MaxWong

For the square one, construct two radii to two consectuive vertices.

Let O be the center, and A, B be the vertices.

Let r be the radii of the circle on the left.

$\angle AOB = 90^\circ$

By Pythagoras theorem,

$r^2 + r^2 = AB^2\\ AB = \sqrt2 r$

Perimeter of square = $4 \sqrt2 r$

Do the same thing for the circle on the right, but notice that this time $\angle AOB = 120^\circ$.

Let R be the radii of the circle on the right.

$AB^2 = R^2 + R^2 - 2(R)(R)\cos 120^\circ\\ AB = \sqrt3R$

Perimeter of triangle = $3\sqrt 3 R$

If the perimeters are the same, then

$3\sqrt 3 R = 4\sqrt2 r\\ \dfrac{R}{r} = \dfrac{4\sqrt6}{9} --- (1)$

We proceed to calculate the areas.

Using the notations in the problem,

$A = \left(\sqrt2 r\right)^2 = 2r^2$

$B = \dfrac{\sqrt3}{4} \left(\sqrt3R\right)^2 = \dfrac{3\sqrt3}{4}R^2$

By (1), we get $\dfrac AB = \dfrac{2r^2}{\dfrac{3\sqrt 3}{4}R^2} = \dfrac{8}{3\sqrt3} \left(\dfrac{R}{r}\right)^{-2} = \dfrac{8\sqrt3}{9}\left(\dfrac{27}{32}\right) = \boxed{\dfrac{3\sqrt 3}{4}}$

To solve this problem, I will first introduce a interesting theorem.

Theorem ABT:

If Y is a point on AC such that BY bisects $\angle ABC$, then $\dfrac{AY}{YC} = \dfrac{AB}{BC}$.

Proof of Theorem ABT:

Let $\angle CYB = \alpha$, $\angle ABY = \beta$.

Then $\angle AYB = 180^\circ - \alpha$

$\sin \angle CYB = \sin \angle AYB = \sin \alpha$

Applying Law of Sines on $\triangle AYB$,

$\dfrac{AY}{AB} = \dfrac{\sin \beta}{\sin \alpha}$

Applying Law of Sines again on $\triangle CYB$,

$\dfrac{YC}{BC} = \dfrac{\sin \beta}{\sin \alpha}$

Therefore, 

$\dfrac{AY}{AB} = \dfrac{YC}{BC}\\ \dfrac{AY}{YC} = \dfrac{AB}{BC}$

Back to the problem. We apply Theorem ABT.

$\dfrac{AY}{YC} = \dfrac{AB}{BC}\\ \dfrac{6}{4} = \dfrac{8}{BC}\\ BC = \dfrac{16}{3}$

Area of figure = Area of rectangle - 2 (Area of semicircle)

Two semicircle of the same radius, say r, has the same area as a circle with that radius r.

Notice that the area of a circle with diameter d is $\dfrac{\pi d^2}{4}$.

   Area of figure

= Area of rectangle - (Area of circle with diameter 6)

= $15(6) - \dfrac{\pi \cdot 6^2}{4}$

= $90 - 9\pi$

$\approx 61.7$

b)

Number of students who studies Spanish = 1060

Total number of students = 870 + 800 = 1670

Number of students who DOES NOT study Spanish = 1670 - 1060 = 610

Probability that a random student DOES NOT study Spanish

= (Number of students who DOES NOT study Spanish)/(Total number of students)

= 610 / 1670

= 61/167

a)

2/3 of the boys studies Spanish, so the number of boys who studies Spanish is (870)(2/3) = 580.

3/5 of the girls studies Spanish, so the number of girls who studies Spanish is (800)(3/5) = 480.

Total number of studies who studies Spanish 

= (Number of boys who studies Spanish) + (Number of girls who studies Spanish)

= 580 + 480

= 1060

$\overline{AB} \cong \overline{AD}\\ \overline{CB} \cong \overline{CD}\\ \overline{AC} \cong \overline{AC}\\ $

Therefore by SSS postulate, $\triangle ABC \cong \triangle ADC$.

By congruent triangles, we know that $m\angle BAE = m\angle DAE$ --- (1)

Also, $\overline{AB} \cong \overline{AD}$ --- (2)

and $\overline{AE} \cong \overline{AE}\\$ --- (3)

By (1), (2), (3), and SAS postulate, $\triangle BAE \cong \triangle DAE$ --- (4)

Then, by (4), $\angle BEA = \angle DEA$ --- (5)

Also, by sum of angles on a straight line, $\angle BEA + \angle DEA = 180^\circ$ --- (6)

Solving (5) and (6) gives $\angle BEA = \angle DEA = 90^\circ$

Therefore, $\triangle ABE$ is a right triangle with $\angle BEA = 90^\circ$.

Slope of line = $\dfrac{-3 - (-12)}{1 - (-2)} = 3$

$y + 3 = 3(x - 1)\\ y = 3x - 6\\ \text{When }y = 0,\\ 3x - 6 = 0\\ x = 2\\ \text{So the x-intercept is }2.$

Hello :)

$\text{Let }x = \sqrt{10 +\sqrt{10 +\sqrt{10 + \cdots}}}\\ x = \sqrt{10 + x}, x \geq 0\\ x^2 = 10 + x\\ x^2 - x - 10 = 0\\ x = \dfrac{1\pm\sqrt{1^2 - 4(1)(-10)}}{2}\\ x = \dfrac{1\pm\sqrt{41}}{2}\\ \text{Rejecting the negative root, we have }\boxed{x = \dfrac{1+\sqrt{41}}{2}}$

Observing the recurrence relation, $\{b_n\}$ is a quadratic sequence.

Let $b_n = pn^2 + qn + r\qquad p,q,r\in \mathbb R$

Obviously, $a_n = 4 + 2n$.

$b_1 = 3\\ p(1)^2 + q(1) + r = 3\\ p + q +r = 3 \phantom{-----}(1)$

$b_n = b_{n - 1} + a_{n - 1}\\ pn^2 + qn + r = p(n - 1)^2 + q(n - 1) + r +(4+2(n-1))\\ pn^2 +qn + r = pn^2 - 2pn + p + qn - q + r + 4 + 2n - 2\\ 0=-2pn+p-q+2+2n\\ (2 - 2p)n+(p-q+2)=0n+0\\ \text{Comparing both sides,}\\ 2-2p = 0\\ 2p = 2\\ \boxed{p = 1}\\ p -q+2=0\\ 1-q+2=0\\ \boxed{q=3}\\ \text{Substitute }p=1\text{ and }q = 3\text{ into equation }(1),\\ 1+3+r=3\\ \boxed{r = -1}\\ \therefore b_n = n^2 + 3n - 1$