How do you find the area of a triangle? 6th grade math problem .

The inner angle of the triangle and 180 degrees, according to the proportional distribution (6 years of mathematical knowledge), is divided into three angles according to 1 to 1 to 2, two 45 degrees, one 90 degrees, is an isosceles right triangle, and the two short sides are the right angled sides, which are the base and height. Column equation 4 times 4 divided by 2 equals 8

I wasn't even six years old when I was studying, how to use the sixth-grade hunger method to solve it, as long as I can solve it, really.

It is an isosceles triangle with a high line in the middle, and then divided into two right-angled triangles.

Text: The base is multiplied by the product of the high divided by 2

Letters: (a+b)h divided by 2

Right triangles (only right triangles are one angle plus one angle equals another). Because it is a right-angled triangle, the base and height are the two short sides, so the area is 4*4 2=8 (square centimeters).

It is known that the ratio of the number of three angles is 1:1:2, and the two short sides are 4 cm, find the area of the triangle, (explain what the short side is), it is a (isosceles right angle) triangle, because the sum of the 3 inner angles of the triangle is 180 degrees, then it is 180 divided by 1 plus 1 plus 2 is 45 degrees, and then multiply them separately, one of the angles is a right angle, according to the triangle with a right angle in the triangle is a right triangle, and their 2 right angles have the same, so this is an isosceles right triangle.

I'm also in 6th grade, and this one is very simple, as our teachers have talked about before).

Expressed in words: base times height divided by two.

Represented by letters: a h 2

Then calculate the area of the circle, and then according to the law of :2, calculate the area of the square 4 4

Hello dear, I'm glad to hear your question: you can't find the area just by knowing the perimeter, but you can find the area by knowing the length of the three sides of a triangle. Suppose that in the plane, there is a triangle with side lengths a, b, and c, and the area s of the triangle can be obtained by the following formula:

s= p(p-a)(p-b)(p-c) and p in the formula is half the circumference (half of the circumference): p=(a+b+c) 2 A triangle is a closed figure composed of three line segments in the same plane that are not on the same straight line, which are connected sequentially by 'head and tail', which has applications in mathematics and architecture. Common triangles are divided into ordinary triangles (the three sides are not equal) according to the sides, isosceles triangles (isosceles triangles with unequal waists and bottoms, isosceles triangles with equal waists and bottoms, Qingyou is equilateral triangles); According to the angle, there are right triangles, acute triangles, obtuse triangles, etc., of which acute triangles and obtuse triangles are collectively referred to as oblique triangles.

The landlord can use the Helen Mountain Pat-Qin Jiu Teasing Xian Shao formula to calculate: known triangles.

The three-sided vertical mu is a, b, c, let p=(a+b+c) 2, then s= [p·( p-a)· (p-b)· (p-c)] a=4,b=5,c=6

So s=(5, 7) 4

The perpendicular line of the point P and D as AP forms a right-angled triangle respectively, and the four-point co-circle, and the right-angled triangle with equal angles is similar.

They have equal hypotenuses, so they are congruent, and we get h=ap=2, s apd=

Let the opposite angles of the triangle a, b, and c be a, b, and c, respectively, then the cosine theorem is.

cosc(a^2+b^2-c^2)/2ab

s=1/2*ab*sinc

1/2*ab*√(1-cos^2

c)=1/2*ab*√[1-(a^2+b^2-c^2)^2/4a^2*b^2]

1/4*√[4a^2*b^2-(a^2+b^2-c^2)^2]

1/4*√[2ab+a^2+b^2-c^2)(2ab-a^2-b^2+c^2)]

1/4*√[a+b)^2-c^2][c^2-(a-b)^2]

1/4*√[a+b+c)(a+b-c)(a-b+c)(-a+b+c)]

Let s=(a+b+c) 2

Then s-a=(-a+b+c) 2,s-b=(a-b+c) 2,s-c=(a+b-c) 2, the above = [(a+b+c)(a+b-c)(a-b+c)(-a+b+c) 16].

[s(s-a)(s-b)(s-c)]

Therefore, the triangular ABC area s = [s(s-a)(s-b)(s-c)].

Proof is complete.

You need to have a diagram, otherwise you don't know which direction your BC is extended, and AB is extended in which direction, such as the next two figures, and AB is extended in the direction of A and it is different.

There is no solution to this problem, may I ask you from **, according to the title of the extended triangle there are countless ones.

6 times 1/3 equals 2

2 plus 6 equals 8 (which is high).

8 times 1/4 equals 2

8 plus 2 equals 10 (this is the bottom).

8 times 10 divided by 2 equals 40

The area of the new triangle is 40

Since the area ratio of QSR to RST is equal to QS TS 5 4 (the two triangles are equal in height, and the area ratio is equal to the ratio of the base), we get: The area of QSR is 10

In the same way, the area ratio of QSR to QOR is equal to SR OR 1 3, resulting in an area of 30 for QOR

The area of the opq is equal to twice the area of the qor, so that the area of the opq is equal to 60

Because qt=1 5qs

So qsr=five times rst=10

The same goes for sr=1 3or

So Qor=three times QSR=15

So opq=15*2=30