Question

ABC is a right angled triangle with \[\angle C = 90^\circ \], \[BC = a\], \[AC = b\], \[CD \bot AB\], and \[CD = P\]. Show that \[\dfrac{1}{{{P^2}}} = \dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}\].

Answer 1

Hint: Here, we need to prove the given expression. We will use Pythagoras’s theorem and the formula for area of triangle to form an equation, and simplify it to prove the given expression. Pythagoras theorem can be applied only on a right-angled triangle.

Formula Used:
We will use the area of a triangle is given by the formula \[\dfrac{1}{2}bh\], where \[b\] is the base of the triangle, and \[h\] is the height of the triangle.

Complete step-by-step answer:
First, we will draw the figure.

Here, angle C is of measure 90 degrees.
Therefore, triangle ABC is a right angled triangle.
Now, we will use the Pythagoras’s theorem in right angled triangle ABC.
In triangle ABC, AB is the hypotenuse, BC is the base, and AC is the height.
Therefore, in triangle ABC, we get
\[ \Rightarrow A{B^2} = B{C^2} + A{C^2}\]
Substituting \[BC = a\] and \[AC = b\] in the equation, we get
\[ \Rightarrow A{B^2} = {a^2} + {b^2}\]
Now, it is given that \[CD \bot AB\].
Therefore, we can observe that CD is the height of triangle ABC.
The area of a triangle is given by the formula \[\dfrac{1}{2}bh\].
Therefore, we get
Area of triangle ABC \[ = \dfrac{1}{2} \times AB \times CD\]
Substituting \[CD = P\] in the equation, we get
\[ \Rightarrow \] Area of triangle ABC \[ = \dfrac{1}{2} \times AB \times P\]
Simplifying the expression, we get
\[ \Rightarrow \] Area of triangle ABC \[ = \dfrac{P}{2} \times AB\]……………………..\[\left( 1 \right)\]
Also, it is given that \[\angle C = 90^\circ \].
Therefore, we can observe that AC is the height of triangle ABC.
Therefore, we get
\[ \Rightarrow \] Area of triangle ABC \[ = \dfrac{1}{2} \times BC \times AC\]
Substituting \[BC = a\] and \[AC = b\] in the equation, we get
\[ \Rightarrow \] Area of triangle ABC \[ = \dfrac{1}{2} \times a \times b\]
Simplifying the expression, we get
\[ \Rightarrow \] Area of triangle ABC \[ = \dfrac{{ab}}{2}\] ……………………..\[\left( 2 \right)\]
From equations \[\left( 1 \right)\] and \[\left( 2 \right)\], we get
\[ \Rightarrow \dfrac{P}{2} \times AB = \dfrac{{ab}}{2}\]
Multiplying both sides by 2, we get
\[ \Rightarrow P \times AB = ab\]
Squaring both sides, we get
\[ \Rightarrow {\left( {P \times AB} \right)^2} = {\left( {ab} \right)^2}\]
Therefore, we get
\[ \Rightarrow {P^2} \times A{B^2} = {a^2}{b^2}\]
Substituting \[A{B^2} = {a^2} + {b^2}\] in the equation, we get
\[ \Rightarrow {P^2}\left( {{a^2} + {b^2}} \right) = {a^2}{b^2}\]
Dividing both sides of the equation by \[{a^2}{b^2}{P^2}\], we get
\[ \Rightarrow \dfrac{{{P^2}\left( {{a^2} + {b^2}} \right)}}{{{a^2}{b^2}{P^2}}} = \dfrac{{{a^2}{b^2}}}{{{a^2}{b^2}{P^2}}}\]
Simplifying the expression, we get
\[ \Rightarrow \dfrac{{{a^2} + {b^2}}}{{{a^2}{b^2}}} = \dfrac{1}{{{P^2}}}\]
Rewriting the expression by splitting using the L.C.M., we get
\[ \Rightarrow \dfrac{{{a^2}}}{{{a^2}{b^2}}} + \dfrac{{{b^2}}}{{{a^2}{b^2}}} = \dfrac{1}{{{P^2}}}\]
Simplifying the expression, we get
\[ \Rightarrow \dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}} = \dfrac{1}{{{P^2}}}\]
Hence, we have proved that \[\dfrac{1}{{{P^2}}} = \dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}\].
Note: We used the Pythagoras’s theorem in the triangle ABC. The Pythagoras’s theorem states that the square of the hypotenuse of a right angled triangle is equal to the sum of squares of the other two sides, that is \[{\rm{Hypotenus}}{{\rm{e}}^2} = {\rm{Bas}}{{\rm{e}}^2} + {\rm{Perpendicula}}{{\rm{r}}^2}\]. The hypotenuse of a right angled triangle is its longest side.