On the first day of junior high school, the problem of finding the area of a coordinate triangle is

Such questions can be cut and patched.

Combine the numbers to make the graph, then place the triangle in a rectangle (the three vertices of the triangle are on the sides of the rectangle), and subtract the other small triangles from the rectangle to get the required triangle area.

For this question.

First, determine that the length of this rectangle is 5-(-4)=9 and the width is 5-(-5)=10, so the area of this rectangle is 90Then subtract the 3 small triangles, i.e. 90-(2 7+9 8+2 10) 2=37

ps: For this kind of question about functions, it is recommended to use the method of combining numbers and patterns, construct the graph, and then observe whether there are basic graphs (congruence, similarity, etc.), and then solve them according to their properties. There are 2 ways to solve an area problem like this:

Direct sum indirect finding. Direct finding is to use the formula to find it directly, such as the area of the triangle is divided by the base height by 2, etc., but this problem is obviously inappropriate (note, it's not that you can't do it, you can still use the distance formula between two points and the point-to-straight line distance formula to do it, that is, 1 junior high school didn't learn, 2 is too annoying). Therefore, indirect search is used.

Commonly used are the cutting method, etc. (and some forgot...) In short, you can ask your teacher about such questions.

Hope it helps.

You can place the triangle in a triangle (all three vertices of the triangle are on the edge of the rectangle) and label the rectangle as a rectangle defg

At this point, you can see that the rectangle is divided into four triangles, three of which are right triangles, first find the area of these three right triangles, and then find the area of the rectangle defg.

Finally, we can find the area of abc by using the area of the rectangle defg - the area of the three triangles on the edge.

，0）b（2，0）

c（0，3）

s=2*3/2=3

2）b（4，-1）

c（0，-6）

First, find the linear equation for point b and point c, and let it be.

b（4，-1）

The linear equation for c(0,-6) is y=ax+b.4a+b=-1,b=-6, a=5 4, y=5 4*x-6, when y=0, x=24 5 intersection coordinates d(24 5,0) s abc=s doc-s dab-s aoc

Do the math yourself, it's easy to calculate. I'm sorry, I can't give you any formula, just to analyze the situation on a case-by-case basis.

By the title: ap is the angular bisector of bac, so pac= a 2=( 2

cp is the angular bisector of bca, so pca = c 2 = ( 2

So in APC, APC=180°- pac- PCA

There are often several steps:

First, first determine whether it is a right triangle, and if so, simplify the problem, the Pythagorean theorem, etc. If not, move on to the second step.

Second, there are several common methods.

1. Use the sine theorem to obtain the area.

2. Use similar triangles to find, which should be combined with plane geometry.

3. If the two triangles share one side and the height is the difference between the two intersection points, the substitution method is used to solve the line segment according to the root finding formula of the one-element quadratic equation.

(1) Because AOB is a right-angled triangle and AO=BO, according to the triangle area formula, 8=AO*BO2, AO=BO=4, BC=BO+OC, OC=12-4=8, A, B, C are on the coordinate axis, so A(0,4),B(-4,0),C(8,0).

2) As the perpendicular Y axis of PE, the trapezoidal area s=(a+4)*6 2,s and other triangles of PAB, AOB, APE are summed, because p(a,6), so the area of APE = A*2 2=A; Then the area of the triangle PAB = (A+4)*6 2-8-A=2A+4;

If the area is pab=abo, then 2a+4=8, a=2, and p is in the second quadrant, so p(-2,6);

3) Since starting from b and moving in the positive direction, and ACQ is an isosceles triangle, then qa=qc, and Q is in the first quadrant; Let q(x,0), and a(0,4),c(8,0), get (x-0) 2+(0-4) 2=(x-8) 2+(0-0) 2, and get x=3;then qb=4+3=7;Because of 3 units per second, the elapsed time is t=7 3

Because the area of the triangle is 8Area = (1 2)*(oa)*(ob) = 8, so oa = ob = 4

The a coordinate is (0,4), and the b coordinate is (-4, 0)The c coordinate is (8, 0).

The area of a triangular PAB is ((A+4) *6) 2 - A)*2 2 - 8 = -3A+12+A-8 = 4-2A

Let the q coordinate be (x, 0), 8-x) 2 = 4 2 + x 2Solving the equation yields x=3So after 7-3 seconds, ACQ is an isosceles triangle.

(1)∵s=1/2*oa*ob=8,oa=ob∴oa=ob=4

oc=bc-ob=12-4=8

a(0,4), b(-4,0), c(8,0)(2) as pd perpendicular to y-axis at point d

s pab=s trapezoidal pboc-s apd-s aob=1 2(oa+pd)*op-1 2pd*ad-8=1 2(4+lal)*6-1 2*lal*(op-oa)-8=1 2(4+lal)*6-1 2*lal*(6-4)-8=3(4+lal)-lal-8

4+2lal (due to a<0).

4-2a, so s pab = 4-2a

s△pab=4-2a ,s△abc=1/2*bc*oa=1/2*12*4=24

s△pab=s△abc

4-2a=24

a=-10p(-10,6)

3) When AC=CQ, AC= (OA2+OC2)= (4 2+8 2)=4 5 CQ=AC=4 5

bq=bc-cq=12-4√5

t=bq/v=(12-4√5)/3

When AQ=CQ, AC bisector intersects OC at point Q, let OQ=X, then AQ=CQ=8-X

oq^2+oa^2=aq^2

x^2+4^2=(8-x)^2

solution, x=3

oq=3bq=ob+oq=4+3=7

t=bq/v=7/3

When ac=cq

bq=bc+cq=12+4√5

t=bq/v=(12+4√5)/3

In summary, t=(12-4 5) 3 , or , (12+4 5) 3, or 7 3

Drawing, the triangle area is equal to the area of the rectangle minus the area of the three right triangles.

Triangle area = (5+4)x(4+3) -1 2 x9 x4 - 1 2 x 2x7 - 1 2 x3x7=

Triangle area = 55 2

If the calculation process is complicated, I won't write it.

Find the length of the 3 edges, and then find the sine value of the angle between any 2 edges.

Let the intersection point of ac and the transverse axis be d, and the linear equation where the points a and c are located is y=kx+b, and substitute a(-4,3) and c(-2,-1) to solve k=-2,b=-5, so the linear equation where a and c are located is y=-2x-5, and when y=0, x=-5 2 is obtained, so the intersection point of ac and the transverse axis is d(-5 2,0).

s△ado=

s△cdo=

s△abc=s△ado+s△cdo=15/4+5/4=5

Find the sum of an edge, find the length of the height and the edge, and then use the formula.

Right triangle, right?

Just set an x.

Let the line segment in the upper left corner be x

Then there is [(x+7) 2]+[x+8) 2]=15 2 [according to the theorem of circles, tangents, and right-angled triangles Pythagorean].

Then there is the area calculation: [(x+8)*(x+7)] 2 [Divide the base by two. 】