Question

I need help with these three problems please​

Answer 1

Answer:

7) 8sqrt(2) in

8) 68sqrt(2) ft

9) 21sqrt(3) cm

Step-by-step explanation:

7) sqrt(8² + 8²) = 8sqrt(2) in

sqrt is square root

8) sqrt(x² + x²) = 34

2x² = 34²

x = 17sqrt(2)

Perimeter = 4x = 68sqrt(2) ft

9) 3s = 126

s = 42

Altitude² + 21² = 42²

Altitude² = 1323

Altitude = 21sqrt(3) cm

Answer 2

Answer:

Q7. 11.3 inches (3 s.f.)

Q8. 96.2 ft

Q9. 36.4cm

Step-by-step explanation:

Q7. Please see attached picture for full solution.

Q8. Let the length of a side of the square be x ft.

Applying Pythagoras' Theorem,

$34^{2} = {x}^{2} + {x}^{2} \\ 2 {x}^{2} = 1156 \\ {x}^{2} = 1156 \div 2 \\ {x}^{2} = 578 \\ x = \sqrt{578}$

Thus, the perimeter of the square is

$= 4( \sqrt{578} ) \\ = 96.2 ft\: \: \: (3 \: s.f.)$

Q9. Equilateral triangles have 3 equal sides and each interior angle is 60°.

Since the perimeter of the equilateral triangle is 126cm,

length of each side= 126÷3 = 42 cm

The green line drawn in picture 3 is the altitude of the triangle.

Let the altitude of the triangle be x cm.

sin 60°= $\frac{x}{42}$

$\frac{ \sqrt{3} }{2} = \frac{x}{42} \\ x = \frac{ \sqrt{3} }{2} \times 42 \\ x = 21 \sqrt{3} \\ x = 36.4$

(to 3 s.f.)

Therefore, the length of the altitude of the triangle is 36.4cm.