Cut out a square as large as possible, the largest square, from an isosceles right triangle cardboard with an inclined edge of 18cm long What is the side length of the?

There are two painting methods. Click my account for graphics, go to my baidu photo album, and collect photos on December 24, 2012
In the first type, the edges of the square coincide with the beveled edges
Side length of square x = 18 / 3 = 6
Second
One vertex of the square is on the hypotenuse
Let the company commander be y
According to isosceles right triangle
2√2y=18
y=9√2/2
9√2/2=6.363>6
So the second kind grows up, and the side length is 9 √ 2 / 2

In triangle ABC, B is known ²- bc-2c ²= 0 and a = 7 under the root sign, cosa = 7 / 9, then the area of triangle ABC is?

b ²- bc-2c ²= 0 = (B + C) (b-2c) because a > 0, b > 0, C > 0 (B + C) = 0, B = 2ccosa = 7 / 9 = (b ^ 2 + C ^ 2-A ^ 2) / (2BC) C ^ 2 = 63 / 17sina = 4 root 2 / 9 s = bcsina / 2 = C ^ 2sina = 63 / 17 * 4 root 2 / 9 = 28 root 2 / 17

stay Δ In ABC, if B ^ 2-bc-2c ^ 2 = 0, a = root 6, cosa = 7 / 8, then Δ The area of ABC is

A ^ = B ^ 2 + C ^ 2-2bccosa, so 6 = B ^ 2 + C ^ 2-7 / 4bc (1) because B ^ 2-bc-2c ^ 2 = 0, that is, (b-2c) (B + C) = 0, and B + C > 0, so B = 2C. Substitute into formula (1) to get 6 = 4C ^ 2 + C ^ 2-7 / 2C ^ 2, so C = 2, B = 4, s △ ABC = 1 / 2bcsina = 4sina = 4 √ [1 - (7 / 8) ^ 2] = 1 / 2 √ 15

It is known that the opposite sides of the three internal angles a, B and C of △ ABC are a, B and C respectively. A is an acute angle, and the root number 3B = 2asinb (1) Find a (2) If a = 7, △ ABC area is 10 root 3, find B ²+ c ² It's root three B, B is outside

1. Because the radical 3B = 2asinb, we can get B / SINB = 2A / radical 3. Using the regular Xuan theorem of triangle, B / SINB = A / Sina. Combined with the previous equation, we can get a = 60 degrees
2. The triangle area s = 1 / 2 times bcsina to obtain BC = 40. Using the cosine theorem, square a = square B + square C - 2bccosa, bring in the value to obtain square B + square C = 89

In the acute angle △ ABC, the opposite sides of the inner angles a, B and C are a, B and C respectively, and 2asinb = (√ 3) B (I) find the size of angle a; (II) if a = 6, B + C = 8, find the area of △ ABC. Find the detailed process of formula simplification

(Ⅰ) 2asinb = (√ 3) B, get a / (√ 3 / 2) = B / SINB, because a / Sina = B / SINB, Sina = √ 3 / 2, a = π / 3 (Ⅱ) B + C = 8, get B ²+ 2bc+c ²= 64 and because of B ²+ c ²- a ²= The simultaneous solution of 2bccosa shows that the area of 64-2bc-36 = 2bccos π / 3, BC = 28 / 3 △ ABC =

In the acute triangle ABC, the opposite sides of the inner angles a, B and C are a, B and C respectively, and 2asinb = root 3B If a = 6, B + C = 8, find the area of the triangle

(1) ∵ 2asinb radical 3B = 0 according to the sine theorem ∵ 2sinasinb - √ 3sinb = 0 ∵ SINB > 0 ∵ 2sina - √ 3 = 0 ∵ Sina = √ 3 / 2, and a is an acute angle, ∵ a = π / 3 (2) is obtained from the cosine theorem: A ^ 2 = B ^ 2 + C ^ 2-2bc • cosa, that is 36 = B ^ 2 + C ^ 2-bc = (B + C) ^ 2-3bc = 64-3bc, ∵ BC = 28 / 3, and s

Cut out a square as large as possible, the largest square, from an isosceles right triangle cardboard with an inclined edge of 18cm long What is the side length of the?