1 ) If $$ x $$ is the side length of the equilateral triangle whose area is $$ 9 \sqrt{3 \mathrm{~cm}^{2} ext {. }} $$ , then $$ x=\ldots \ldots \ldots \ldots \ldots \mathrm{cm} $$. $$ \begin{array}{llll} ext { (a) } 6 & ext { (b) } 6 \sqrt{3} & ext { (c) } 3 \sqrt{3} & ext { (d) } 3\end{array} $$ (2) The radius length of the circle of the circular sector whose area is $$ 45 \mathrm{~cm}^{2} $$. and the length of its arc is 10 cm . equals $$ \cdots \ldots \ldots \ldots \ldots \mathrm{cm} $$. $$ \begin{array}{llll} ext { (a) } 4.5 & ext { (b) } 3 & ext { (c) } 9 & ext { (d) } 6\end{array} $$ (3) The area of the regular pentagon whose side length is 12 cm. $$ \begin{array}{llll} ext { (a) } 131 & ext { (b) } 991 & ext { (c) } 50 & ext { (d) } 248\end{array} $$

Question

1 ) If $$ x $$ is the side length of the equilateral triangle whose area is $$ 9 \sqrt{3 \mathrm{~cm}^{2} 	ext {. }} $$ , then $$ x=\ldots \ldots \ldots \ldots \ldots \mathrm{cm} $$. $$ \begin{array}{llll}	ext { (a) } 6 & 	ext { (b) } 6 \sqrt{3} & 	ext { (c) } 3 \sqrt{3} & 	ext { (d) } 3\end{array} $$ (2) The radius length of the circle of the circular sector whose area is $$ 45 \mathrm{~cm}^{2} $$. and the length of its arc is 10 cm . equals $$ \cdots \ldots \ldots \ldots \ldots \mathrm{cm} $$. $$ \begin{array}{llll}	ext { (a) } 4.5 & 	ext { (b) } 3 & 	ext { (c) } 9 & 	ext { (d) } 6\end{array} $$ (3) The area of the regular pentagon whose side length is 12 cm. $$ \begin{array}{llll}	ext { (a) } 131 & 	ext { (b) } 991 & 	ext { (c) } 50 & 	ext { (d) } 248\end{array} $$

UpStudy AI · Accepted Answer

**(1) Finding the side length $$x$$ of the equilateral triangle:**

The area $$A$$ of an equilateral triangle with side $$x$$ is given by
$$
A = \frac{\sqrt{3}}{4} x^2.
$$
We are told that
$$
\frac{\sqrt{3}}{4} x^2 = 9\sqrt{3}.
$$
Divide both sides by $$\sqrt{3}$$:
$$
\frac{1}{4} x^2 = 9.
$$
Multiply both sides by 4:
$$
x^2 = 36.
$$
Taking the positive square root (since side lengths are positive):
$$
x = 6.
$$
Thus, the answer is $$\boxed{6}$$.

---

**(2) Finding the radius $$r$$ of the circular sector:**

Let the central angle (in radians) be $$\theta$$. We are given:
- The area of the sector is
  $$
  \text{Area} = \frac{1}{2}r^2\theta = 45.
  $$
- The length of its arc is
  $$
  \text{Arc Length} = r\theta = 10.
  $$

Solve the second equation for $$\theta$$:
$$
\theta = \frac{10}{r}.
$$

Substitute $$\theta$$ into the area formula:
$$
\frac{1}{2} r^2 \cdot \frac{10}{r} = 45.
$$
Simplify:
$$
\frac{1}{2} \cdot 10 \cdot r = 45 \quad \Longrightarrow \quad 5r = 45.
$$
Thus,
$$
r = \frac{45}{5} = 9.
$$

So, the radius is $$\boxed{9}$$ cm.

---

**(3) Finding the area of the regular pentagon with side length $$s = 12$$ cm:**

The area $$A$$ of a regular pentagon with side length $$s$$ can be calculated using the formula
$$
A = \frac{1}{4} \sqrt{5(5+2\sqrt{5})} \, s^2.
$$
Substitute $$s = 12$$:
$$
A = \frac{1}{4} \sqrt{5(5+2\sqrt{5})} \times 144 = 36\sqrt{5(5+2\sqrt{5})}.
$$

To estimate, first compute the expression under the square root:
$$
5(5+2\sqrt{5}) \approx 5\left(5 + 2(2.236)\right) = 5(5 + 4.472) = 5(9.472) \approx 47.36.
$$
Then,
$$
\sqrt{47.36} \approx 6.888.
$$
Thus,
$$
A \approx 36 \times 6.888 \approx 248.
$$

So, the area of the pentagon is approximately $$\boxed{248}$$ cm$$^2$$.
**(1) Side length of the equilateral triangle:** - $$ x = 6 $$ cm **(2) Radius of the circular sector:** - $$ r = 9 $$ cm **(3) Area of the regular pentagon:** - $$ \approx 248 $$ cm²

1 ) If is the side length of the equilateral triangle whose area is
, then .

(2) The radius length of the circle of the circular sector whose area is .
and the length of its arc is 10 cm . equals .

(3) The area of the regular pentagon whose side length is 12 cm.

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1 ) If is the side length of the equilateral triangle whose area is , then . (2) The radius length of the circle of the circular sector whose area is . and the length of its arc is 10 cm . equals . (3) The area of the regular pentagon whose side length is 12 cm.