Given that the length of the hypotenuse of a right triangle is 2 and the circumference is 2 + root 6, the area of the triangle is calculated

one-second
Let the unknown number solve
X + y = root 6
X + y squared = 4
Solvable xy = 1
The area is 1 / 2 XY

How to find the area and the length of the oblique side of a right triangle with a 75 degree angle and a known hypotenuse? RT

This is a special method; given that RT △ ABC, ∠ C = 90 ° and B = 75 ° is known, take the point D on the opposite side AC of the angle of 75 ° so that Da = dB, then ∠ BDC = 30 ° and set BC = m, then Da = DB = 2m. DC = m of radical 3, of course, is easy to find

In RT △ ABC, ∠ ACB = 90, center line CD = 1 on hypotenuse AB, perimeter of △ ABC is 2 + √ 6, and area of △ ABC is calculated

Because CD is the height on the hypotenuse of a right triangle
So 2CD = AB = 2
Because AC + BC = 2 + √ 6-2 = √ 6
And AC 2 + BC 2 = AB 2 = 4
（AC+BC）²-（AC²+BC²）=2AC×BC=2
So AC = 1 × 1
S△=1/2AC×BC=1/2

Given that the circumference of RT △ ABC is 5 and the length of hypotenuse AB is 2, then the area of RT △ ABC is

Let x, x ^ 2 + (3-x) ^ 2 = 4
2x^2-6x+5=0
Discriminant

Given that the hypotenuse of the RT triangle ABC is 2%, and the perimeter is 2 + √ 6, calculate its area

Given that the hypotenuse of the RT triangle ABC is 2%, and the perimeter is 2 + √ 6, calculate its area
a+b=2+√6-2=√6
a^2+b^2=2^2
4S=2ab=(a+b)^2-(a^2+b^2)=2
s=1/2

The area of the triangle in the figure is 12 square centimeter

A vertical line passing through the vertex of the triangle intersects at 0,
The area of the triangle is 2R × R △ 2 = R2 = 12,
Semicircle area π R2 △ 2,
=3.14×12÷2，
=18.84 (square centimeter),
So the area of the shadow is:
84-12 = 6.84 (square centimeter);
A: the shadow area is 6.84 square centimeters

As shown in the figure, the area of the right triangle is 12 square centimeters, then the area of the shadow part is ------ the detailed process

[analysis] obviously, the shadow area is the area of a semicircle minus a right triangle. In isosceles RT △ ABC, ∠ a = 45 ° and point D is the point where the semicircle with BC diameter intersects the hypotenuse AB, BDC = 90 °, DCB = 45 °, △ CDB is also isosceles RT △ and ∵ CD is △ CDB and △ CDA

Given that the area of a right triangle is 12 square centimeters, calculate the shadow area

Let the right angle side length of a right triangle be a, half of the square of a = 12, the square of half of 24 A = 6
Let the arcuate area of a right triangle be s, then the shadow area = 3S
If the arch of the lower left corner is removed from the semicircle, the area of the semicircle - half of the area of the right triangle = 2S
The square of half of 3.14xa x0.5-6 = 2S 3.14x6x0.5-6 = 2S s = 1.71
Shadow area = 3S = 5.13

Given that the area of the triangle ABC is 12 square centimeters, calculate the area of the shadow part?

Let the right sides of the isosceles right triangle ABC be a, 12a2 = 12, A2 = 24; the area of sector abd: 18 π A2 = 3 π = 9.42 (square centimeter), the area of blank part BCD: 12-9.42 = 2.58 (square centimeter), the area of semicircle: 12 π (a △ 2) 2 = 12 × 3.14 × 14a2 = 18 × 3.14 × 24 = 9.42 (square)

It is known that the area of the right triangle ABC is 12 square centimeters. Find the area of the shadow part. Don't use letters, It must be a formula. It must be a formula. It must be a formula

Analysis: the current trefoil shadow is actually composed of two small shadows,
Two small shadow areas = semicircle area - small RT triangle area
If the side length of the large right triangle is 6 cm, the side length of the small right triangle is 3 √ 2 cm
S shadow = 2 [semicircular area small RT triangle area] = 2 [π 3 2 / 2 - (3 √ 2) 2 / 2]
=2[π9/2-18/2]=π9-18=28.26-18=10.26 cm² 10.26÷2=5.13cm²

Given that the length of the hypotenuse of a right triangle is 2 and the circumference is 2 + root 6, the area of the triangle is calculated