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If the square of a plus the square of B equals twice the square of C, then the minimum value of COSC is
Let A.B.C satisfy the conditions B ^ 2 + C ^ 2-bc = a ^ 2 and C / b = 1 / 2 + = √ 3 to find angles a and Tana
It is C / b = 1 / 2 + √ 3
According to the cosine theorem: A ^ 2 = B ^ 2 + C ^ 2-2bc * cosa
Because of the known condition: A ^ 2 = B ^ 2 + C ^ 2-bc
So 2cosa = 1
So cosa = 1 / 2
Because a is the inner angle of a triangle
So a = 60 degrees
tanA=√3
How to prove that the radii of inscribed circle of right triangle (∠ C = 90 °) are (a + B-C) / 2 and AB / (a + B + C)!
Let ∠ C = 90 ° in △ ABC, the radius of inscribed circle O is r, which is tangent to points D, e and f respectively with AB, BC and ac. it is proved that OECF is a square, then EC = FC = R, BD = be = bc-e-c = A-R, ad = AF = ac-fc = B-R, so C = AB = AD + BD = A-R + B-R, that is, C = a + b-2r, so r = (a + B-C) / 2
Proof of radius of inscribed circle of right triangle r = 2S / C
Such as the title
This
Prove by triangle congruence
The relationship between the radius of the inscribed circle and the three sides of a right triangle
R = (a + B-C) / 2 or r = AB / (a + B + C)
Exploration: in Figures 1 to 3, the area of △ ABC is known as a. (1) as shown in Figure 1, extend the edge BC of △ ABC to point D, so that CD = BC, connecting da. If the area of △ ACD is S1, then S1=______ (expressed by the algebraic formula containing a) (2) as shown in Figure 2, extend the edge BC of △ ABC to point D, extend the edge CA to point E, so that CD = BC, AE = Ca, connect de. if the area of △ Dec is S2, then S2=______ (expressed by the algebraic formula containing a) (3) on the basis of Figure 2, extend AB to point F, make BF = AB, connect FD and Fe, and get △ def (as shown in Figure 3). If the area of the shadow part is S3, then S3=______ (expressed by the algebraic formula containing a), and use the conclusion of (2) to write the reason. It is found that: as above, the edges of △ ABC are extended by one time in turn, and the end points are connected to get △ def (as shown in Figure 3). At this time, we call △ ABC expanded outward once. We can find that the area of △ def obtained after one expansion is the same as that of the original △ ABC______ Application: to plant flowers in a large enough open space, the engineers designed the following patterns: first, plant red flowers in the open space of △ ABC, and then expand △ ABC outward three times (the patterns of the first two expansion are shown in Figure 4). Plant yellow flowers in the first expansion area, purple flowers in the second expansion area, and blue flowers in the third expansion area The area of △ ABC is 10 square meters. Please use the above conclusion to calculate: (1) the area of purple flower; (2) the area of blue flower
(1) The height of BC and CD is equal, BC = CD, according to the equal area of the triangle with equal base and equal height, S1 = s △ ACD = a, so the answer is: A. (2) connect ad, similar to (1), according to the equal area of the triangle with equal base and equal height, s △ ACD = s △ ade = a, ∧ S2 = 2A, so the answer is: 2A. (3) similar to (2): s △ AFE = s △ BFD = s △ CDE = 2A, ∧ S3 = 2A + 2A+ 2A = 6a, so the answer is: 6A. (3) the area of yellow flower area is 6 × 10 = 60 square meters, the area of purple flower area is 6 × (60 + 10) = 420 square meters; the area of blue flower area is 6 × (420 + 60 + 10) = 2940 square meters
In △ ABC, the bisector of ∠ A is equal to 60 °, the bisector of ∠ B and ∠ C intersects P, extends BC to D, and the bisector of ∠ ACD intersects the extension of BP to e, so as to find ∠ BPC and angle E
∵∠ABC + ∠ACB = 120
∴∠PBC + ∠PCB = ∠ABC/2 + ∠ACB/2 = 60
∴∠BPC = 180 - 60 = 120
&Similarly, PCE = & nbsp; - ACB / 2 & nbsp; + & nbsp; - ACD / 2 & nbsp; = & nbsp; 90
From the external angle theorem, we know that ∠ BPC & nbsp; = & nbsp; ∠ PCE & nbsp; + & nbsp; ∠ E
∴∠E = 120 - 90 = 30
In the triangle ABC, D is the midpoint of AB and E is the midpoint of AC. the area of the triangle ade is 8 square meters. Find the surface of the shaded part bec
Why is s (ABC) equal to 32?
S(ADE)= 8
S(ABC)= 32
S(DEB)=8
S(BEC)=S(ABC)-S(ADE)-S(DEB)=16
The area of triangle ABC is 24 square centimeters, and be = 2ec, D and F are the midpoint of AB and CD respectively. What is the area of the shadow
Good score!
24 / 12 * 5 = 10 square centimeter
In △ ABC, BD = 13bc, the area of △ abd is 30cm2, and the area of △ ade is 10cm2. What is the area of shadow △ def?
If a is used as ah ⊥ BC over h, then s △ abd = 12bd · ah, s △ ADC = 12DC · ah, so s △ abd: s △ ADC = BD · ah: DC · ah = BD: DC (the ratio of triangle area with equal height = the ratio of bottom) because BD = 13bc, so BD = 12DC, so s △ abd: s △ ADC = 12, and because s △ abd = 30cm2, so s △ ADC = 60cm2. According to the ratio of triangle area with equal height = the ratio of bottom, s △ ade: s △ ADC = AE : AC because s △ ade = 10cm2, because s △ ade: s △ ADC = 16, so AE: AC = 16, so AE = 16ac, so AE = 15ce, if DG ‖ AC intersects g through D, then DG: CE = BD: BC = 13, so DG = 13ce, and because AE = 15ce, so DG: AE = 53, because DG ‖ AC, that is DG ‖ AE, so DF: AF = DG: AE = 53, so DF: ad = 58, so according to the ratio of triangle area with equal height = bottom Compared with s △ def: s △ ade = DF: ad = 58, because s △ ade = 10 cm 2, s △ def = 254 (cm 2). A: the area of shadow △ DEF is 254 cm 2
RELATED INFORMATIONS
- 1. It is known that: as shown in the figure, the circle O with the diameter of AB on one side of the equilateral triangle ABC intersects with the sides AC and BC at points D and e respectively, DF ⊥ BC is made through point D, and F is the perpendicular foot (1) Proof: DF is tangent of circle O (2) If the side length of equilateral triangle ABC is 4, find the length of DF (3) Finding the area of the shadow in a graph The first question is not, the shadow area is DFE
- 2. In △ ABC, D and E are on the lines AB and AC respectively, ad = 2, BD = 5, EA = 4, EC = 10, Ed: BC=
- 3. It is known that CD is the height of AB side in △ ABC, AC + BC = 8, CD = 2, AE is the diameter of circumscribed circle of △ ABC, (1) try to explain △ CBD ∽ AEC, (2) let AC = x, AE = y, find the functional relation of y with respect to x, and (3) find the maximum value of Y
- 4. As shown in the figure, in equilateral △ ABC, CD intersects be at point F, ∠ BFC = 120?: ad = CE
- 5. In 3-angle ABC, the angle BAC is 90 degrees, ad is perpendicular to BC, EF is perpendicular to BC, FM is perpendicular to AC, the perpendicular foot is D, F, m respectively, angle 1 = angle 2, verification: FM = FD
- 6. As shown in the figure, it is known that CD is perpendicular to ab at point D, be is perpendicular to AC at point E, and CD intersects be at point o (1) If OC = ob, prove that point O is on the bisector of angular BAC (2) If point O is on the bisector of angle BAC, OC = ob is proved
- 7. In the triangle ABC, ab = AC, D is a point on BC, and
- 8. As shown in the figure, in RT △ ABC, ab = AC, ∠ BAC = 90 °, ∠ 1 = ∠ 2, CE ⊥ BD is extended to E. verification: BD = 2ce
- 9. As shown in the figure, in the triangle ABC, the angle ACB = 90 °, CD is perpendicular to AB, the perpendicular foot is D, and the angle a = 60 ° to prove BD = 3aD
- 10. As shown in the figure, in the triangle ABC, the angle ACB is 90 degrees, the angle CD is high, and the angle a is 30 degrees
- 11. In the triangle ABC, a = 120 °, ab = 5, BC = 7, then the value of sina / SINB
- 12. In the right triangle ABC, C is equal to 90 degrees, AC is equal to twice BC, then the value of sina is?
- 13. In the right triangle ABC, angle c is right angle, angle B ∶ angle c = 1 ∶ 4, find the degree of angle A The middle equation is still missing
- 14. If the angle c = 90 degrees, (a + b) * (a + b) = 12 and C = 2 in the right triangle ABC, then the s triangle ABC =?
- 15. The three sides of triangle are equal difference series, the circumference is 36, and the circumference of inscribed circle is 6 π Some people do that ∵ the three sides grow into an arithmetic sequence with a circumference of 36 Let the three sides be 12-n, 12, 12 + n respectively The circumference of the inscribed circle is 6 π --- radius of the inscribed circle r = 3 --- triangle area = (36 / 2) * 3 = 54 From Helen's formula: 54 & sup2; = 18 * 6 * (6 + n) * (6-n) --->27=36-n²--->n²=9--->n=3 --->The lengths of the three sides are 9, 12 and 15 respectively ∵ 9 & sup2; + 12 & sup2; = 15 & sup2;, ∵ the triangle is a right triangle Then why must one side be 12?
- 16. Three straight lines PA, Pb, PC, ∠ APC = ∠ APB = 60 ° and ∠ BPC = 90 ° in space are used to calculate the dihedral angle b-pa-c
- 17. As shown in the figure, in △ ABC, ab = AC, e is on AC, and the extension line of ad = AE, de intersects BC at point F
- 18. If alpha is equal to - 3, then which quadrant is the end of alpha?
- 19. In known △ ABC, a = 30 °, C = 45 °, a = √ 2, solve triangle There are two more questions, please master solution, online and so on. 2. It is known that in △ ABC, a = √ 2, B = 2, a = 30 ° to solve the triangle. 3. In △ ABC, a = 1, B = √ 3, B = 60 ° to solve the triangle.
- 20. As shown in the figure, ∠ ABC = ∠ CDB = 90 °, AC = a, BC = B, when BD and a, B meet what kind of relationship, △ ABC ∽ Δ CDB?