If the central line and the high line on the hypotenuse of a right triangle are 6cm and 5cm respectively, then its area is

If the central line and the high line on the hypotenuse of a right triangle are 6cm and 5cm respectively, then its area is

Let ABC, AC be the hypotenuse, BD the middle line on the hypotenuse, be the height on the hypotenuse, then:
AC=2BD
S=AC×BE/2
=BD×BE=30

Given that the high line and the middle line on the hypotenuse of a right triangle are 5cm and 6cm respectively, the area of a right triangle is ()
And please explain why

The center line on the hypotenuse is half the length of the hypotenuse
So the hypotenuse is 12cm long
The area is 1 / 2 * 5 * 12 = 30cm ^ 2

The length of the hypotenuse of a right triangle is 12cm, the length of one right triangle is 8cm, and the length of the other right triangle is 8cm_______ cm?

4 times the root 5

Finding the circumference of right triangle according to Pythagorean theorem
If one right side of a right triangle is 11 and the other two sides are integers, the perimeter of the triangle is?

Let the hypotenuse be x and the other right angle be y
According to Pythagorean theorem: x ^ 2-y ^ 2 = 11 ^ 2
That is, (x + y) * (X-Y) = 121
If C = x + y + 11, you must know X-Y
According to the triangle theorem x > y and X-Y

It is known that the perimeter of a right triangle is 12cm, and the sum of two right sides is 7cm. What is the area of the triangle

The girth is 12, the sum of right angles is 7, and the hypotenuse is 5. Then let the two right angles X and y be equal to x + y = 7, the square of X + the square of y = 25. The solution is x = 4, y = 3, area = 6

What is the area of a right triangle with a circumference of 12cm and a right side of 4cm

Let the other right angle side be x cm
x^2+4*4=(12-4-x)^2
x=3
The area is 4 × 3 △ 2 = 6 square centimeters

If the perimeter of a right triangle is 12cm, and the length of a right side is 4cm, what is its area

Let the other right angle side be x x + 4 = (12-4-x) x + 16 = 64-16x + X 16x = 48 x = 33 * 4 / 2 = 6, so the area is 6 square centimeters

The perimeter of a right triangle is 12cm, and the area of a corner is 30 degrees

Let the small right angle side be x, then the hypotenuse side be 2x, and the other right angle side be √ 3x,
According to the meaning of the title:
X+2X+√3X=12
X=12÷(3+√3)
X=6-2√3,
∴S=1/2X*√3X=√3/2*(6-2√3)^2=√3/2(48-24√3)=24√3-36.

Rectangular ABCD and point P, when point P is in the position of Figure 1, then there is a conclusion: s △ PBC = s △ PAC + s △ PCD reason: point P is an ef vertical BC, and AD and BC are at two points E and f respectively. \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\p AC + s △ PCD. Please refer to the above information. When point P is located in Figure 2 and figure 3 respectively, what is the quantitative relationship among s △ PBC, s △ PAC and s △ PCD? Please write down your conjecture about the above two cases and choose one of them to prove

Conjecture results: Fig. 2 Conclusion s △ PBC = s △ PAC + s △ PCD Fig. 3 conclusion s △ PBC = s △ PAC-S △ PCD (2 points) proves that: as shown in Fig. 2, through point P for EF vertical ad, ad, BC at two points E and F, respectively, ∵ s △ PBC = 12bc · PE + 12bc · EF & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; (1 point) = 12ad · PE + 12bc · EF = s △ Pad + 12s rectangular ABCD (2 points) ∵ s △ PAC + s △ PCD = s △ Pad + s △ ADC = s △ Pad + 12s rectangular ABCD (2 points) ∵ s △ PBC = s △ PAC + s △ PCD (1 point). If the conclusion of Figure 3 is proved, the score can be given according to the above scoring standard

In p-abc, M. n is the side edge PB.PC If the section amn is perpendicular to the side PBC, calculate the tangent of the angle of the side edge

Therefore, PA = AE = √ 3 / 2 through P as po ⊥ AE to o, then Po is the height of pyramid, from OA = √ 3 / 3 to Po = √ [(√ 3 / 2) 2 - (√ 3 / 3) 2] = √ 15 / 6, ∠ Pao is the angle between PA and plane ABC ⊥ Tan}