Let the distance of gate P from point B be x m. Since the distance of P from A is 35 m more than its distance from B, we have AP = (x + 35) m. Since AB is a diameter of the circle, the angle subtended by diameter AB at point P on the circumference is a right angle, meaning APB = 90^°.

Let the distance of gate P from point B be x m. Since the distance of P from A is 35 m more than its distance from B, we have AP = (x + 35) m. Since AB is a diameter of the circle, the angle subtended by diameter AB at point P on the circumference is a right angle, meaning APB = 90^°.

Applying Pythagoras theorem in right-angled triangle APB, we have AP² + BP² = AB². Substituting AP = x + 35, BP = x, and AB = 65, we expand (x + 35)² = x² + 70x + 1225 and evaluate 65² = 4225. Combining like terms gives 2x² + 70x - 3000 = 0, which simplifies to x² + 35x - 1500 = 0 upon dividing throughout by 2.

Applying Pythagoras theorem in right-angled triangle APB, we have AP² + BP² = AB². Substituting AP = x + 35, BP = x, and AB = 65, we expand (x + 35)² = x² + 70x + 1225 and evaluate 65² = 4225. Combining like terms gives 2x² + 70x - 3000 = 0, which simplifies to x² + 35x - 1500 = 0 upon dividing throughout by 2.

17. There is a circular park of diameter $65\text{ m}$ as shown in the following figure, where $AB$ is a diameter. An entry gate is to be constructed at a point $P$ on the boundary of the park such that distance of $P$ from $A$ is $35\text{ m}$ more than the distance of $P$ from $B$ . Find distance of point $P$ from $A$ and $B$ respectively. (CBSE-2024) [Figure Description: A circle with center $O$ has a diameter $AB$ passing through $O$ . A point $P$ lies on the circumference of the circle, forming a triangle $\triangle APB$ inscribed in the semicircle, with line segments $AP$ and $BP$ connected to diameter $AB$ .]

Answer: Distance of point $P$ from $A$ is $60\text{ m}$ and from $B$ is $25\text{ m}$ .

Step-by-step solution

Step 1: Define variables and apply geometry of semicircle

Let the distance of gate $P$ from point $B$ be $x\text{ m}$ . Since the distance of $P$ from $A$ is $35\text{ m}$ more than its distance from $B$ , we have $AP = (x + 35)\text{ m}$ . Since $AB$ is a diameter of the circle, the angle subtended by diameter $AB$ at point $P$ on the circumference is a right angle, meaning $\angle APB = 90^\circ$ .

Step 2: Formulate the quadratic equation using Pythagoras theorem

Applying Pythagoras theorem in right-angled triangle $\triangle APB$ , we have $AP^2 + BP^2 = AB^2$ . Substituting $AP = x + 35$ , $BP = x$ , and $AB = 65$ , we expand $(x + 35)^2 = x^2 + 70x + 1225$ and evaluate $65^2 = 4225$ . Combining like terms gives $2x^2 + 70x - 3000 = 0$ , which simplifies to $x^2 + 35x - 1500 = 0$ upon dividing throughout by $2$ .

Step 3: Solve the quadratic equation by factorisation

To solve $x^2 + 35x - 1500 = 0$ by splitting the middle term, we find two numbers whose product is $-1500$ and sum is $35$ . These numbers are $60$ and $-25$ . Factorising by grouping gives $(x - 25)(x + 60) = 0$ , leading to roots $x = 25$ or $x = -60$ .

Step 4: Find the required distances

Since distance cannot be negative, we reject $x = -60$ and accept $x = 25$ . Therefore, the distance of point $P$ from $B$ is $25\text{ m}$ , and the distance of $P$ from $A$ is $25 + 35 = 60\text{ m}$ .

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