Question

\[ABC\] is a right triangle, right-angled at $ C $ . If $ BC = a $ , $ CA = b $ , $ AB = c $ and $ p $ be the length of perpendicular from $ C $ on $ AB $ , then
i) $ cp = ab $
ii) $ \dfrac{1}{{{p^2}}} = \dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}} $

A) True
B) False

Answer 1

Hint: In this question, it is given that \[ABC\] is a right triangle which is right-angled at $ C $ . And also given that $ BC = a $ , $ CA = b $ , $ AB = c $ and $ p $ be the length of perpendicular from $ C $ on $ AB $ . We need prove $ cp = ab $ and $ \dfrac{1}{{{p^2}}} = \dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}} $ is true or false. For this, we will construct the right triangle with given conditions. And, as we know the base and height of the triangle, we will use the area of the triangle formula to prove the condition (i). Then, by using Pythagoras theorem we will prove the condition (ii).

Complete step-by-step answer:
i) Consider a right triangle \[ABC\], right-angled at $ C $ .
 Let $ CD \bot AB $ . Then, $ CD = p $

Here, $ BC = a $ , $ CA = b $ , $ AB = c $ and $ CD = p $ .
We know that, area of triangle= $ \dfrac{1}{2} \times base \times height $
Therefore, area of triangle\[ABC\] with base $ AB $ ,
 $ = \dfrac{1}{2} \times AB \times CD $
 $ = \dfrac{1}{2} \times c \times p $
 $ = \dfrac{1}{2}cp $
Also, area of triangle\[ABC\] with base $ AC $ ,
 $ = \dfrac{1}{2} \times BC \times AC $
 $ = \dfrac{1}{2} \times a \times b $
 $ = \dfrac{1}{2}ab $
Both the areas are the same. Therefore, on equating, we get,
 $ \dfrac{1}{2}cp = \dfrac{1}{2}ab $
 $ \Rightarrow cp = ab $
Therefore, the condition (i) is satisfied.
ii) Since, $ \Delta ABC $ is right triangle, right-angled at $ C $ .
Now, using Pythagoras theorem in triangle \[ABC\], we have,
  $ A{B^2} = B{C^2} + A{C^2} $
 $ \Rightarrow {c^2} = {a^2} + {b^2} $
Where $ c $ is the hypotenuse side, $ a $ is the perpendicular side and $ b $ is the base
We know that, from result (i),
  cp = ab
 $ \Rightarrow c = \dfrac{{ab}}{p} $
Therefore, by substituting the value of $ c $ , we get,
 $ {\left( {\dfrac{{ab}}{p}} \right)^2} = {a^2} + {b^2} $
 $ \Rightarrow \dfrac{{{a^2}{b^2}}}{{{p^2}}} = {a^2} + {b^2} $
By cross multiplying, bringing $ {a^2}{b^2} $ to the denominator of the RHS,
 $ \dfrac{1}{{{p^2}}} = \dfrac{{{a^2} + {b^2}}}{{{a^2}{b^2}}} $
 $ \dfrac{1}{{{p^2}}} = \dfrac{{{a^2}}}{{{a^2}{b^2}}} + \dfrac{{{b^2}}}{{{a^2}{b^2}}} $
 $ \dfrac{1}{{{p^2}}} = \dfrac{1}{{{b^2}}} + \dfrac{1}{{{a^2}}} $
 $ \Rightarrow \dfrac{1}{{{p^2}}} = \dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}} $ .
Therefore, condition (ii) is also satisfied.
Hence, the given statement is true.
So, the correct answer is “Option A”.

Note: In this question, it is important to note here that to apply the area of the triangle formula, we have to know the base ( $ b $ ) and height ( $ h $ ) of it. This is applicable to all types of triangles, whether it is scalene, isosceles or equilateral. Here, the base and height are perpendicular to each other. However, Pythagoras theorem can be only used for right angled triangles. Be careful while considering the sides.