Three points D, e and F are obtained by extending the three sides of the triangle ABC by 1, 2 and 3 times, respectively. The area of the triangle ABC is 10 square centimeters, and the formula of the triangle DEF is obtained

Three points D, e and F are obtained by extending the three sides of the triangle ABC by 1, 2 and 3 times, respectively. The area of the triangle ABC is 10 square centimeters, and the formula of the triangle DEF is obtained

10+10*（2+1）*1+10*2*（3+1）+10*3*（1+1）
=180

The edge of triangle ABC is extended one time to point D, CB is extended two times to point F, and AC is extended three times to point e. the triangle DEF is obtained by connecting ed, EF and FD,
How many times of triangle ABC is the area of triangle def?

The length of the three sides is 2a, 3b, 4C
S-original = (1 / 4) √ [(a + B + C) (a + B-C) (a + C-B) (B + C-A)] Helen theorem
After s = (1 / 4) √ (2a + 3B + 4C) (2a + 3b-4c) (2a + 4c-3b) (3b + 4c-2a)
Compare yourself, it's OK, it's troublesome

In triangle ABC, extending Ba makes Ba = ad, extending AC makes AC = CE, extending CB makes CB = BF, then the following conclusions are obtained: 1
In triangle ABC, extending Ba makes Ba = ad, extending AC makes AC = CE, extending CB makes CB = BF, then the following conclusions are obtained: 1. If the area of triangle ABC is equal to 1, then the area of triangle CFE is equal to 2, 2. The areas of triangle DFB, CEF and AED are equal, 3. Triangle ABC is similar to triangle def?

The correct ones are 1, 2
1. Connect be
In △ BCE and △ ABC, make high H 1, H 2 on BC
∵AC=CE
We can get H1 = H2
∴S△BCE=S△ABC=1
∵BC=BF
∴S△BCE=S△EFB=1
That is s △ EFC = 2
Similarly, s △ FBD = s △ ade = 2

In triangle ABC, C is a right angle. Given AC = 2, CD = 2, CB = 3, am = BM, what is the area of triangle amn (shadow part)?

We can see from the meaning that BD = bc-cd = 3-2 = 1, because am = MB, & nbsp; so gmbd = 12, GM = 12, so gmcd = 0.52 = 14, because △ NGM ∽ ndcmncn = gmcd = 14, s △ ABC = 2 × 3 △ 2 = 3, so s △ ACM = 12S △ ABC = 32

Define triangle class to calculate perimeter and area in C + +
A triangle like triangle that defines a triangle is composed of three vertices (the vertex is a point object). The perimeter and area of the triangle are calculated according to the values of the three vertices. If the three vertices are on a straight line, it will be prompted that the triangle is illegal. Whether the size of the triangle is isosceles triangle or right triangle can be compared

Point object should have two parameters to generate three point objects. First, calculate the length of line segments between two points according to the coordinates of two points, and then calculate the triangle area according to the comparison of line segments. There is a formula (it is said to be Helen's theorem, but I remember that formula was first invented by a high school teacher in Huanggang. How can it become a foreigner's Theorem

Java defines a triangle class. The member variable is the length of three sides. Define the method perimeter () to find the perimeter
(1) Define a triangle class. The member variable is the length of three sides. Define the method perimeter () to calculate the perimeter. Create the main class triangletest to test it

class Triangle{
inta,b,c,l;
public static void perimeter(){
l=a+b+c;
}
}
public class TriangleTest{
public static void main(String[] args){
Triangle t1= new Triangle();
t1.a=x;
t1.b=y;
t1.c=z;
t1.perimeter();
System.out.println (the perimeter of triangle T1 is + T1. L);
}
}

Write a triangle class, data members for three sides, member function calculation perimeter, area, define two overloaded construction C++

class Triangle{private:\x05double a;\x05double b;\x05double c;public:\x05Triangle():a(0),b(0),c(0){}\x05Triangle(double x,double y,double z):a(x),b(y),c(z){}\x05double cirlce(){return a+b+c;}\x05doubl...

Take one point on each side of the triangle, and the perimeter of the triangle is the shortest. How to find such a point
Please give reasons
By the way, what is the nature of the intersection of the heights of three sides

The intersection point (vertical center) of the three sides will be divided into two segments with a length ratio of 2:1 (the distance from the vertical center to the vertex is 2)
(1) Let d be a fixed point on BC and find the inscribed triangle with the shortest perimeter
Let d be the symmetric points D1 and D2 of AB and AC, and let d 1 and D2 intersect AB and AC with E and F, then △ DEF is obtained. In fact, for any inscribed △ de ′ f ′ of △ ABC, there are
DE′+E′F′+F′D=D1E′+E′F′+F′D2
≥D1D2=D1E+EF+FD2
=DE＋EF＋FD.
That is, the perimeter of △ DEF is less than the perimeter of △ def
Therefore, we only need to find the shortest inscribed triangle def corresponding to the point D on each BC, and find the shortest one among these triangles
(2) Because AD1 = ad, ad2 = ad, so △ ad1d2 is an isosceles triangle. And because ∠ 1 = ∠ 2, ∠ 3 = ∠ 4, so the vertex angle of △ ad1d2 is a fixed value, so only when its waist AD1 is the shortest, D1D2 is the shortest. At this time, there must be the shortest ad. so when ad is the height of △ ABC, the perimeter of the inscribed triangle DEF is the shortest
(3) When ad is the height of △ ABC, it can be proved that among the inscribed triangles of △ ABC, the perimeter of the perpendicular triangle DEF is the shortest

If the two sides of the triangle are 3 and 5, then the perimeter of the triangle connected to the midpoint of the three sides of the triangle may be ()
A. 5.5B. 5C. 4.5D. 4

Let the three sides of a triangle be a, B and C respectively, so that a = 3, B = 5, 2 < C < 8, 10 < perimeter of triangle < 16, 5 < perimeter of midpoint triangle < 8

Java to write an application program, create a circle class, in the class to define the radius member variable, calculate the area and perimeter method,
Define the circle object in the main class, calculate its area and perimeter according to the radius of the input circle

import java.util.Scanner ;class Round{\x09public double perimeter(double radius){\x09\x09return radius*2;\x09}\x09public double area(double radius){\x09\x09return radius*radius*3.14;\x09}\x09public voi...