NCERT Class X Chapter 4: Quadratic Equation Example 8

Question:

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?.

Given:

A circular park with diameter AB = 13 metres.

A pole (P) is erected on the boundary.

The difference of its distances from two diametrically opposite fixed gates A and B is 7 metres.

                               

To Find:

1. Is it possible to erect the pole under these conditions?

2. If yes, at what distances from the two gates should the pole be erected?

Formula:

WKT, an angle inscribed in a semicircle is a right angle (90°). Therefore, if a point P lies on the circumference and AB is the diameter, then triangle APB is a right-angled triangle with the right angle at P.

WKT, for a right-angled triangle, the Pythagorean theorem states: AP² + BP² = AB² (where AB is the hypotenuse).

WKT, for a quadratic equation of the form ax² + bx + c = 0, real solutions exist if and only if its discriminant (Δ) is greater than or equal to zero.

Δ = b² - 4ac ≥ 0

WKT, the roots of a quadratic equation can be found using the quadratic formula:

x = -b ± √Δ 2a

Solution:

Let the position of the pole be P.

Since gates A and B are diametrically opposite on the boundary and the pole P is also on the boundary, the triangle APB forms a right-angled triangle with the right angle at P (angle APB = 90°).

This is because an angle inscribed in a semicircle is a right angle.

Let the distance from gate A to the pole be AP (in metres) = x

Let the distance from gate B to the pole be BP (in metres) = y

The diameter of the park, AB, is the hypotenuse of triangle APB, so AB = 13 metres.

Given that the difference of its distances from A and B is 7 metres:

|x - y| = 7

Let's assume x > y (we can swap if y > x, the result will be consistent).

⇒ x - y = 7

⇒ x = y + 7 (Equation 1)

Apply the Pythagorean theorem to the right-angled triangle APB:

⇒ AP² + BP² = AB²

⇒ x² + y² = 13²

⇒ x² + y² = 169 (Equation 2)

Substitute (Equation 1) into (Equation 2):

⇒ (y + 7)² + y² = 169

Expand (y + 7)²:

⇒ (y² + 14y + 49) + y² = 169

Combine like terms and move 169 to the left side:

⇒ 2y² + 14y + 49 - 169 = 0

⇒ 2y² + 14y - 120 = 0

Divide the entire equation by 2 to simplify:

⇒ y² + 7y - 60 = 0

This is a quadratic equation of the form ay² + by + c = 0, where:

a = 1, b = 7 and c = -60

To determine if it's possible to erect the pole (i.e., if real roots for y exist), calculate the discriminant (Δ):

WKT, Δ = b² - 4ac

⇒ Δ = (7)² - 4(1)(-60)

⇒ Δ = 49 + 240

⇒ Δ = 289

Since Δ = 289 > 0, real roots exist, which means it is possible to erect the pole under the given conditions.

Now, find the values of 'y' (distance BP) using the quadratic formula:

⇒ y = -b ± √Δ 2a

⇒ y = -7 ± √289 2 × 1

⇒ y = -7 ± 17 2

We get two possible values for y:

y₁ = -7 + 17 2 = 10 2 = 5

y₂ = -7 - 17 2 = -24 2 = -12

Since distance cannot be negative, we discard y = -12.

Therefore, BP (y) = 5 metres.

Now, find the distance AP (x) using Equation 1 (x = y + 7):

⇒ x = 5 + 7

⇒ x = 12 metres.

Verification:

Check with Pythagorean theorem:

5² + 12² = 25 + 144 = 169

5² + 12² = 13².

The distances are consistent.

Result:

Yes, it is possible to erect the pole under the given conditions.

The pole should be erected at a distance of

• 12 metres from gate A and
• 5 metres from gate B.

Next Question Solution:

NCERT Class X Chapter 4: Quadratic Equation Example 9.

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