![](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>)
In △ ABC and △ def, ab = De, BC = EF, and ∠ B and ∠ e complement each other
AB = De, BC = EF, and ∠ B and ∠ e complement each other, and ∠ B and ∠ e complement each other
Then the heights on BC and EF are equal
Equal base, equal height, equal area
Extend the three sides of triangle ABC by 1, 2 and 3 times to form a new triangle. How many times is the area of the new triangle?
The answer: 18:1
&Good topic:
The answer is: 18 times
If it is a choice or fill in the blank, you can consider using equilateral triangle to solve the problem
I use analytic geometry to prove the general case
Proof: suppose that the coordinates of triangle ABC are:
A（a,b）
B（c,0）
C（0,0）
Where C is the origin;
After extending the original triangle ABC by 1, 2, 3 times, the new triangle is a1b1c1
According to the fixed score point formula, calculate the coordinates of A1, B1 and C1 points
First calculate A1 & nbsp; (2a, 2b)
Recalculation of B1 & nbsp; (4c-3a, - 3b)
Finally, C1 (- 2C, 0) is calculated
Then calculate the linear a1a1 equation: y = 5B / (5a-4c) & nbsp; * x-8bc / (5a-4c)
Then let x = 0 and calculate the coordinate of point D as (8C / 5,0)
Then | C1d | = 8C / 5 - (- 2C) = 18C / 5
S△A1B1C1=S△A1C1D+S△C1B1D
=1/2*(C1D*(2b-(-3b)))
=1/2*18c/5*5b
=9bc
S△ABC=1/2*BC*b=bc/2
So s △ a1b1c1 / s △ ABC = 9bc / BC / 2 = 18
Get it!
The two sides of a triangle are respectively extended 2 times to form a new triangle. How many times is the area of the new triangle ABC? 1
The area of the new triangle is 2x2 = 4 times that of ABC
Given the point a (3,1), there are points m and N on the straight line X-Y = 0 and y = 0 respectively, so that the perimeter of △ amn is the shortest, the coordinates of points m and N can be obtained
As shown in the figure, a (3,1) is symmetric point A1 (1,3) with respect to y = x, a (3,1) is symmetric point A2 (3, - 1) with respect to y = 0, the minimum perimeter of △ amn is | A1A2 |, | A1A2 | = 25, and the equation of A1A2 is: 2x + Y-5 = 0. The intersection point of A1A2 and X-Y = 0 is m, from 2x + Y-5 = 0x-y = 0 {m (53,53), and the intersection point of A1A2 and y = 0 is n, from 2x + Y-5 = 0y = 0} n (52,0)
Coordinate formula of * center of triangle
1. Triangle ABC, a (x1, Y1) B (X2, Y2) C (X3. Y3), what is the perpendicular formula?
On this issue,
The answer given in it is No
Given such an answer, is it really so?
2. Triangle ABC, a (x1, Y1) B (X2, Y2) C (X3. Y3), what is the formula of outer center?
Is the answer to this question all wrong? Is there no direct formula to be derived?
The formula of the center of gravity is: let the center of gravity H (x0, Y0), then x0 = (x1 + x2 + x3) / 3y0 = (x1 + x2 + x3) / 3. There seems to be no formula for the center of gravity and the center of perpendicularity
Triangle inner coordinate formula?
The heart is the intersection of the bisectors of the angles, with equal distances to the three sides
Let: in the triangle ABC, the coordinates of the three vertices are: a (x1, Y1), B (X2, Y2), C (X3, Y3), BC = a, CA = B, ab = C
The heart is m (x, y)
M（（aX1+bX2+cX3)/(a+b+c),（aY1+bY2+cY3)/(a+b+c)）
The inner coordinate formula of triangle
The heart is the intersection of the bisectors of the angles, with equal distances to the three sides
Let: in the triangle ABC, the coordinates of the three vertices are: a (x1, Y1), B (X2, Y2), C (X3, Y3), BC = a, CA = B, ab = C
The heart is m (x, y)
M（（aX1+bX2+cX3)/(a+b+c),（aY1+bY2+cY3)/(a+b+c)）
Given point a (0,3), point B (- 1,0), point C is on the x-axis, and the area of triangle ABC is 6, the coordinates of point C are obtained
Point C is on the x-axis, so the height of the triangle is OA = 3 and the area is 6
According to the area formula, the bottom BC = 4
So the coordinates of point C are (3,0) or (- 5,0)
A (1, - 2,11) B (4,2,3) C (6, - 1,4), find the area of triangle ABC,
The vertices of triangle ABC are a (1, - 2,11), B (4,2,3), C (6, - 1,4)
AB=√[(1-4)^2+(-2-2)^2+(11-3)^2]=√89
AC=√[(1-6)^2+(-2+1)^2+(11-4)^2]=√75
BC=√[(4-6)^2+(2+1)^2+(3-4)^2]=√14
So AC ^ 2 + BC ^ 2 = 75 + 14 = 89, AB ^ 2 = 89
AC^2+BC^2= AB^2
From the Pythagorean inverse theorem, it is concluded that
∠ACB=90°
So s △ ABC = 1 / 2 * ac * BC = 1 / 2 * √ 75 * √ 14 = 5 √ 42 / 2
Given points a (1,3), B (3,1), C (- 1,0), find the area of triangle ABC
Solve the equation of the line L with the same distance between two parallel lines L1: 3x + 4y-10 = 0 and L2:: 3x + 4y-12 = 0 (both questions are done with the "distance from point to line" in the second compulsory course of senior one) in detail!
| 1 1 3 |
| 1 3 1 |
| 1 -1 0 |＝-10 ∴ S⊿ABC＝5
RELATED INFORMATIONS
- 1. In ABC, AC = 15, BC = 13 and CD = 12, what is the area?
- 2. In the triangle ABC, AC = 13, BC = 15, height CD = 12, then its area is?
- 3. In △ ABC, ∠ C = 60 °, BC = a, AC = B, AC + BC = 8, write the functional relationship between the area s and side length a of △ ABC and the value range of independent variable a
- 4. Given the length of the three sides of a triangle, how to find the height of the triangle?
- 5. Given the length of three sides of a triangle, how to find the area of the triangle? Such as the title
- 6. Given the length of the three sides of a triangle, can we find the area? How? And we don't give the degree of the angle
- 7. In triangle ABC, edge B = 2, diagonal B = 30 degrees, find the maximum area of triangle?
- 8. The area of a triangular wheat field is 1620 square meters. The length of its bottom is 36 meters. How high is the field?
- 9. Mathematics of grade five in primary school: enlarge the bottom of a triangle by 8 times and reduce the height by 2 times, then the new triangle will be 90 square centimeters larger than the original triangle. What is the original triangle's area?
- 10. The base and height of a triangle are increased by 1 / 3. Now the area of the triangle is equal to the area of the original triangle Please explain why,
- 11. If the side lengths of ABC and def are 6,5,8 and 30,25,40 respectively, So this triangle - yes, or no - is similar to a triangle, according to-
- 12. 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)?
- 13. Java defines a rectangle, including method, perimeter and area
- 14. 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
- 15. For a trapezoid, if the bottom and height remain unchanged, the area of the trapezoid will increase by 1.2 square meters if the top and bottom increase by 0.6 meters; if the top and bottom remain unchanged, the height will decrease by 0.8 meters, and the area of the trapezoid will decrease by 2 square meters?
- 16. The bottom of the trapezoid is 12 meters, 2 meters more than the top, and the height is 4 meters. How many square meters is the area of the trapezoid
- 17. There is a trapezoid. If the upper bottom is increased by 2 meters, the lower bottom and height will not change. Its area will increase by 4.8 square meters. If the upper bottom and the lower bottom are both unchanged, and the height is increased by 2 meters, its area will increase by 8.5 square meters. Find the area of the original trapezoid
- 18. In △ ABC, ab = 10cm, BC = 16cm, ad = 6cm, AC
- 19. If both sides of a right triangle are 5 cm and 12 cm long, what is the area of the triangle? There are two cases
- 20. The bottom of a parallelogram is 11cm and its height is 4cm. How many square centimeters is the area of a triangle with the same height as its bottom?