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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?
If △ ABC ∽ CDB, then accb = bcdb, that is ab = BBD, ∽ BD = B2A, when BD = B2A, ∽ ABC ∽ CDB
All isosceles right triangle ABC is similar to isosceles right triangle a'b'c ', the similarity ratio is 3:1, the hypotenuse AB = 5cm
1 find the length of the hypotenuse a'b'of the triangle a'b'c '. 2 find the height of the hypotenuse a'B'
1
AB/A'B'=3/1
A'B'=AB/3=5/3
two
The height of isosceles right triangle ABC is 2.5
And then according to the similarity ratio
The height of similar triangle is 2.5 / 3 = 5 / 6
In the right triangle ABC, the angle c = 90 degrees, CD is perpendicular to AB and D, ask CD ^ 2 =?
AC ^ 2 =?, BC ^ 2 =?, AC times BC =?
It should be seen from the proportional line segments of a right triangle
CD^2=AD×BD
AC^2=AD×AB
BC^2=BD×AB
It can be deduced from similar triangles
AC×BC=CD×AB
Equal area
For the inscribed circle of triangle ABC, the radius of the circle is r, and the circumference of the circle is 1?
Set heart as I, connect IB, IA, IC
Then s △ ABC = 1 / 2 (AB + BC + AC) r = 1 / 2 * LR
Triangle ABC is isosceles right triangle, waist length is 1, find the inscribed circle radius of triangle ABC
The radius of the inscribed circle of the triangle is r = 2S △ C, where s is the area of the triangle and C is the perimeter of the triangle
r=2*0.5*1*1/（1+1+√2）=（2-√2）/2
Please accept
If the two right sides of a right triangle are A.B, the radius of circumscribed circle is r, and the radius of inscribed circle is r, then the relationship between a.b.r.r is ()
A.R+r=1/2（a+b） B.a+b=1/2（R+r） C.R+r=a+b D.R+r＞a+b
If the inscribed circle of RT △ cab is tangent to D, e and f respectively with AC, AB and BC, and the center of the circle is I, then the quadrilateral CDIE is a square, CD = CF = R, ad = AE, be = BF, (the outer circle of the circle is equal to the two tangent lines), let the perimeter = P = a + B + C, 2R + AD + AE + be + BF = P, 2R + 2C = P = a + B + C, ∧ r = (a + B-C) / 2, ∧ RT △ circumscribed circle center is oblique
If the two right sides of a right triangle are 3 and 4, what is the radius of its inscribed circle? What is the distance between the inner and outer center?
R * (3 + 4 + 5) / 2 = 3 * 4 / 2 r = 1 with another formula: r = (3 + 4-5) / 2 (only for right triangle) r outer = 5 / 2 r outer - r = 1.5
As shown in figure 5-z-15, the triangle DEF is obtained by moving the triangle ABC 2.5cm horizontally to the right. Please make the triangle ABC,
Find the perimeter of quadrilateral abed
As shown in the figure: abed is a parallel polygon
It's 2.5 on one side and 3 on the other
The circumference is (2.5 + 3) × 2 = 11cm
Triangle DEF is obtained by moving triangle ABC 2.5cm horizontally to the right. Please make triangle ABC. If ed = 3cm, find the perimeter of quadrilateral abed
11.
In the triangle ABC, the following price adjustments are known: a = 45 ° C = 30 ° C = 10m; (2) a = 60 ° B = 45 ° C = 20cm
B = 180 ° - a-c = 105 ° by sine theorem B / SINB = C / sinc = A / Sina, so 10 / (1 / 2) = A / [(radical 2) / 2] = B / [(radical 6 + radical 2) / 4] so a = [10 * radical 2] m b = [5 * (radical 6 + radical 2)] MC = 180 ° - A-B = 75 ° by sine theorem B / SINB = C / sinc = A / Sina
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