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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
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)?
From the meaning of the question, we can see 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, according to a certain height, the area of triangle is proportional to the bottom: s △ amn: sacm = Mn: MC = 1: (1 + 4) = 1:5, so yin = 15s △ ACM = 32 × 15 = 310, a: the area of triangle amn (shadow part) is 310
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
Given that the circumference of a triangle is 15, the sum of the first and second sides is equal to twice of the third side, and their difference is equal to
Given that the perimeter of a triangle is 15, the sum of the first side and the second side is equal to twice of the third side, and the difference between them is equal to four-thirds of the third side?
If you can't solve it, just one system of equations
Let the length of the first side be x and the length of the second side be y
x+y+1/2(x+y)=15
x-y=4/3*【1/2（x+y）】
The perimeter of a triangle is 15cm, a: B = 2:3, C-A = 1cm, then B=
Another question: in triangle ABC, angle a-c = 25 degrees, angle B-A = 10 degrees, then angle B=
First question:
A: B = 2:3, so B = 3 / 2a, C-A = 1, so C = a + 1,
A + B + C = 15, that is: a + 3 / 2A + (a + 1) = 7 / 2A + 1 = 15, the solution is a = 4, B = 3 / 2A = 6, C = 4 + 1 = 5
Let me talk about the second question if the angle B = 135 degrees
Know that each triangle is 360 degrees
It is known that a + B + C = 360
A-C=25
B-A=10
Seeking B
C = A-25, B = a + 10, a + B + C = 360,
A+（A+10)+(A-25)=360
A=125
Then B = 135
The two sides of a triangle are 15 cm and 7 cm respectively. What is the maximum perimeter of the triangle
The third side is larger than 15-7 = 8 cm
Less than 15 + 7 = 22 cm
If it's an integer
So the maximum is 21 centimeters
So the maximum circumference is 15 + 7 + 21 = 43 cm
The problem of finding the shortest circumference of triangle
E. F is the two fixed points on the side AB AC of triangle ABC. Find a point m on BC to make the perimeter of triangle MEF shortest?
(please write down the practice)
Make the symmetric point of point e about BC, denote it as e ', and then connect Fe' to BC at M. at this time, the perimeter of triangle MEF is the shortest
Given that the perimeter of a triangle is p, and the length of one side is twice that of the other side, the range of the shortest side can be obtained
2x+x>y 3x>y
2x-x
What conditions should be satisfied to make the perimeter of triangle shortest
Equilateral △ shortest perimeter
Given that the perimeter of a triangle is 15cm, and the length of both sides is equal to twice of the third side, then the shortest side of the triangle is ()
Let the shortest side be x, then the other two sides are 2x, right?
Lie equation: x + 2x + 2x = 15, 5x = 15, x = 3
So the shortest side is 3cm
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