Direction: 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.
The plot is a semicircle, and we are told that its area equals its perimeter. Since the pole sits at the center O and the rope stretches out to point B on the curved edge, the rope's length BO is simply the radius of the semicircle. Let's call this radius r and check which option satisfies the given condition.
\(2+\pi\): Substituting a radius of this size into the area and perimeter expressions of the semicircular plot does not produce a matching pair of values, so this radius does not satisfy the condition that area equals perimeter.
\(2+\pi/2\): This value is also too large to balance the area and perimeter equation correctly for a semicircular plot of this shape.
\(2+2/\pi\): The area of the semicircular plot is \( \frac{1}{2}\pi r^2 \), and its boundary length, made up of the curved arc and the straight edge OB, is \( \pi r + r \). Setting these equal, \( \frac{1}{2}\pi r^2 = \pi r + r \), and dividing both sides by \(r\), gives \( \frac{1}{2}\pi r = \pi + 1 \), so \( r = \frac{2(\pi+1)}{\pi} = 2 + \frac{2}{\pi} \). This satisfies the condition exactly.
Blank option: No value is given here, so it cannot be evaluated or selected.
Solving the area equals perimeter condition directly for the radius gives a clean closed form expression matching the third option.
Therefore, the correct answer is \(2+\dfrac{2}{\pi}\).
Now that we know the radius of the plot from the earlier part, BO equals \(2+\dfrac{2}{\pi}\), we can use it to work out the plot's area. Let's check each option against the value this radius gives.
18.13: This is lower than the value obtained when the radius is squared and multiplied by pi, so it does not match.
21.83: With \( r = 2+\dfrac{2}{\pi} \approx 2.637 \), squaring gives \( r^2 \approx 6.952 \). Multiplying by pi gives \( \pi r^2 \approx 3.1416 \times 6.952 \approx 21.83 \). This matches the area precisely.
28.13: This overshoots the correct figure and does not match the computed area for this radius.
11.83: This is roughly half the correct value and would only arise from a miscalculation, not the actual area for this radius.
Substituting the radius found earlier into the area expression gives a value that lines up exactly with one of the options.
Therefore, the correct answer is 21.83.
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Question: 3
If a triangle is created by joining points A, B, and C. What will be the length of AB?
Points A and C mark the two ends of the semicircle's flat base, with B lying on the curved edge directly above the center O. Joining A, B, and C forms a triangle, and we can use a standard circle property to find AB.
3.73: Since AC is the diameter of the semicircle and B lies on the arc, the angle ABC is a right angle, a standard property of any triangle inscribed in a semicircle with the diameter as one side. Because B sits directly above the center O, triangle ABC is also isosceles with AB equal to BC. Using \( AB^2+BC^2=AC^2 \) with \( AC=2r \) and \( AB=BC \), we get \( 2AB^2=(2r)^2 \), so \( AB=r\sqrt{2} \). With \( r=2+\dfrac{2}{\pi}\approx 2.637 \), this gives \( AB\approx 2.637\times 1.414\approx 3.73 \). This matches the computed length.
7.33: This is roughly double the correct value and does not match \( r\sqrt{2} \) for the radius found earlier.
1.86: This is roughly half the correct value, too small to satisfy the right angle relationship between the three sides.
Can't be calculated with the given data: Since the radius is already known from the earlier part, and B's position on the arc is fixed by the figure, there is enough information to compute AB exactly, so this option does not apply.
Using the right angle formed at B and the isosceles symmetry of the triangle gives an exact value for AB.
Using the right angle at B established earlier, along with AB and BC as the two legs of the triangle, we can find the area of triangle ABC directly.
6.96: Since angle ABC is 90 degrees, AB and BC act as the base and height of the triangle. With \( AB=BC=r\sqrt{2} \), the area is \( \frac{1}{2}\times AB\times BC=\frac{1}{2}\times(r\sqrt{2})^2=\frac{1}{2}\times 2r^2=r^2 \). With \( r\approx 2.637 \), this gives \( r^2\approx 6.95 \), which rounds to 6.96. This matches the computed area.
6.96: This repeats the same figure as above and is consistent with the same correct calculation of \( r^2 \).
3.66: This is roughly half the correct area and would only arise if the two legs were not both used in the area formula, which is not the case here since AB and BC form the right angle.
Can't be calculated with the given data: Since both AB and BC, along with the right angle between them, are already established from the earlier parts, there is enough information to calculate the area exactly, so this option does not apply.
Squaring the radius directly gives the area of the right triangle formed at B, since the two equal legs simplify the standard area formula.
Continuing from the triangle ABC formed by joining the three points, we now need its full perimeter, using the side lengths already established, AB and BC, together with the straight distance AC.
40.4: This is far larger than the true perimeter of a triangle whose two known legs are each only about 3.73 units long.
30.3: This also greatly overstates the perimeter for a triangle of this size.
20.2: This is still too large compared to the actual sum of the three sides.
10.1: With \( AB=BC\approx 3.73 \) and the straight segment \( AC \), which measures the same as the rope length found earlier, \( \approx 2.64 \), the perimeter is \( AB+BC+AC\approx 3.73+3.73+2.64\approx 10.1 \). This matches the computed perimeter.
Adding the two equal legs of the triangle to the third side gives the full perimeter of triangle ABC.
