1 Marks Question

Question 11 Mark

In the figure, $POQ$ is a line. Ray $OR$ is perpendicular to line $PQ. OS$ is another ray lying between rays $OP$ and $OR.$ Prove that $\angle ROS={1\over2}(\angle QOS-\angle POS)$

Answer

Ray $OR$ is perpendicular to line $PQ$
$ \therefore \angle QOR = \angle POR = 90^o . . . .(1) $
$ \angle QOS = \angle QOR + \angle ROS. . . .(2) $
$ \angle POS = \angle POR – \angle ROS. . . .(3)$
From $(2)$ and $(3),$
$\therefore \angle QOS – \angle POS = ( \angle QOR – \angle POR) + 2 \angle ROS = 2 \angle ROS . . . [$Using$ (1)]$
$ \therefore \angle ROS={1\over2}(\angle QOS-\angle POS)$

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Question 21 Mark

In figure, if $x + y = w + z,$ then prove that $AOB$ is a line.

Answer

Given that, $x+y=w+z ... (1)$
As the sum of all angles round a point is equal to $360^{\circ}$
$\therefore \mathrm{x}+\mathrm{y}+\mathrm{w}+\mathrm{z}=360^{\circ}$
$\therefore \mathrm{x}+\mathrm{y}+\mathrm{x}+\mathrm{y}=360^{\circ}$
$\therefore 2(\mathrm{x}+\mathrm{y})=360^{\circ}$
$\therefore x+y=\frac{360^{\circ}}{2}$
$\therefore \mathrm{x}+\mathrm{y}=180^{\circ}$
$\therefore A O B$ is a line.

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Question 31 Mark

In Fig., if $QT \perp PR, \angle TQR = 40^\circ $ and $\angle SPR = 30^\circ ,$ find $x$ and $y$

Answer

In $\triangle TQR, 90^\circ + 40^\circ + x = 180^\circ ($Angle sum property of a triangle$)$
Therefore, $x = 50^\circ $
Now, $y = \angle SPR + x ($If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.$)$
Therefore, $y = 30^\circ + 50^\circ = 80^\circ $

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