + 2 times the root sign 3. Find: (1) the area of the triangle; (2) the height CE on the hypotenuse I'm not sure the problem is right Can you tell me how to calculate it
(1) Because in the RT triangle ABC, the center line CD on the hypotenuse is root 3
So AB = 2CD = 2 √ 3
Because + 3 is perimeter
So AC + BC = 4 + 2 √ 3-2 √ 3 = 4
So let AC be x, then BC is (4-x)
So AC ^ 2 + BC ^ 2 = AB ^ 2
So x ^ 2 + (4-x) ^ 2 = (2 √ 3) ^ 2
2x^2-8x=-4
x^2-4x=-2
x^2-4x+2^2=-2+2^2
(x-2)^2=2
x-2= √2
x=2+√2
So AC = 2 + √ 2, BC = 4 - (2 + √ 2) = 2 - √ 2
So the area is: (AC × BC) △ 2 = (2 + √ 2) (2 - √ 2) △ 2 = (2 ^ 2 - (√ 2) ^ 2) △ 2 = 2 △ 2 = 1
(2) (equal product method)
Because AB = 2 √ 3 and area = 1
So the area = (AB × CE) △ 2 = 2 √ 3ce △ 2 = √ 3ce = 1
So CE = 1 ÷ 3 = 3 / 3 √ 3
PS：
Given that the circumference of the right triangle ABC is 4 + 2 times the root sign 3, and the center line CD on the hypotenuse is 3, calculate the area of the right triangle
Oblique side length = 2CD = 2 * root 3
Let the length of two right angles be a, B
A + B = 4 + 2 * radical 3-2 * Radix 3 = 4 -------- (1)
A ^ 2 + B ^ 2 = (2 * radical 3) ^ 2 -------- (2)
Solve the two equations,
(1)^2-(2)=2ab=4
ab=2
Area = 1 / 2 * AB = 1
Given that the circumference of the right triangle ABC is 4 + 2, the root sign 2, find the maximum area of the triangle
Let the right angles of RT ⊿ ABC be a, B (a, B ∈ R ＋), then the hypotenuse is √ a ⊿ B 2
From the meaning of the title: a + B + √ a 2 + B 2 = 4 + 2 √ 2
If and only if a = B, the above equation takes the equal sign
4+2√2≥（2+√2）√ab
√ab≤2
ab≤4
ab／2≤2
To sum up, the maximum area of a right triangle is 2, and a = b = 2
As shown in the figure, if the circumference of right triangle ABC is 4 + 2 root sign 3, and the length of hypotenuse AB is 2 radical 3, then the area of right triangle ABC is?
Let two right angles be a and B respectively
Then a + B + 2 √ 3 = 4 + 2 √ 3 A + B = 4 (1)
From the Pythagorean theorem, a 2 + B 2 = (2 √ 3) 2 = 12 (2)
(1)²-(2) 2ab=4
AB = 2
So the area of the right triangle ABC = (1 / 2) AB = 1
RELATED INFORMATIONS
- 1. If the perimeter of a right triangle is 4 + 2, the root is 6, and the length of the central line on the hypotenuse is 2, then the area of the triangle is
- 2. In the triangle ABC, D is the midpoint of BC. If angle a is equal to 120 degrees and ab vector multiplied by AC vector equals negative 1, then the minimum value of Ad vector modulus is?
- 3. Any line passing through the center of gravity of △ ABC intersects AB and AC at points D and e respectively AD＝x AB， AE＝y AC, XY ≠ 0, then 1 X+1 The value of Y is______ .
- 4. In the triangle ABC, D is a point on the edge of AB, and vector ad = 2 vector dB, vector CD = 1 / 3 vector Ca + n vector CB, then n What is the value of N?
- 5. In △ ABC, it is known that BD and CF are high, M is the midpoint of BC, and N is the midpoint of DF Yes,
- 6. In the triangle ABC, points D, e, f are the midpoint of sides BC, Ca and ab respectively. Then AB + AD + BC + be + CF (all vectors)=
- 7. As shown in the figure, △ ABC is an isosceles right triangle, ∠ ACB = 90 ° and ad is the center line on the side of BC. Through C, the vertical line of ad is made, which intersects AB at point E and ad at point F. it is proved that ∠ ADC = ∠ BDE
- 8. As shown in the figure, points a, B and E are on the same straight line, △ ABC and △ BDE are equilateral triangles. Ad intersects BC at f and CE intersects BD with AD at g and h respectively
- 9. As shown in the graph, △ ABC, ab = AC, points D, e, f are on edges BC, AB, AC respectively, and BD = CF, ∠ EDF = ∠ B, is there any triangle congruent with △ BDE in the graph? And explain the reason
- 10. As shown in the figure, in △ ABC, ab = AC, ad ⊥ BC is at point D, de ∥ AB intersects AC at point E. is △ ade an isosceles triangle? Why?
- 11. As shown in the figure, in circle O, AB is the diameter of circle O, CD is a chord, and CD is perpendicular to point P, connecting BC and AD
- 12. In ⊙ o, the chord ab ⊥ CD is in M. the radius of ⊙ o is 5, BM = 6, am = 2. Find the size of dm-cm
- 13. Find a proof question: AB is the diameter of circle O, CD is a chord in the circle, AE ⊥ CD is in E, BF ⊥ CD is in F, and CE = DF
- 14. It is known that in the triangle ABC of RT, the angle ACB = 90 degrees, D and E are the midpoint of AB and BC respectively, and the point F is on the extension line of AC, and CF = De
- 15. If AB is 10, Pb is 4 and OP is 5, then find the radius of the circle
- 16. 0
- 17. It is known that, as shown in the figure, the diameter ab of the circle O intersects with the chord CD at the point E, AE = 1, be = 5, ∠ AEC = 30 ° and find the length of CD I can't draw a picture
- 18. If the lengths of two parallel chords are 6cm and 8cm respectively in a circle with a radius of 5cm, the distance between the two chords is______ .
- 19. As shown in the figure, in ▱ ABCD, AE bisects ∠ bad to intersect DC at point E, ad = 5cm, ab = 8cm, and calculate the length of EC
- 20. In circle O, the chord Mn / / EF, P is a point on circle O. the radius of circle O is known to be 10cm, Mn = 12cm, EP = 16cm. Find the distance between chord Mn and ef