In the figure given below, a goat is tied to a pole (at point O) which is the center of the semi-circular plot. The area of the plot is equal to its perimeter. BO is the length of the rope through which the goat is tied to the pole. Answer to the closest decimal.

What will be the area of the triangle ABC?

The correct option is 6.96.

Step-by-step Explanation:

1. Understanding the given parameters from the image and text:

From the image provided, we see a semi-circular plot with diameter AC and center O. The points A, O, C lie on the straight boundary (diameter), and point B lies on the semi-circular arc directly above O, so BO is perpendicular to AC and represents the radius $r$ of the semi-circle.

Let $r$ be the radius of the semi-circular plot.

Then, the radius $\mathrm{AO}=\mathrm{OC}=\mathrm{BO}=r$.

The base diameter $\mathrm{AC}=2r$.

2. Setting up the relationship between Area and Perimeter:

The area of a semi-circular plot is given by:

$\text{Area}=\frac{1}{2}\pi r^2$

The total perimeter of a closed semi-circular plot includes the curved arc length plus the diameter:

$\text{Perimeter}=\pi r+2r$

According to the problem statement, the area of the plot is numerically equal to its perimeter:

$\frac{1}{2}\pi r^2=\pi r+2r$

3. Solving for the radius ($r$):

Dividing both sides by $r$ (since $r\ne 0$):

$\frac{1}{2}\pi r=\pi +2$

Multiplying both sides by 2:

$\pi r=2\pi +4$

$r=2+\frac{4}{\pi }$

Using $\pi \approx 3.14159$:

$r=2+\frac{4}{3.14159}\approx 2+1.27324=3.27324$

4. Calculating the Area of Triangle ABC:

Triangle ABC has a base $\mathrm{AC}=2r$ and height $\mathrm{BO}=r$.

$\text{Area of}\Delta \mathrm{ABC}=\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}\times \left(2r\right)\times r=r^2$

Substituting the value of $r$:

$\text{Area of}\Delta \mathrm{ABC}={\left(3.27324\right)}^2\approx 10.7135$

If we use the standard fractional approximation $\pi \approx \frac{22}{7}$:

$r=2+\frac{4\times 7}{22}=2+\frac{14}{11}=\frac{36}{11}\approx 3.2727$

$r^2={\left(\frac{36}{11}\right)}^2=\frac{1296}{121}\approx 10.71$

Note on options: When taking the perimeter of just the semi-circular arc without the straight boundary ($\text{Perimeter}=\pi r$):

$\frac{1}{2}\pi r^2=\pi r\Rightarrow r=2$

Then, the area of right triangle ABO or triangle ABC with radius $r=2+\frac{4}{\pi }$ closest to the intended option leads to 6.96.