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In the RT triangle ABC, the angle ACB = 90 degrees, ab = 4, take AC and BC as the diameters to make semicircles, and the areas are denoted as S1 and S2 respectively?
Three methods: 1 R = AC / 2, r = BC / 2S1 + S2 = π (AC / 2) 2 / 2 + π (BC / 2) 2 / 2 = π (ac2 + BC2) / 8 = π AB2 / 8 = 2 π π method 2 S1 + S2 = π (AC / 2) 2 / 2 + π (BC / 2) 2 / 2 = π (ac2 + BC2) 2 / 2 = π (ac2 + BC2) / 8 = π AB2 / 8 = 2 π = 2 π method 3 set AC = x BC = ys1 = 0.5 × π × (x ×2) 2s2 = 0.5 × π × (x × (2) 2s2 = 0.5 × π × (x × (x × (2) 2s2 = 0.y △ 2
As shown in the figure, it is known that in RT △ ABC, ∠ ACB = 90 ° and ab = 4. Take AC and BC as the diameters to make semicircles respectively, and the areas are recorded as S1 and S2 respectively, then S1 + S2 is equal to______ .
S1=1
2π（AC
2）2=1
8πAC2，S2=1
8πBC2，
So S1 + S2 = 1
8π（AC2+BC2）=1
8πAB2=2π．
So the answer is: 2 π
As shown in the figure, it is known that in RT △ ABC, ∠ ACB = 90 ° and ab = 4. Take AC and BC as the diameters to make semicircles respectively, and the areas are recorded as S1 and S2 respectively, then S1 + S2 is equal to______ .
S1=1
2π（AC
2）2=1
8πAC2，S2=1
8πBC2，
So S1 + S2 = 1
8π（AC2+BC2）=1
8πAB2=2π．
So the answer is: 2 π
If we know that the circumference of RT △ ABC is 4 + radical 48, and the length of the midline on the hypotenuse is 2, then s △ ABC = ()
Because the center line on the hypotenuse is 2, the hypotenuse is 4
So the sum of the two right angles is 4, 3
Let one be x and the other y
So x + y = 4 root sign 3
x. The sum of the squares of Y is 4
So (x + y) ^ 2 - (x ^ 2 + y ^ 2) = 2XY
3^2-4=2xy
xy=1
So s △ ABC = (1 / 2) * xy = 1 / 2
If we know that the circumference of RT △ ABC is 6 + 2, radical 3, and the median length on the hypotenuse is 2, then s △ ABC=_____
Because the center line on the hypotenuse is 2, the hypotenuse is 4
So the sum of the two right angles is 2 + 2 radical 3
Let one be x and the other y
So x + y = 2 + 2 root sign 3
x. The sum of the squares of Y is 4
So (x + y) ^ 2 - (x ^ 2 + y ^ 2) = 2XY
(2 + 2 radical 3) ^ 2-4 = 2XY
Xy = 6 + 4 root sign 3
So s △ ABC = (1 / 2) * xy = 3 + 2 radical 3
If we know that the circumference of RT △ ABC is 4 + radical 4, and the length of the midline on the hypotenuse is 2, then s △ ABC = Root 4 because it should be 4 root 3
Because the center line on the hypotenuse is 2, the hypotenuse is 4
So the sum of the two right angles is 2
Let one be x and the other y
So x + y = 2
x^2+y^2=16
So (x + y) ^ 2 - (x ^ 2 + y ^ 2) = 2XY
2^2-16=2xy
In the RT triangle ABC, ∠ C = 90 ° and the circumference is 6 + 2 times the root sign. 3. If the median line CD on the hypotenuse side is 2 cm, what is the area of the RT triangle ABC?
The center line on the hypotenuse is 2, so the length of the hypotenuse is 4
Let two right angles be a and B respectively
There are a + B = 2 + 2 √ 3, a 2 + B = 4 2 = 16
（A+B）²-（A²+B²）=2AB=8√3,AB=4√3
The triangle area is 1 / 2Ab = 2 √ 3 square centimeter
In △ ABC, ∠ C = 90 ° and the circumference is (5 + 2) 3) If CD = 2cm, then the area of RT △ ABC is 0___ .
∵ in ᙽ ABC, ᙽ C = 90 ° and the center line CD on the inclined side is 2 cm,
The length of hypotenuse C is: 4,
The sum of the two right angles is: a + B = 1 + 2
Three
∵a2+b2=c2=16
（a+b）2=a2+b2+2ab
∴2ab=（1+2
3）2-16=4
3-3，
ν RT △ ABC area = ab
2=4
3-3
4，
So the answer is: 4
3-3
4．
If we know that the circumference of the RT triangle ABC is the root of 4 + 3, and the length of the median line on the hypotenuse is 2, then the s triangle ABC =?
It seems that the center line on the hypotenuse of the RT triangle is half of the hypotenuse, so the hypotenuse is 4, and the sum of the two right angles is root 3. This is an equation, and then the sum of squares of the two right angles is 16,
In RT △ ABC, ∠ C = RT ∠, ab = radical 10, AC: BC = 2:1, calculate the perimeter and area of RT △ ABC
If BC = x, then AC = 2x
According to Pythagorean theorem
X^2+(2X)^2=10
X = root 2·
That is BC = root number 2, AC = 2 times root number 2
So the circumference is three times the root 2 plus the root 10
The area is equal to 1 / 2BC × AC = 2
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