If the ratio of the two right sides of a right triangle is 3:4 and the length of the hypotenuse is 20, then its area is______ .

If the ratio of the two right sides of a right triangle is 3:4 and the length of the hypotenuse is 20, then its area is______ .

According to the meaning of the title, let the two right angles be 3x and 4x,
Then (3x) 2 + (4x) 2 = 202,
X = 4, so the two right angles are 12, 16;
One
2×12×16=96，
So its area is 96

The three sides of a right triangle are 6cm, 8cm and 10cm respectively. The height of the longest side of this triangle is () cm

The area of the triangle is
6 × 8 △ 2 = 24 (square centimeter)
The height on the longest side of this triangle is
24 × 2 △ 10 = 4.8 (CM)

The length of the three sides of a right triangle is 6cm.8cm.10cm. What is the area of the triangle? What is the height of the hypotenuse? Use the knowledge of proportion

6 * 8 / 2 = 24cm square
Let the height on the bevel be xcm
6*8=10X
X=4.8

The two right sides of a right triangle are 6cm and 8cm respectively, and the hypotenuse is 10cm A. 2.4 cm B. 4.8 cm C. 6 cm D. 1.2 cm

Let the height on its bevel be x cm,
10x÷2=6×8÷2，
   10x=48，
     x=4.8；
A: its height on the bevel is 4.8 cm;
Therefore, B

The length of three sides of a right triangle is 10 cm, 8 cm and 6 cm, and the height on the inclined side is______ Centimeter

Area of triangle: 6 × 8 △ 2,
=48÷2，
=24 (square centimeter);
Height on bevel: 24 × 2 △ 10,
=48÷10，
=8 (CM)
A: the height on the bevel is 4.8 cm
Therefore, the answer is: 4

a. B, C are the length of the three sides of a right triangle, h is the height of the hypotenuse C. prove a + B

Triangle area 2S = a * b = C * h
So a * b = C * h; - - 1
C^2

If the length of the two right angles of a right triangle is a, B, the length of the hypotenuse is C, and the height of the hypotenuse is h, it is proved that a + B < C + H We also want quality

It is proved that the Pythagorean theorem gives a 2 + B 2 = C 2
Area s = 1 / 2Ab = 1 / 2CH ν AB = Ch
∴(a+b)²-(c+h)²=a²+2ab+b²-c²-2ch-h²=-h²

In the right triangle ABC, the side lengths of the three squares are a, B, C respectively. It is proved that a + C = B See search for details,

Your question is: in RT △ ABC, the angle a = 90 ° with BC (oblique side) as the bottom of the inscribed square (that is, a square is inscribed on the hypotenuse), intersects AB with E, AC with H, and the intersection with BC side from left to right is f, G. let the length of this square be a, and then in RT △ EBH and RT △ HGC, the intersection point of the square is f and G from left to right, Let two squares be inscribed on their right angles respectively. Let the side lengths of these two squares be B and C respectively. The relationship between them is proved
Proof: using similar triangles
It is easy to prove that all the small right triangles in this large right triangle are similar
(if your mathematical foundation is OK, I will omit the proofs of similar triangles and use them as conditions directly.)
Let the intersection of the inscribed square in RT △ EBF and EB be m and N respectively
Let the intersection point of the inscribed square in RT △ HGC and Hg be p and Q respectively
Then RT △ EMN ∽ RT △ qhp
MN/HP=EN/QP
That is, B / (A-C) = (a-b) / C
(a-c)(a-b)=bc
a^2-(b+c)a=0
a(a-b-c)=0
a≠0
∴a=b+c

Let the hypotenuse of a right triangle be C and the length of the two right angles is a, B respectively. It is proved that a + B ≥ 2 times C under the root

A + B and √ (2) C
(a + b) ^ 2 and 2C ^ 2
A ^ 2 + 2Ab + B ^ 2 and 2C ^ 2
2Ab and C ^ 2
2Ab and a ^ 2 + B ^ 2
So, 2Ab = so, a + B = < √ (2) C
Note: because (a-b) ^ 2 > = 0
So, a ^ 2 + B ^ 2 > = 2Ab
Ask me if you don't understand

If a, B, C are the lengths of the three sides of a right triangle, C is the length of the oblique side, and the height of the hypotenuse is h, it is proved that C + H > A + B

If the square of both sides, plus C2 = A2 + B2, then ch = AB can be obtained from the equal area, and H 2 > 0 can be obtained from the reduction of both sides