Respuesta :

gmany

Answer:

[tex]\large\boxed{Q5.\ x=45\sqrt2}\\\boxed{Q6.\ x=8\sqrt2,\ y=4\sqrt6}[/tex]

Step-by-step explanation:

Q5.

x it's a diagonal of a square.

The formula of a length of diagonal of a square:

[tex]d=a\sqrt2[/tex]

a - side of a square

We have a = 45.

Substitute:

[tex]x=45\sqrt2[/tex]

Q6.

Look at the first picture.

In a triangle 45° - 45° - 90°, all sides are in ratio 1 : 1 : √2.

In a triangle 30° - 60° - 90°, all sidea are in ratio 1 : √3 : 2.

Look at the second picture.

from the triangle 45° - 45° - 90°

[tex]a\sqrt2=8[/tex]        multiply both sides by √√2  (use √a · √a = a)

[tex]2a=8\sqrt2[/tex]      divide both sides by 2

[tex]a=4\sqrt2[/tex]

from the triangle 30° - 60° - 90°

[tex]x=2a\to x=2(4\sqrt2)=8\sqrt2[/tex]

[tex]y=a\sqrt3\to y=(4\sqrt2)(\sqrt3)=4\sqrt6[/tex]

Ver imagen gmany
Ver imagen gmany
BasketballEinstein

Answer:

6. [tex]\displaystyle 4\sqrt{6} = y \\ 4\sqrt{2} = x[/tex]

5. [tex]\displaystyle 45\sqrt{2} = x[/tex]

Step-by-step explanation:

30°-60°-90° Triangles

Hypotenuse → 2x

Short Leg → x

Long Leg → x√3

45°-45°-90° Triangles

Hypotenuse → x√2

Two identical legs → x

6. You solve the shorter triangle first:

[tex]\displaystyle a^2 + b^2 = c^2 \\ \\ \\ x^2 + x^2 = 8^2 \\ \\ \frac{2x^2}{2} = \frac{64}{2} → \sqrt{x^2} = \sqrt{32} \\ \\ 4\sqrt{2} = x[/tex]

Now that we know our x-value, we can solve the larger triangle:

[tex]\displaystyle 4\sqrt{6} = 4\sqrt{2}\sqrt{3} \\ \\ 4\sqrt{6} = y[/tex]

5. This exercise is EXTREMELY SIMPLE since two congruent isosceles right triangles form that square, so all you have to do, according to the rules for 45°-45°-90° triangles, is attach [tex]\displaystyle \sqrt{2}[/tex]to 45, giving you [tex]\displaystyle 45\sqrt{2}.[/tex]

I am joyous to assist you anytime.

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