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m. 6. As shown in the figure, CD is the height on the hypotenuse of the right triangle ABC. Obviously △ ACD is similar to △ CBD. Given that ad = 9cm, BD = 4C, the length of CD is cm
Let CD = X
Because △ ACD is similar to △ CBD
So ad / CD = CD / BD
Because ad = 9cm, BD = 4cm
So 9 / x = x / 4
X = 6
In a right triangle, ∠ C = 90 °, a = 30 ° BD is an angular bisector BC = 30 of ∠ ABC. Find the length of AD
∵∠A=30° ∠C=90°
∴∠B=60°
And ∵ BD is the angular bisector of ∵ ABC
﹤ abd = ∠ DBC = 30 ° and ∠ BDC = 60 °
∴AD=BD
∵BD=30
∴Sin∠BDC=BC／BD
∵ sin ∠ BDC = sin60 ° = (root 3) / 2
ν (radical 3) / 2 = 30 / BD
/ / ad = BD = 20 × (root 3)
As shown in the figure, RT triangle ABC ~ RT triangle def, cm and en are the midlines on the hypotenuse AB and DF respectively, Given that AC = 9cm, CB = 12cm, de = 3cm, find: (1) the length of CM and en; (2) what relationship do you find between cm / en and similarity ratio? What conclusions can be drawn
AB ^ 2 = AC ^ 2 + CB ^ 2 = 15 ^ 2, so AB = 15, because the length of the center line on the hypotenuse of a right triangle is half of the length of the hypotenuse, CM = 15 / 2
RT triangle ABC ~ RT triangle def, corresponding side is proportional, EF = 4, DF = 5. En = 5 / 2
Cm / en = 3 = AC / de = BC / EF = AB / DF
As shown in the figure, rotate the RT triangle ABC by 40 degrees anticlockwise around point a to get the right triangle AB, C, and point C, which fall right on side ab As shown in the figure, rotate the RT triangle ABC by 40 degrees anticlockwise around point a to get the right triangle AB, C, and point C, which fall right on side AB and connect BB, then what are the angles BB, C?
Because the point C falls on AB, so ∠ bad = ∠ bab ′ = 40 °∠ ab ′ C ′ = 90 ° - 40 ° = 50 °
Ab ′ = AB triangle ABB ′ is isosceles triangle ∠ ab ′ B = (180 ° - ∠ bab ′) / 2 = (180 ° - 40 °) / 2 = 70 °
∠BB′C=∠AB′B-∠AB′C′=70°-50°=20°
As shown in the figure, in RT △ ABC, ∠ C = 90 ° and three isosceles right triangles are made with the three sides of RT △ ABC as the oblique sides, where ∠ h, ∠ E and ∠ f are right angles. If the hypotenuse AB = 3, the area of the shadow part in the graph is () A. 1 B. 2 C. 9 Two D. Thirteen
In the right angle △ ABC, ∠ C = 90 °, AB2 = ac2 + BC2,
According to the area calculation method of isosceles right triangle, the area of △ AEB is 1
2×AB•1
2AB=AB2
4，
The area of △ AHC is 1
2×AC•1
2AC=AC2
4，
The area of △ BCF is 1
2×BC•1
2BC=BC2
4，
The shadow area is 1
4（AB2+AC2+BC2）=1
2AB2，
∵AB=3，
The shadow area is 1
2×32=9
2．
Therefore, C
In RT △ ABC, ∠ C = 90 ° is known as AC = 6cm, BC = 8cm (1) Find the length of the center line cm on AB edge; (2) Point P is a moving point on the line cm (point P does not coincide with point C and point m). The functional relationship between the area y (square centimeter) of △ APB and the length x (centimeter) of CP is obtained, and the definition domain of the function is obtained; (3) Is there such a point P that the area of △ ABP is 3% of that of ACBP 2？ If yes, the length of CP is requested; if not, please explain the reason
(1) C = 90 °, AC = 6cm, BC = 8cm, AB = 62 + 82 = 5 (CM). In a right triangle, according to the property that the length of the center line of the hypotenuse is half of the length of the hypotenuse, ? cm = 12ab = 5 (CM); (2) ? CP = x, CM = am, ? cab = ? sin ? cab = 45, ? s
As shown in the figure, fold the right angle △ ABC paper with right angle sides AC = 6cm, BC = 8cm, so that point B coincides with point a, and the crease is De, then CD is equal to () A. 25 Four B. 22 Three C. 7 Four D. 5 Three
Let CD = x, then de = 8-x,
∵△ BDE is formed by ade turning along the straight line De,
∴AD=BD=8-x，
∵ △ ACD is a right triangle,
/ / ac2 = ad2-cd2, i.e. 62 = (8-x) 2-x2, x = 7
4．
Therefore, C
In a right triangle ABC, the lengths of the two right sides AB and BC are 6cm and 8cm respectively, and the height of the hypotenuse AC is__ CM
Four point eight
Firstly, AC ^ 2 = 6 ^ 2 + 8 ^ 2, AC = 10 is obtained by Pythagorean theorem
Then we know from the area that 1 / 2 ab times BC = 1 / 2 AC times height H
The solution H = 4.8
In the right triangle ABC, if the angle c is equal to 90 degrees, the angle a is equal to 30 degrees, and BC = 6cm, then AB =? AC =?
According to Pythagorean theorem, ab = 12, AC = radical (108)
In the right triangle ABC, the angle c is 90 degrees, AB is equal to 5cm, AC is equal to 4cm, BC is equal to 3cm. Make the inscribed circle of the triangle ABC to find the radius of the inscribed circle
Let the center of the inscribed circle be o and the radius R
S△ABC＝S△AOB+S△BOC+S△AOC
1 / 2 (3 * 4) = 1 / 2 (3R + 4R + 5R)
So r = 1
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