Question
State and prove Py thagoras theorem.
✓
Answer
Pythagoras theorem: In a right triangle, the
Given: $\triangle P Q R$ is a right triangle, right angled at $Q$.
To prove: $P R^2=P Q^2+Q R^2$
Construction: Draw $Q S \perp P R$
Proof: In $\triangle P Q S$ and $\triangle P Q R$,
$\angle P S Q=\angle P Q R=90^{\circ}$
$\angle Q P S=\angle Q P R \ ($common$) $
(by AAA similarity criterion)
$\Rightarrow \frac{P S}{P Q}=\frac{P Q}{P R}$
$\Rightarrow P S \times P R=P Q^2 \ldots \ldots(i)$
Now, In $\triangle Q S R$ and $\triangle P Q R$,
$\angle Q S R=\angle P Q R=90^{\circ}$
$\angle Q R S=\angle Q R P \ ($common$)$
$\therefore \triangle Q S R-\triangle P Q R \ ($by $AA$ similarity criterion$) $
$\Rightarrow \frac{R S}{Q R}=\frac{Q R}{P R}$
$\Rightarrow R S \times P R=Q R^2 \ldots \ldots(ii)$
Adding $(i)$ and $(ii)$
$\Rightarrow P Q^2+Q R^2=P S \times P R+R S \times P R$
$\Rightarrow P Q^2+Q R^2=P R(P S+R S)$
$\Rightarrow P Q^2+Q R^2=P R . P R$
$\Rightarrow P Q^2+Q R^2=P R^2 $
$\Rightarrow P R^2=P Q^2+Q R^2$
Hence proved
Given: $\triangle P Q R$ is a right triangle, right angled at $Q$.
To prove: $P R^2=P Q^2+Q R^2$
Construction: Draw $Q S \perp P R$
Proof: In $\triangle P Q S$ and $\triangle P Q R$,
$\angle P S Q=\angle P Q R=90^{\circ}$
$\angle Q P S=\angle Q P R \ ($common$) $
(by AAA similarity criterion)
$\Rightarrow \frac{P S}{P Q}=\frac{P Q}{P R}$
$\Rightarrow P S \times P R=P Q^2 \ldots \ldots(i)$
Now, In $\triangle Q S R$ and $\triangle P Q R$,
$\angle Q S R=\angle P Q R=90^{\circ}$
$\angle Q R S=\angle Q R P \ ($common$)$
$\therefore \triangle Q S R-\triangle P Q R \ ($by $AA$ similarity criterion$) $
$\Rightarrow \frac{R S}{Q R}=\frac{Q R}{P R}$
$\Rightarrow R S \times P R=Q R^2 \ldots \ldots(ii)$
Adding $(i)$ and $(ii)$
$\Rightarrow P Q^2+Q R^2=P S \times P R+R S \times P R$
$\Rightarrow P Q^2+Q R^2=P R(P S+R S)$
$\Rightarrow P Q^2+Q R^2=P R . P R$
$\Rightarrow P Q^2+Q R^2=P R^2 $
$\Rightarrow P R^2=P Q^2+Q R^2$
Hence proved
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