The problem states that a sector of radius $r$ is cut from a circular plate. This sector is used to form a cone with a base radius $x$ and slant height $r$. (i) Given that the area of the cone's base is $11 cm^2$, show that $x = \frac{1}{4}r$ and $r = 4\sqrt{\frac{11}{\pi}}$. (ii) Using $\pi = 3.141$, find the approximate value of $r$ to the first decimal place using logarithmic tables.

1. Problem Description

The problem states that a sector of radius $r$ is cut from a circular plate. This sector is used to form a cone with a base radius $x$ and slant height $r$.

(i) Given that the area of the cone's base is $11 cm^2$, show that $x = \frac{1}{4}r$ and $r = 4\sqrt{\frac{11}{\pi}}$.

(ii) Using $\pi = 3.141$, find the approximate value of $r$ to the first decimal place using logarithmic tables.

2. Solution Steps

(i)

The area of the base of the cone is given as $11 cm^2$. The formula for the area of a circle is $A = \pi x^2$. Thus,

$\pi x^2 = 11$

The circumference of the base of the cone is $2\pi x$. The arc length of the sector is $2\pi r \times \frac{\theta}{2\pi} = r \theta$, where $\theta$ is the angle subtended by the sector at the center of the circle.

Since the arc length of the sector forms the circumference of the base of the cone, we have:

$2\pi x = r\theta$

The area of the sector is $\frac{1}{2} r^2 \theta$. When the sector is formed into the cone, the arc length becomes the circumference of the base of the cone. The fraction of the original circle that is used becomes $\frac{x}{r}=\frac{\theta}{2\pi}$. The slant height of the cone is then $r$.

Using the fact that $x = \frac{1}{4} r$ is an area relationship and that surface area = base radius * slant height * $\pi$

$\pi r x = \pi x^2 + \pi x^2 ...$ is incorrect. It is area of the lateral surface.

Consider the circumference relationship: $2\pi x = \frac{1}{4}(2 \pi r)$. Then, x = r/

4.

Since $\pi x^2 = 11$, we substitute $x = \frac{1}{4}r$ into this equation:

$\pi (\frac{1}{4}r)^2 = 11$

$\pi \frac{1}{16} r^2 = 11$

$r^2 = \frac{16 \times 11}{\pi}$

$r = \sqrt{\frac{16 \times 11}{\pi}}$

$r = 4\sqrt{\frac{11}{\pi}}$

(ii)

$r = 4\sqrt{\frac{11}{\pi}}$

Given $\pi = 3.141$, we have

$r = 4\sqrt{\frac{11}{3.141}}$

$r = 4\sqrt{3.502}$

Using log tables:

Let $y = \sqrt{3.502} = (3.502)^{1/2}$

$\log y = \frac{1}{2} \log 3.502$

$\log 3.502 \approx 0.5443$

$\log y = \frac{1}{2}(0.5443) = 0.27215$

$y = 10^{0.27215} \approx 1.871$

Then, $r = 4y = 4(1.871) = 7.484$

Rounding to the first decimal place, $r \approx 7.5$

3. Final Answer

(i) $x = \frac{1}{4}r$ and $r = 4\sqrt{\frac{11}{\pi}}$

(ii) $r \approx 7.5$