Question

2 Geometry Questions thank you guys :):)

Answer 1

Answer:

$\large\boxed{Q5.\ x=45\sqrt2}\\\boxed{Q6.\ x=8\sqrt2,\ y=4\sqrt6}$

Step-by-step explanation:

Q5.

x it's a diagonal of a square.

The formula of a length of diagonal of a square:

$d=a\sqrt2$

a - side of a square

We have a = 45.

Substitute:

$x=45\sqrt2$

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°

$a\sqrt2=8$        multiply both sides by √√2  (use √a · √a = a)

$2a=8\sqrt2$      divide both sides by 2

$a=4\sqrt2$

from the triangle 30° - 60° - 90°

$x=2a\to x=2(4\sqrt2)=8\sqrt2$

$y=a\sqrt3\to y=(4\sqrt2)(\sqrt3)=4\sqrt6$

Answer 2

Answer:

6. $\displaystyle 4\sqrt{6} = y \\ 4\sqrt{2} = x$

5. $\displaystyle 45\sqrt{2} = x$

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:

$\displaystyle a^2 + b^2 = c^2 \\ \\ \\ x^2 + x^2 = 8^2 \\ \\ \frac{2x^2}{2} = \frac{64}{2} → \sqrt{x^2} = \sqrt{32} \\ \\ 4\sqrt{2} = x$

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

$\displaystyle 4\sqrt{6} = 4\sqrt{2}\sqrt{3} \\ \\ 4\sqrt{6} = y$

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 $\displaystyle \sqrt{2}$to 45, giving you $\displaystyle 45\sqrt{2}.$

I am joyous to assist you anytime.