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In the right triangle ABC, ∠ C = 90 ° cm ⊥ AB is in M In the right triangle ABC, ∠ C = 90 °, cm ⊥ AB is m, ab = 10, AC = 8, then BC =? Cm =? Area of triangle ABC =? Cm?
According to Pythagorean law
AC square + BC square = AB square
64 + BC squared = 100
BC square = 36
So BC = 6
Then the triangle area s = AC × BC = 8 × 6 = 48
And S = ab × cm
48= 10×CM
So: cm = 4.8
In the right triangle ABC, ∠ C = 90, cm and BN intersect at g, and cm is perpendicular to BN. If G is the center of gravity of ABC, BC = 2, find the length of BN
The key is to use the center of gravity to divide the center line into 2:1
Let BM = X
AC^2=4CN^2=4 NG.NB= (4/3)NB^2=(4/3)x^2
AB^2=4CM^2=4.((3/2)CG)^2=9CG^2=9 NG.BG=9 (1/3)(2/3)x^2=2x^2
2X ^ 2 = (4 / 3) x ^ 2 + 4 (Pythagorean theorem)
X = root 6
Given that the right triangle ABC, hypotenuse AB = 2, the minimum value of the sum of the distances from a moving point P to three vertices in the triangle is root 7, and find the size of two acute angles
∵ AC ⊥ BC, ᙽ point P coincides with point C
∴AP+BP+CP=b+a+0=√7
In addition, it is also found that B? A? 2 = C? = = > (a + b)? - 2Ab = C? = = > 7-2ab = 4 = = > AB = 3 / 2
/ / A, B are the two roots of the equation x 2 - √ 7x + 3 / 2 = 0
The solution is: x = (√ 7 ± 1) / 2
∴sinA=a/c=(√7+1)/4=0.9114===>∠A=65.7º,∠B=90-65.7=24.3º
Or ∠ B = 65.7 °, a = 24.3 °
It is known that the length of hypotenuse ab of right triangle ABC is 10cm, and the sine value of its two acute angles is Two roots of equation m (x ^ 2-2x) + 5 (x ^ 2 + x) + 12 = 0 (1) Find the value of M (2) Find the area of the inscribed circle of △ ABC
(1) M (x ^ 2-2x) + 5 (x ^ 2 + x) + 12 = 0 = = > (M + 5) * x ^ 2 - (2m-5) * x + 12 = 0 two X1 + x2 = (2m-5) / (M + 5) and X1 ^ 2 + x2 ^ 2 = 1, so there are - (2m-5) * (2m-5) / (M + 5) + 2 * 12 = 0 = = = > m = - 2 or M = 20 and (2m-5) ^ 2 > 4 * (M + 5) * 12, so m = 20 (2) Sina = 3 / 5, SINB = 4 / 5 or Si
(2011 · Anshun) in RT △ ABC, the oblique side AB = 4, ∠ B = 60 °, rotate △ ABC around point B by 60 ° and the path length of vertex C is () A. π Three B. 2π Three C. π D. 4π Three
Length of arc CC '= 60 π × 2
180＝2
3π．
Therefore, B
The angle of a B, O, which is parallel to the angle of a B, O, and intersects with the sharp edge of a triangle Is AE equal to BF? What is the reason?
EM ? BC crosses BC in M, connected with OM, ∵ EM ? EM ≓ ad, ? DOM = emo, CE is the bisector of C, AE = me, ≌ 87808780 8780≌ ome = Ooe, and ome = mod, OAE, 57; OAE, ? AEE ? EM
The probability that △ APC is an acute triangle is () A. 1 B. 1 Two C. 1 Three D. 1 Six
In the isosceles right triangle ABC, if AC length is 1, then AB length is
2，
If point P is on segment dB, then the condition is satisfied: △ APC is an acute triangle
∵|DB|=
Two
2，|AB|=
2，
The probability that △ APC is an acute triangle is 1
2．
Therefore, B
As shown in the figure, in △ ABC, ∠ BAC = 90 degrees, ab = AC, point E is on AB, CE is the hypotenuse of an isosceles right triangle DCE, and point D and point a are on the same side of CD The question is (1) is △ ACD similar to △ BCE? Why? (2) Is ad parallel to BC? Why?
(1) Δ ACD is similar to △ BCE
∵ △ ABC and △ Dec are isosceles right triangles
∴BC∶EC=AC∶CD
∠ACB=∠DCE=45°
∴∠ACB-∠ACE=∠DCE-∠ACE
∴∠BCE=∠ACD
∴△BEC∽△ADC
（2）AD‖BC
∵△BEC∽△ADC
∴∠DAC=∠B=∠ACB
∴AD‖BC
As shown in the figure, take the hypotenuse ab of the isosceles right triangle ABC as the edge to make the equilateral △ abd, connect DC, and make the equilateral triangle DCE with DC as the edge. Points B and E are on the same side of points c and D. if BC = 1, then be=
Solution ∵ ∵ abd is a regular triangle △ DCE is a regular triangle
∴AD=BD CD=ED
∵∠ADC+∠CDB=60°
∠CDB+∠BDE=60°
∴∠ADC=∠BDE
In △ ADC and △ BDE
AD=BD
∠ADC=∠BDE
CD=ED
∴△ADC≌△BDE（SAS）
∴AC=BE
∵ △ ABC is an isosceles right triangle
∴BC=AC=BE=1
As shown in the figure, in △ ABC and △ CDE, ab = AC = CE, BC = DC = De, AB > BC, ∠ BAC = ∠ DCE = ∠ α, points B, C and D are on the straight line L, and the drawing shall be made according to the following requirements (the drawing trace shall be reserved); (1) Draw the symmetry point e 'of point e with respect to line L, and connect CE' and de '; (2) Taking point C as the rotation center, rotate the △ CDE ′ obtained in (1) in a counter clockwise direction so that CE ′ coincides with CA to obtain △ CD ′ e ″ (a). Draw △ CD ′ e ″ (a) ① The position relation between line AB and line CD 'is______ . ② Find the degree of ∠α
(1) The drawing is as follows:; (2) the drawing is as follows:;; draw △ CD ′ e ″ (a), ① parallel, reason: ? DCE = ∠ DCE ′ = ∠ D ′ CA = ∠ α, ∵ BAC = ∠ D ′ CA = ∠ α, ∵ ab ∵ CD ′. ② ∵ quadrilateral ABCD ′ is isosceles trapezoid, ? ABC = ∠ D ′ AB = 2 ∠ BAC = 2 ∠ α, ∵ AB = AC
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