A right-angled triangle has sides of length \(2k\), \(k^2 - 1\) and \(k^2 + 1\), where \(k > 1\). Show that \((2k)^2 + (k^2 - 1)^2 = (k^2 + 1)^2\). (3) Find...
Question 1 Report
A right-angled triangle has sides of length \(2k\), \(k^2 - 1\) and \(k^2 + 1\), where \(k > 1\).
- Show that \((2k)^2 + (k^2 - 1)^2 = (k^2 + 1)^2\). (3)
- Find the lengths of the three sides when \(k = 3\). (1)
Answer Details
This question tests algebraic manipulation to verify a Pythagorean identity, and substituting a value into the general expressions.
- (a) [3] Expand each square on the left-hand side: \[(2k)^2 = 4k^2\] \[(k^2-1)^2 = k^4 - 2k^2 + 1\] Adding these: \[4k^2 + (k^4 - 2k^2 + 1) = k^4 + 2k^2 + 1\] Expanding the right-hand side: \[(k^2+1)^2 = k^4 + 2k^2 + 1\] Since both sides simplify to \(k^4 + 2k^2 + 1\), the two expressions are equal, so \((2k)^2 + (k^2-1)^2 = (k^2+1)^2\) for every value of \(k\), meaning these three lengths always satisfy Pythagoras' theorem.
- (b) [1] Substituting \(k=3\): \[2k = 6, \qquad k^2 - 1 = 8, \qquad k^2 + 1 = 10\] The sides are \(6\), \(8\) and \(10\) cm.
This expression is a standard way of generating Pythagorean triples for any \(k>1\); the case \(k=3\) reproduces the familiar \(6\text{-}8\text{-}10\) triple.
Download The App On Google Playstore
Everything you need to excel in your exams
Personalized AI Learning Chat Assistant
200,000+ Exam Questions Across IGCSE, JAMB, WAEC & NECO
Over 3,900 Lesson Notes
Offline Support - Learn Anytime, Anywhere
Green Bridge Timetable
Literature Summaries & Potential Questions
Track Your Performance & Progress
In-depth Explanations for Comprehensive Learning