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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
Join a1f, C1F, the geometry is divided into two pyramids, f-a1b1c1, volume is 32 / 3, f-a1c1ce, volume is 28 / 3, the result is 20
It is known that the triangle ABC is similar to the triangle a1b1c1, the similarity ratio is K (k is greater than 1), and the three side lengths of triangle ABC are a, B and C (a > b > C), and the three side lengths of triangle a1b1c1 are a1b1c1 If C = A1, try to give a pair of triangles ABC and a1b1c1 which meet the conditions. Yes, A.B.C and a1b1c1 are positive integers and are explained
∵ A / A1 = k, C = A1 ? A / C = k a = kc2. C = A / KC1 = C / k = A / k? A / K and a / k? Are positive integers. For example, a = 27, k = 3 ? C = A1 = A / k = 9, C1 = A / K ? 327-9 ﹤ B ﹤ 27 + 918 ﹤ B ? B ? a 18 ﹤ B ? B M a
It is known that △ ABC ∽ a1b1c1, the similarity ratio is K (k > 1), and the three side lengths of △ ABC are a, B, C (a > b > C). The side lengths of △ a1b1c1 are A1, B1, C1 (1) If C = A1, prove that a = KC (2) If C = A1, we try to give a pair of △ ABC and △ a1b1c1, so that a, B, C and A1, B1, C1 are all positive integers, and prove them (3) If B = a, C = B, is there △ ABC and △ a1b1c1 such that k = 2? Please explain the reason
(1) The results show that: Q △ ABC ∽ A1 b1c1, and the similarity ratio is K (k > 1), a = k, a = KA1 ﹤ A1 and Q C = A1, a = KC. Take a = 8, B = 6, C = 4, and at the same time take A1 = 4, B1 = 3, C1 = 2
Given the obtuse angle △ ABC, a = k, B = K + 2, C = K + 4, find the value range of K______ .
From the meaning of the title, C is the largest edge, that is, C is an obtuse angle
According to the cosine theorem, (K + 4) 2 = (K + 2) 2 + k2-2k (K + 2) · COSC ＞ = (K + 2) 2 + K2
That is (K + 2) 2 + K2 ＜ (K + 4) 2, the solution is - 2 ＜ K ＜ 6,
∵a+b＞c，
ν K + (K + 2) > K + 4, the solution is k > 2
To sum up, the value range of K is (2,6)
So the answer is: (2,6)
As shown in the figure, △ ABC ∽ a1b1c1 is known, the similarity ratio is K (K ﹥ 1), and the three side lengths of △ ABC are a, B, C (a > b > C) The three side lengths of △ ABC are a, B, C (a > b > C). The side lengths of △ a1b1c1 are A1, B1, C1 (1) If C = A1, prove that a = KC (2) If C = A1, we try to give a pair of △ ABC and △ a1b1c1, so that a, B, C and A1, B1, C1 are all positive integers, and prove them (3) If B = a, C = B, is there △ ABC and △ a1b1c1 such that k = 2? Please explain the reason
(1) The results show that: Q △ ABC ∽ A1 b1c1, and the similarity ratio is K (k > 1), a = k, a = KA1 ﹤ A1 and Q C = A1, a = KC. Take a = 8, B = 6, C = 4, and at the same time take A1 = 4, B1 = 3, C1 = 2
It is known that the triangle a1b1c1 is similar to the triangle ABC, and the vertex ABC corresponds to a1b1c1 respectively, ab = 6cm, BC = 9cm, CA = 12cm, and the circumference of triangle a1b1c1 is 81cm. Calculate the length of each side of triangle a1b1c1
Triangle ABC perimeter = 6 + 9 + 12 = 27 (CM), A1 / a = P1 / P, (P1, P is the perimeter of two triangles)
27/81=9/a1,a1=27,27/81=b/b1=12/b1,b1=36,27/81=c/c1=6/c1,c1=18,
A1B1=18(cm),B1C1=(27(cm),A1C1=36(cm).
In the triangle ABC, a (- 4, - 2), B (- 1, - 3), (- 2, - 1), if △ ABC is first shifted to the right by 4 unit length, and then up by 3 unit length, then corresponding The coordinates of points a ', B' and C 'are (- 2, - 1) is the coordinate of C
After translation, a '= (0,1), B' = (3,0), C '= (2,4)
As shown in Figure 10, shift the triangle ABC to the right by 2 unit length, and then down by 3 unit length to obtain the corresponding triangle a'b'c ', and calculate the area of triangle ABC
ABC area
S△ABC=8×7-1/2×2×7-1/2×8×5-1/2×2×6
=56-7-20-6
=56-33
=23．
In △ ABC, a (- 4, - 2), B (- 1, - 3), C (- 2, - 1), if △ ABC is first shifted to the right by 4 unit lengths and then up by 3 unit lengths, then the coordinates of corresponding points a ', B', and C 'are______ 、______ 、______ .
The coordinates of point a 'are (0, 1), B (- 1, - 3) after 3 units of length up, 4 unit lengths of point B' are (3, 0), C (- 4, - 1), 4 units of length are shifted to the right, and the coordinates of point B 'are (3, 0), C (- 4, - 1)
As shown in the figure, the area of triangle ABC is known to be 16, BC = 16. Now, translate the triangle ABC along the straight line BC by a unit length to the position of triangle def. When the area swept by triangle ABC is 32, calculate the value of A
Let the height of △ ABC be h, ∵ area ᙽ 16h = 16, so h = 2. ∵ abed area of parallelogram ah = 16, ᙽ a = 8
RELATED INFORMATIONS
- 1. 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'
- 2. 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' = °
- 3. 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
- 4. 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
- 5. As shown in the figure, the vertex a (- 2 √ 3,0) of the equilateral triangle ABC, B, C are on the Y axis.)
- 6. In triangle ABC, COSC:COSB= (2sina-sinc) / SINB for CoSb
- 7. In △ ABC, a and B are acute angles, and the corresponding edges of angles a, B and C are a, B and C respectively, and cos2a = 3 / 5, SINB = 10 / 10 under the quadratic root sign If a + B = 2-1 under the square root sign, find the value of a, B, C [find detailed explanation]
- 8. It is known that in △ ABC, ∠ C = 90, ∠ a = 60, a + B = 3 + radical 3, then a is equal to
- 9. As shown in the figure, it is known that in △ ABC, ad is the height on the edge of BC, AE is the angular bisector of ∠ BAC. If ∠ C = 40 ° and ∠ B = 64 °, find out The degree of DAE
- 10. As shown in the figure, it is known that the triangle ABC is an equilateral triangle. D and E are on sides BC and AC respectively, and CD = CE. Connect de and extend to point F, so that EF = AE, connect AF, be and Cf. if AB = 6, BD = 2dc, find the area of abef of quadrilateral~
- 11. If BC = 3, s △ pb1c= 3, then BB1=______ .
- 12. If we know that in RT △ ABC, ∠ C = 90 ° a + B = 14cm, C = 10cm, then the area of RT △ ABC is equal to______ .
- 13. 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
- 14. 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
- 15. 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,
- 16. 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
- 17. 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?
- 18. 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
- 19. 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
- 20. As shown in the figure, ∠ a = 35 °, C = 60 ° in △ ABC, BD bisection ∠ ABC, de ‖ BC intersecting AB with E, then ∠ BDE=______ ，∠BDC=______ .