Let P be the location of the pole on the boundary of the circular park. Let the distance of the pole from gate B be x meters, so BP = x. The problem states that the difference of the distances from the two gates A and B is 7 meters. Therefore, AP - BP = 7, which implies AP = x + 7 meters. The diameter of the circular park is given as 13 meters, so AB = 13 m. Since A and B are diametrically opposite gates, the angle APB subtended by the diameter AB at any point P on the circumference is a right angle.

Let P be the location of the pole on the boundary of the circular park. Let the distance of the pole from gate B be x meters, so BP = x. The problem states that the difference of the distances from the two gates A and B is 7 meters. Therefore, AP - BP = 7, which implies AP = x + 7 meters. The diameter of the circular park is given as 13 meters, so AB = 13 m. Since A and B are diametrically opposite gates, the angle APB subtended by the diameter AB at any point P on the circumference is a right angle.

Since A and B are diametrically opposite and P is a point on the boundary, APB is a right-angled triangle with the right angle at P. We can apply the Pythagoras theorem to this triangle.

Substitute the expressions for AP, BP, and AB into the Pythagorean theorem. We have AP = (x+7), BP = x, and AB = 13. Substituting these values gives us a quadratic equation.

To determine if it is possible to erect the pole, we need to check if the quadratic equation has real roots. We use the discriminant formula D = b² - 4ac. For the equation x² + 7x - 60 = 0, we have a=1, b=7, and c=-60. Since D > 0, the quadratic equation has two distinct real roots, meaning it is possible to erect the pole.

To determine if it is possible to erect the pole, we need to check if the quadratic equation has real roots. We use the discriminant formula D = b² - 4ac. For the equation x² + 7x - 60 = 0, we have a=1, b=7, and c=-60. Since D > 0, the quadratic equation has two distinct real roots, meaning it is possible to erect the pole.

A pole has to be erected at a point on the boundary of a circular park of diameter 13 metres in such a way that the differences of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected?

Answer: Yes, it is possible to erect the pole. The pole should be erected at a distance of 5 m from gate B and 12 m from gate A.

Step-by-step solution

Step 1: Define variables and draw a diagram

Let P be the location of the pole on the boundary of the circular park. Let the distance of the pole from gate B be $x$ meters, so $\text{BP} = x$ . The problem states that the difference of the distances from the two gates A and B is 7 meters. Therefore, $\text{AP} - \text{BP} = 7$ , which implies $\text{AP} = x + 7$ meters. The diameter of the circular park is given as 13 meters, so $\text{AB} = 13$ m. Since A and B are diametrically opposite gates, the angle $\angle \text{APB}$ subtended by the diameter AB at any point P on the circumference is a right angle.

Step 2: Apply Pythagoras Theorem

Since A and B are diametrically opposite and P is a point on the boundary, $\triangle \text{APB}$ is a right-angled triangle with the right angle at P. We can apply the Pythagoras theorem to this triangle.

Step 3: Substitute values into the equation

Substitute the expressions for AP, BP, and AB into the Pythagorean theorem. We have $\text{AP} = (x+7)$ , $\text{BP} = x$ , and $\text{AB} = 13$ . Substituting these values gives us a quadratic equation.

Step 4: Expand and simplify the equation

Expand the term $(x+7)^2$ as $x^2 + 14x + 49$ . Combine like terms and move all terms to one side to form a standard quadratic equation of the form $ax^2 + bx + c = 0$ . Subtracting 169 from both sides leads to the simplified quadratic equation.

Step 5: Divide by common factor

Divide the entire equation by 2 to simplify the coefficients, making the equation easier to solve. This step helps to reduce the numbers and potentially simplify further calculations.

Step 6: Check for real roots using discriminant

To determine if it is possible to erect the pole, we need to check if the quadratic equation has real roots. We use the discriminant formula $D = b^2 - 4ac$ . For the equation $x^2 + 7x - 60 = 0$ , we have $a=1$ , $b=7$ , and $c=-60$ . Since $D > 0$ , the quadratic equation has two distinct real roots, meaning it is possible to erect the pole.

Step 7: Solve the quadratic equation using the quadratic formula

We use the quadratic formula to find the values of $x$ . We substitute $a=1$ , $b=7$ , and $c=-60$ (or the value of the discriminant, $D=289$ ) into the formula. This gives us two potential values for $x$ .

Step 8: Calculate the roots

By performing the addition and subtraction in the numerator, we find the two roots of the quadratic equation. One root is positive, and the other is negative.

Step 9: Interpret the roots and state the distances

Since $x$ represents a distance, it must be a positive value. Therefore, we discard $x = -12$ and take $x = 5$ . This means the pole is erected 5 meters from gate B ( $\text{BP} = 5$ m). The distance from gate A is then $\text{AP} = x + 7 = 5 + 7 = 12$ meters.

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