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

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प्रश्न

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

योग
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उत्तर

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.

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 13: Pythagoras Theorem [Proof and Simple Applications with Converse] - Exercise 13 (B) [पृष्ठ १६३]

APPEARS IN

सेलिना Concise Mathematics [English] Class 9 ICSE
अध्याय 13 Pythagoras Theorem [Proof and Simple Applications with Converse]
Exercise 13 (B) | Q 1 | पृष्ठ १६३

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