# Suppose you have a traingle with sides: a, b and c. Using pythagorean theorem what can you deduce from the following inequality? i) a^2+b^2 = c^2 ii) a^2+b^2 lt c^2 iii) a^2+b^2 gt c^2

##### 1 Answer
Dec 26, 2016

Please see below.

#### Explanation:

(i) As we have ${a}^{2} + {b}^{2} = {c}^{2}$, which means that sum of the squares of the two sides $a$ and $b$ is equal to square on the third side $c$. Hence, $\angle C$ opposite side $c$ will be right angle.

Assume, it is not so, then draw a perpendicular from $A$ to $B C$, let it be at $C '$. Now according to Pythagoras theorem, ${a}^{2} + {b}^{2} = {\left(A C '\right)}^{2}$. Hence, $A C ' = c = A C$. But this is not possible. Hence, $\angle A C B$ is a right angle and $\Delta A B C$ is a right angled triangle.

Let us recall the cosine formula for triangles, which states that ${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cos C$.

(ii) As range of $\angle C$ is ${0}^{\circ} < C < {180}^{\circ}$, if $\angle C$ is obtuse $\cos C$ is negative and hence ${c}^{2} = {a}^{2} + {b}^{2} + 2 a b | \cos C |$. Hence, ${a}^{2} + {b}^{2} < {c}^{2}$ means $\angle C$ is obtuse.

Let us use Pythagoras theorem to check it and draw $\Delta A B C$ with $\angle C > {90}^{\circ}$ and draw $A O$ perpendicular on extended $B C$ as shown. Now according to Pythagoras theorem

${a}^{2} + {b}^{2} = B {C}^{2} + A {C}^{2}$

= ${\left(B O - O C\right)}^{2} + A {C}^{2}$

= $B {O}^{2} + O {C}^{2} - 2 B O \times C O + A {O}^{2} + O {C}^{2}$

= $B {O}^{2} + A {O}^{2} - 2 O C \left(B O - O C\right)$

= $A {B}^{2} - 2 O C \times B C = {c}^{2} - O C \times B C$

Hence ${a}^{2} + {b}^{2} < {c}^{2}$

(iii) and if $\angle C$ is acute $\cos C$ is positive and hence ${c}^{2} = {a}^{2} + {b}^{2} - 2 a b | \cos C |$. Hence, ${a}^{2} + {b}^{2} > {c}^{2}$ means $\angle C$ is acute.

Again using Pythagoras theorem to check this, draw $\Delta A B C$ with $\angle C < {90}^{\circ}$ and draw $A O$ perpendicular on $B C$ as shown. Now according to Pythagoras theorem

${a}^{2} + {b}^{2} = B {C}^{2} + A {C}^{2}$

= ${\left(B O + O C\right)}^{2} + A {O}^{2} + O {C}^{2}$

= $B {O}^{2} + O {C}^{2} + 2 B O \times C O + A {O}^{2} + O {C}^{2}$

= $A {B}^{2} + 2 O C \left(C O + O B\right)$

= ${c}^{2} + 2 a \times O C$

Hence ${a}^{2} + {b}^{2} > {c}^{2}$