Question

What is the ratio of the area of a square to the area of a circle of the same perimeter?

5, A square of length l, a rhombus is out of this triangle making it have an angle of 60⁰ on a single diagonal. What is the area of this rhombus?

6, Solve the inequality tan(x+ ∏/3) ≥ √3

7, Given a triangle with sides 7m, 8m and 3m. What is 50% the area of this triangle?

8, tan (22.5) can be expressed as

A.     (1+2^0.5)1/2

B.     (-1+2^0.5)

C.     (2+2^0.5)1/2

D.     None

Answer 1

Area of Square $= s \times s = s\\^2$

Area of Circle $= \pi r\\^2$

Perimeter of Square = $4\times s$

Perimeter of Circle = $2\pi r\\$

Perimeter of Square = Perimeter of Circle

= $4s = 2 \pi r$

$= s =0.5 \pi r$

$= r = \frac{2s}{\pi}$

Area of Square $= (0.5 \times \pi \times r) \\ ^2 = (0.5 \pi r) \\^2$

Area of circle $= \pi \times (\frac{2s}{\pi})\\^2$ $= \frac{4 (s) \\^2}{\pi}$

Substituting $= s =0.5 \pi r$ in $= \frac{4 (s) \\^2}{\pi}$

$= \frac{4 (0.5 \pi r) \\^2}{\pi}$

$= \pi r\\^2$

The ratio of the area of a square to the area of a circle is: $= (0.5 \pi r) \\^2 : \pi r\\^2$

$= 0.25 \pi\\^2 r\\^2 : \pi r\\ ^2$

$=0.25 \pi : 1$

5, All squares are rhombuses, but not all rhombuses are squares.

Therefore, the area of the given rhombus is $l \times l$ since they have the same area as that of the square.

6,

$\arctan \left(\sqrt{3}\right)+\pi n\le \frac{x+1}{3}<\frac{\pi }{2}+\pi n$

$\arctan \left(\sqrt{3}\right)+\pi n\le \frac{x+1}{3}\quad \mathrm{and}\quad \frac{x+1}{3}<\frac{\pi }{2}+\pi n$

$x\ge \:\pi +3\pi n-1\quad \mathrm{and}\quad \:x<\frac{3\pi -2}{2}+3\pi n$

Answer is: $\pi -1+3\pi n\le \:x<\frac{3\pi -2}{2}+3\pi n$

7) Area of the triangle: $=\sqrt{(s(s-a) (s-b) (s-c)}$

$= s = \frac{p}{2} = \frac{7 + 8 +3}{2} = 9$

$=\sqrt{(9(9-7) (9-8) (9-3)}$

$= \sqrt{108} = 10.39$

50% of the area of the triangle is: $0.5 \times 10.39 = 4.156$

8) The correct answer is: option B :$= (-1 + 2\\ ^\frac{1}{2})$

Explanation:

$=\sqrt{3-2\sqrt{2}}$

$=\sqrt{2}-1$