In the Figure, Given Below, Ad ⊥ Bc. Prove That: C2 = A2 + B2 - 2ax - Mathematics

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Sum

In the figure, given below, AD ⊥ BC.
Prove that: c2 = a2 + b2 - 2ax.

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Solution

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

First, we consider the ΔABD and applying Pythagoras theorem we get,
AB2 = AD2   + BD
c2  = h2  + ( a - x )2  
h2  = c - ( a - x )2                      ......(i)
First, we consider the ΔACD and applying Pythagoras theorem we get,
AC2 = AD2 + CD2 
b2  = h2 + x 
h2  = b2 - x2                              ......(ii)

From (i) and (ii) we get,
c2  - ( a - x )2 = b2 - x2  
c - a- x2  + 2ax = b2 - x
c2 = a2 + b2  - 2ax
Hence Proved.

  Is there an error in this question or solution?
Chapter 13: Pythagoras Theorem [Proof and Simple Applications with Converse] - Exercise 13 (B) [Page 163]

APPEARS IN

Selina Concise Mathematics Class 9 ICSE
Chapter 13 Pythagoras Theorem [Proof and Simple Applications with Converse]
Exercise 13 (B) | Q 1 | Page 163

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