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In RT △ ABC, ∠ ACB = 90 ° and CD is the height on the edge of AB, ab = 13cm BC = 12cm AC = 5cm 1. Calculate the area of △ ABC 2. The length of CD
Δ ABC area = 12 × 5 △ 2 = 30
CD=12×5/13=60/13
As shown in the figure, in the right triangle ABC, ∠ C = 90 °, AC = 12, BC = 5, then the area of the semicircle with ab as its diameter is___ .
∵∠C=90°，AC=12，BC=5，
∴AB=
AC2+BC2=
122+52=13，
The area of semicircle with ab as diameter = 1
2π（AB
2）2=1
2π（13
2）2=169
8π．
So the answer is: 169
8π．
In the RT triangle ABC, the angle c = 90 degrees, AC = 5, ab = 7. Draw a semicircle with BC as the diameter and calculate the area of the semicircle
Does Pythagorean theorem calculate BC length? BC is diameter, you are lazy
If the perimeter of a right triangle is 2 + radical 6, and the length of the hypotenuse is 2, then the area of the triangle is___________ .
A + B = 2 + Radix 6-2 = Radix 6
(a + b) ^ 2 = a ^ 2 + 2Ab + B ^ 2 = 61
A ^ 2 + B ^ 2 = 4 2
Type 1-2
2ab=2
ab=1
The area of triangle is 1 / 2Ab = 1 / 2
In RT △ ABC, ∠ C = 90 °, Sina = 3 / 5, BC = 15, find the perimeter and area of SINB, △ ABC
In RT △ ABC,
Because: ∠ C = 90 °
Therefore, Sina = BC / AB = 3 / 5
Because BC = 15
So: ab = 25
Using Pythagorean theorem, we can get: AC = 20
So: SINB = AC / AB = 4 / 5
Perimeter C = AB + BC + AC = 60
Area s = AC * BC / 2 = 150
In a triangle, ∠ C = 90 °, Sina = 4 / 5, ab = 20, find the perimeter of the triangle ABC and the value of Tana
Because the angle c is 90 degrees
So Sina / 5 = 5
Because AB = 20
So BC = 16
AC²+BC²=AB²
So 20? 16? 2 = 12
Then AB = 12
So the circumference is 12 + 12 = 16
2、tanA=BC/AC=16/12=4/3
If the radius of the inscribed circle of RT △ ABC with 4cm hypotenuse is 2cm, what is the perimeter of the right triangle
According to the formula of radius of inscribed circle of right triangle: r = 1 / 2 * (a + B + C), where a, B, C are two right angle sides and oblique sides, so substituting the data into the formula, a + B-4 = 2 * 4, so a + B = 12, so perimeter = a + B + C = 12 + 4 = 16
Well, it makes sense upstairs. Such a triangle doesn't exist
Upstairs - I just said... Before I did not look at the formula carefully, but I said later that... Such a triangle does not exist
In addition, formulas and deformation formulas are used to cover, as long as the preconditions are considered. Otherwise, there is no significance of existence
In the right triangle ABC, ∠ C = 90 °, ab = 10cm, and the radius of inscribed circle is 2cm, then the circumference of the triangle is_ Please write down the analysis process and answer briefly
The circumference of the circle is 24cm. Make the vertical lines of the three sides through the center of the inscribed circle. The center of the circle O, the vertical line intersects AB to e, CB to F, AC to g, triangle OGA = triangle OEA, triangle OFB = triangle OEB, so Ag = AE, FB = EB, and AE + EB = AB = 10oC bisection ∠ ACB, so the triangle OGC = triangle OFC are isosceles right triangle, so GC = go
Right triangle ABC, find the radius of inscribed circle In the right triangle ABC, the angle c = 90 degrees, AC = 6, BC = 8, to find the radius of inscribed circle; 2. The radius of two equal inscribed circles which are tangent to two right angles and two circles are circumscribed; 3. For N equal inscribed circles, calculate the radius of inscribed circle Where are the people who can do it? You should show up soon! I'm still waiting for you! Looking forward to you!
Take a look at my answer, I was the champion of Science in our county's high school entrance examination. (1) (8-r) + (6-r) = 10, the answer is r = 2 (2) (8-r-2r * 4 / 5) + (6-r-2r * 3 / 5) = 10-2r, r = 10 / 7 (3) [8-r-2 * (n-1) r * 4 / 5] + [6-r-2 * (n-1) r * 3 / 5] = 10-2 (n-1) r = 10 / (2n + 3) the third one can be
If the radius of the inscribed circle of a right triangle ABC is r tangent (a + b) square = C square + 10 △ ABC perimeter is 6, then r =?
Because the circumference of △ ABC is 6, so a + B + C = 6, so a + B = 6-C, if a + B = 6-C, is taken into (a + b) 2 = C 2 + 10, the solution is: C = 13 / 6, so a + B = 6-C = 6-13 / 6 = 23 / 6, so r = (a + B-C) / 2 = (23 / 6-13 / 6) / 2 = (10 / 6) / 2 = 5 / 6
RELATED INFORMATIONS
- 1. In the RT triangle ABC, the right angle side AB, the oblique side C, and a, B, C form the arithmetic sequence, area = 12, find the perimeter?
- 2. As shown in the figure, in the triangle ABC, BC = 8 cm, ad = 6 cm, e and F are the midpoint of AB and AC respectively______ Square centimeter
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