English
Tamil Nadu Board of Secondary EducationSSLC (English Medium) Class 10

A pole has to be erected at a point on the boundary of a circular ground of diameter 20 m in such a way that the difference of its distances from two diametrically opposite fixed gates P and Q on the - Mathematics

Advertisements
Advertisements

Question

A pole has to be erected at a point on the boundary of a circular ground of diameter 20 m in such a way that the difference of its distances from two diametrically opposite fixed gates P and Q on the boundary is 4 m. Is it possible to do so? If answer is yes at what distance from the two gates should the pole be erected?

Sum

Solution

Let “R” be the required location of the pole

Let the distance from the gate P is “x” m : PR = “x” m

The distance from the gate Q is (x + 4)m

∴ QR = (x + 4)m

In the right ∆PQR,

PR2 + QR2 = PQ2 (By Pythagoras theorem)

x2 + (x + 4)2 = 202

x2 + x2 + 16 + 8x = 400

2x2 + 8x – 384 = 0


x2 + 4x – 192 = 0  ...(divided by 2)

(x + 16) (x – 12) = 0

x + 16 = 0 or x – 12 = 0 ...[negative value is not considered]

x = – 16 or x = 12

Yes it is possible to erect

The distance from the two gates are 12 m and 16 m

shaalaa.com
Quadratic Equations
  Is there an error in this question or solution?
Chapter 3: Algebra - Exercise 3.12 [Page 116]

APPEARS IN

Samacheer Kalvi Mathematics [English] Class 10 SSLC TN Board
Chapter 3 Algebra
Exercise 3.12 | Q 5 | Page 116
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×