[2 Mark Question Answer]

Question 12 Marks

In the given figure, $T$ is a point on the side $\text{PR}$ of an equilateral $\triangle PQR$. Show that $\text{RT} < \text{QT}.$

Answer

In $\triangle PQR,$
$\text{PQ} = \text{QR} = \text{PR}$
$\Rightarrow \angle P = Q = \angle = 60^\circ $
In $\triangle TQR,$
$\angle TQR < 60^\circ $
$\therefore \angle TQR < \angle R$
$\therefore \text{RT} < \text{QT}.$

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Question 22 Marks

In $\triangle PQR, \text{PS} \perp \text{QR} ;$ prove that$: \text{PQ} > \text{QS}$ and $\text{PR} > \text{PS}$

Answer

In $ \triangle PQS,$
$\text{PS} \perp \text{QR} ....($Given$)$
$\text{PS} < \text{PR} ....($Of all the straight lines that can be drawn to a given straight line from a point outside it, the perpendicular is the shortest.$)$
I.e. $\text{PR} > \text{PS}.$

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Question 32 Marks

$\text{ABCD}$ is a trapezium. Prove that:

$\text{CD} +\text{DA} +\text{AB} > \text{BC}.$

Answer

In $\triangle ACD,$ we have
$\text{CD} +\text{DA} > \text{CA}$
$\Rightarrow \text{CD} + \text{DA} + \text{AB} > \text{CA} +\text{AB}$
$\Rightarrow \text{CD} + \text{DA} + \text{AB} > \text{BC}. ...[\therefore \text{AB} + \text{AC} > \text{BC}]$

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Question 42 Marks

$\text{ABCD}$ is a trapezium. Prove that$:$

$\text{CD} +\text{DA} +\text{AB} +\text{BC} > 2\text{AC}.$

Answer

In $\triangle ABC,$ we have
$\text{AB} + \text{BC} > \text{AC} ...(i)$
In $\triangle ACD,$ we have
$\text{AD} +\text{CD} > \text{AC} ...(ii)$
Adding $(i)$ and $(ii),$ we get
$\text{AB} + \text{BC} + \text{AD} + \text{CD} > 2\text{AC}.$

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Question 52 Marks

In $\triangle PQR, \text{PR} > \text{PQ}$ and $T$ is a point on $\text{PR}$ such that $\text{PT} = \text{PQ}.$ Prove that $\text{QR} > \text{TR}.$

Answer

In $\triangle PQT,$ we have
$\text{PT} = \text{PQ} ...(1)$
In $\triangle PQR,$
$\text{PQ} + \text{QR} > \text{PR}$
$\text{PQ} + \text{QR} > \text{PT} + \text{TR}$
$\text{PQ} + \text{QR}>\text{PQ} + \text{TR} ...[$Using $(1)]$
$\text{QR} > \text{TR}$
Hence, proved.

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MATHEMATICS [2 Mark Question Answer]

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