![](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>)
An inner angle of the quadrilateral ABCD is 120 ° connecting AC to obtain equilateral △ ABC and right triangle ACD. It is known that the side length of equilateral △ ABC is 2 (1) Find the height on the base BC of △ ABC (2) find the area of △ ACD
The height on the side of BC is the root 3. The area is the root 3 / 2
In trapezoid ABCD, AD / / BC, AB vertical AC. angle B = 45 degrees, ad = root 2, BC = root 32, find the length of DC
AC ? ab ⊙ ab ⊙ ab ? ABC is isowaist, RT ⊙ a ? ab ⊙ ABC is isowaist, RT ⊙ AE ? BC ⊸ e is the midpoint of BC, AE = be = EC = BC / 2 = 2 ∵ 2 ∵ 2 ∵ 2 ⊙ 2 ∵ AD / / BC, AE ⊸ BC, DF ? BC ? DF = DF = 2 ? 2ef = ad = 2 9 2 9 CF = ce-ef = CF = ce-ef = 2 and
In trapezoid ABCD, AD / / BC, AB vertical AC, angle B = 45 °, ad = root 2, BC = 4 times root 2, find the length of DC
According to the fact that AB is perpendicular to AC and the angle B is 45 ° we can see that the triangle ABC is an isosceles right triangle, then AC = AB = 4 can be obtained;
Because the angle bad and angle ABC complement each other, the angle bad = 135 °;
AB is perpendicular to AC, so the angle ABC is 90 ° and the angle cab is 45 °;
Then in the triangle cab, given the angle cab = 45 °, ad = radical 2, AC = AB = 4, DC = 10 can be obtained by using cosine theorem
In RT △ ABC, ∠ C = 90 °, AC = 12, BC = 5. If △ ABC is rotated one circle around the line where AC is located, then the side area of the cone is______ .
It is known that the bus length L = 13, radius r = 5,
The side area of the cone is s = π LR = 13 × 5 × π = 65 π
So the answer is 65 π
As shown in the figure, in the rectangle ABCD, ab = 1, if the side area of the cone obtained by the rotation of the right triangle ABC around AB is equal to the side area of the cylinder obtained by the rotation of rectangular ABCD around AB, calculate the length of BC
∵ S cone side = π· BC · AC, s cylinder side = 2 π· BC · CD,
And ∵ S cone side = s cylinder side,
∴π•BC•AC=2π•BC•CD，
∴AC=2CD，
∵ ABCD is a rectangle,
∴CD=AB=1，∴AC=2CD=2，
In RT △ ABC, BC=
AC2−AB2＝
3，
∴BC=
3．
In RT △ ABC, ∠ C = 90 °, AC = 12, BC = 5. If △ ABC is rotated one circle around the line where AC is located, then the side area of the cone is______ .
It is known that the bus length L = 13, radius r = 5,
The side area of the cone is s = π LR = 13 × 5 × π = 65 π
So the answer is 65 π
In RT △ ABC, ∠ C = 90 °, AC = 12, BC = 5, the side area of the cone is () A. 25π B. 65π C. 90π D. 130π
∵ RT △ ABC, ∵ C = 90 °, AC = 12, BC = 5,
∴AB=
AC2+BC2=13，
The bus length is L = 13 and the radius R is 5,
The side area of the cone is s = π LR = 13 × 5 × π = 65 π
Therefore, B
As shown in the figure, in RT △ ABC, ∠ C = 90 °, AC = 5cm, BC = 12cm. Take the straight line where BC edge is as the axis, rotate △ ABC for one cycle to obtain the conical side area______ .
It is known that the bus length L = 13, radius r = 5,
The side area of the cone is s = π LR = 13 × 5 × π = 65 π
So the answer is: 65 π cm2
As shown in the figure, in RT △ ABC, ∠ ACB = 45 °, BAC = 90 °, ab = AC, point D is the midpoint of AB, AF ⊥ CD intersects with H, BC intersects with F, be ∥ AC, and the extension of AF is at E
It is proved that in △ ADC, ∠ Dah + ∠ ADH = 90 °, ACh + ∠ ADH = 90 °,
∴∠DAH=∠DCA，
∵∠BAC=90°，BE∥AC，
∴∠CAD=∠ABE=90°．
And ∵ AB = ca,
In △ Abe and △ CAD,
∠DAH=∠DCA
∠CAD=∠ABE
AB=AC
∴△ABE≌△CAD（ASA），
∴AD=BE，
And ∵ ad = BD,
∴BD=BE，
In RT △ ABC, ∠ ACB = 45 °, BAC = 90 °, ab = AC,
Therefore, ABC = 45
∵BE∥AC，
∴∠EBD=90°，∠EBF=90°-45°=45°，
∴△DBP≌△EBP（SAS），
∴DP=EP，
The result shows that BC is vertical and bisects De
In RT △ ABC, ∠ ACB = 90 °, CD ⊥ AB in D, ﹤ BAC bisector AF intersects CD in E, BC in F, cm ⊥ AF in M. The results show that EM = FM
It is proved that: ∵ ACB = 90 °, CD ⊥ AB,
∴∠ADC=90°，
∴∠AED+∠DAE=90°，∠CFE+∠CAE=90°，
The bisector AF of BAC intersects with E,
∴∠DAE=∠CAE，
∴∠AED=∠CFE，
And ∵ AED = ∵ CEF,
∴∠CEF=∠CFE，
And ∵ cm ⊥ AF,
∴EM=FM．
RELATED INFORMATIONS
- 1. It is known that in the RT triangle ABC, ∠ ACB = 90 ° CD is perpendicular to AB and D, and AF bisects ∠ cab As shown in the figure, in the RT triangle ABC, it is called ACB = 90 degrees, CD is vertical AB, vertical foot is D, AF bisection is called cab in E, intersection CB in F, and FG parallel AB intersects CB in G. it is proved that CF = GB
- 2. If we know that in RT △ ABC, ∠ C = 90 ° a + B = 14cm, C = 10cm, then the area of RT △ ABC is equal to______ .
- 3. If BC = 3, s △ pb1c= 3, then BB1=______ .
- 4. As shown in the figure, in the triangular prism abc-a1b1c1, each side edge is perpendicular to the bottom and is an isosceles right triangle, ∠ ACB = 90 degrees, AC = BC = 4, Aa1 = 4, e, f are on AC, BC respectively, and CE = 3, CF = 2, calculate the volume of geometry efc-a1b1c1
- 5. Given the vertex a [0,3} B {- 3,0} C {5,0} of the triangle ABC, shift △ ABC up two unit length, and then three unit length upward to obtain a'b'c ' 【1】 Find the coordinates of point a'b'c '[2] find the area of a'b'c'
- 6. As shown in the figure, rotate RT triangle ABC by 40 ° anticlockwise around point a to obtain RT triangle ab'c '. Point C' just falls on the hypotenuse AB and connects BB ', then ∠ bb'c' = °
- 7. Given vector a = (- 1 / 2, root 3 / 2), vector OA = vector a - vector B, vector ob = vector a + vector B, If △ AOB is an isosceles right triangle with o as its right vertex (1) coordinates of vector b (2) the angle between vector a and vector B
- 8. In the plane rectangular coordinate system, it is known that a (- radical sign 3,1) B is a moving point on the X axis, and △ ABC is an equilateral triangle. When B moves, what figure does point C form The analytic formula is obtained
- 9. As shown in the figure, the vertex a (- 2 √ 3,0) of the equilateral triangle ABC, B, C are on the Y axis.)
- 10. In triangle ABC, COSC:COSB= (2sina-sinc) / SINB for CoSb
- 11. In the triangle ABC, the angle cab120 degrees, ab = 4, AC = 2, ad is perpendicular to BC, and the perpendicular foot is d? Students are in the second year of junior high school, just learning Pythagorean theorem. We hope to solve it in other ways,
- 12. In the triangle ABC, the angle BAC is equal to 20 degrees, AB is equal to AC, D is a point on the edge of AB, and ad is equal to BC to find the degree of angle BDC
- 13. In the triangle ABC, ab = AC, angle BAC = 60 degrees. D is a point outside the triangle (under BC), connecting ad, BD, CD. The angle BDC = 120 degrees How to find the relationship between angle BDA and angle CDA?
- 14. As shown in the figure, in △ ABC, ∠ abd = ∠ ACD = 60 ° and ∠ ADB = 90 ° - 1 2∠BDC． It is proved that △ ABC is an isosceles triangle
- 15. In triangle ABC, angle a is equal to angle abd, angle ABC is equal to angle c, angle c is angle BDC, and degree of angle DBC is calculated
- 16. As shown in the figure, ∠ a = 35 °, C = 60 ° in △ ABC, BD bisection ∠ ABC, de ‖ BC intersecting AB with E, then ∠ BDE=______ ，∠BDC=______ .
- 17. As shown in the figure, △ ABC, ab = AC, CD bisection ∠ ACB, intersection AB at point D, ∠ BDC = 150 ° and find the degree of ∠ a
- 18. In the triangle ABC, if edge B + C = radical 2 + 1, angle c = 45 degrees, angle B = 30 degrees, what are the values of edges B and C
- 19. In RT △ ABC, ∠ C = 90 ° s = 18 root sign 3, a
- 20. As shown in △ ABC, the angle ACB = 90 °, the angle a = 60 ° and the height of the hypotenuse CD = the root sign 3. Find the length of ab D is on AB, CD is perpendicular to AB, AC is perpendicular to CB