ABC is a right triangle right angled at C. Let BC = a, CA = b, AB = c and let p be the length of perpendicular from C on AB. Prove that
(i) pc = ab
(ii) 1/p² = 1/a² + 1/b²
Answers
In the attachment I have answered this problem. Pythagoras theorem: In a right angled triangle, square on the hypotenuse is equal to sum of the squares of other two sides. See the attachment for detailed solution.
SOLUTION : GIVEN : BC = a, CA = b, AB = c & CD = p
(i) Area of ΔABC = 1/2 × Base × Height
= 1/2 × AB × CD Area of ΔABC = ½ cp ……………(1)
Area of ΔABC = 1/2 × Base × Height
= 1/2 × BC × AC Area of ΔABC = ½ ab
½ cp = ½ ab
[From eq 1]
ab = cp
Hence proved.
(ii) In right angled ∆ ABC,
AB² = BC² + AC²
[By Pythagoras theorem]
c² = a² + b²
(ab/p)² = a² + b²
[ab = cp, c = ab/p]
a²b²/p² = a² + b²
1/p² = (a² + b²) / a²b²
1/p² = (a² / a²b² + b²/ a²b²)
1/p² = (1/b² + 1/a²)
1/p² = (1/a² + 1/b²)
Hence proved.
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