Statement-1 (A): Let P Q R be a right triangle right angled at Q … - Vidyadip

MCQ

Statement-1 (A): Let $\triangle P Q R$ be a right triangle right angled at $Q$ such that the perpendicular drawn from $Q$ on hypotenuse $P R$ meets $P R$ at $S$. If $P S=4$ cm and $R S=9 cm$, then $Q S=6 cm$.
Statement-2 (R): In a right triangle, the square of the perpendicular drawn from the vertex forming right angle to the hypotenuse is equal to the product of projections of two sides on the hypotenuse.

✓
Statement-1 and Statement-2 are True; Statement-2 is a correct explanation for Statement-1.
B
Statement-1 and Statement-2 are True; Statement- 2 is not a correct explanation for Statement-1.
C
Statement-1 is True, Statement-2 is False.
D
Statement-1 is False, Statement-2 is True.

✓

Answer

Correct option: A.

Statement-1 and Statement-2 are True; Statement-2 is a correct explanation for Statement-1.

(A)Statement-1 and Statement-2 are True; Statement-2 is a correct explanation for Statement-1.

Let $A B C$ be a right triangle right angled at $A$ and $A D$ be perpendicular drawn from $A$ on hypotenuse $B C$.
In $\triangle$ 's $A D B$ and $C D A$, we have
$\angle A D B=\angle C D A$ $\qquad$ [Each equal to $90^{\circ}$ ]
and, $\quad \angle D A B=\angle D C A$
So, by using $A A$-criterion of similarity, we obtain
$
\triangle A D B \sim \triangle C D A \Rightarrow \frac{A D}{C D}=\frac{D B}{D A} \Rightarrow A D^2=B C \times C D
$
Thus, statement-2 is true. Using statement-2, we find that in $\triangle P Q R$.
$
Q S^2=P S \times R S \Rightarrow Q S^2=4 \times 9=36 \Rightarrow Q S=6 cm
$
So, statement- 1 is also true and statement- 2 is a correct explanation for statement- 1 .
Hence, option (a) is correct.

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