3 Marks Question

Question 13 Marks

In figure, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B. The reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD.

Answer

Draw ray BL $\perp$ PQ and ray CM $\perp$ RS.

BL $\perp$ PQ, CM $\perp$ RS and PQ $\parallel$ RS
BL || CM
$\angle$LBC = $\angle$MCB . . . (1)
$\angle$ABL = $\angle$LBC . . . [Angle of incident = Angle of reflection] . . . .(2)
$\angle$MCB = $\angle$MCD . . . [Angle of incident = Angle of reflection] . . . . (3)
$\angle$ABL = $\angle$MCD . . . [From (1), (2) and (3)]
$\angle$LBC + $\angle$ABL = $\angle$MCB + $\angle$MCD . . . [Adding (1) and (4)]
$\angle$ABC = $\angle$BCD are alternate interior angles and are equal.
$\therefore$ AB || CD

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Question 23 Marks

In the given figure, if AB || CD, $\angle$ APQ = 50° and $\angle$PRD = 127°, find x and y.

Answer

We are given that $AB\parallel CD,\angle APQ = {50^\circ }$ and $\angle$PRD = 127°
We need to find the value of x and y in the figure.
$\angle$APQ = x = 50° (Alternate interior angles)
$\angle$PRD = $\angle$APR = 127° (Alternate interior angles)
$\angle$APR = $\angle$QPR + $\angle$APQ.
127° = y + 50°
$\Rightarrow$ y = 77°.
Therefore, we can conclude that x = 55° and y = 77°
Alternatively, 127° = y + x (because exterior angle is equal to the sum of interior opposite angles).
so, 127° = y + 50°
which gives, x = 50° and y = 77°

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Question 33 Marks

In fig lines XY and MN intersect at O If $\angle $POY = 90° and a:b = 2:3 find $\angle $c.

Answer

Lines XY and MN intersect at O.
$\therefore \angle C = \angle XON = \angle MOY$ [vertically opposite angle]
$= \angle b + \angle POY$
But, $\angle$POY = 90°
$\therefore $ $\angle$C = $\angle$b + 90° ...(i)
Also,
$\angle$POX = 180° - $\angle$POY
= 180° - 90°
= 90°
$\therefore$ a + b = 90°
But,
a:b = 2:3 [Given]
$a = \frac{2}{5} \times {90^0}$
= 36° ...(ii)
Thus, From (i) and (ii) we get
b = 90° - 36° = 54°
$\angle$C = 54° + 90° ( From (1) )
= 144°

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Question 43 Marks

In the given figure, OP, OQ, OR and OS are four rays. Prove that
$\angle$POQ + $\angle$ROQ + $\angle$SOR + $\angle$POS = 360°.

Answer

Let us produce a ray OQ backwards to a point M, then MOQ is a straight line.
Now, OP is a ray on the line MOQ. Then, by linear pair axiom, we have
$\angle$MOP + $\angle$POQ = 180^o ......(i)

Similarly, OS is a ray on the line MOQ. Then, by linear pair axiom, we have
$\angle$MOS + $\angle$SOQ = 180^o ....(ii)
Also, $\angle$SOR and $\angle$ROQ are adjacent angles.
$\therefore$ $\angle$SOQ = $\angle$SOR + $\angle$ROQ ...(iii)
On putting the value of $\angle$SOQ from Eq.(iii) in Eq.(ii), we get
$\angle$MOS + $\angle$SOR + $\angle$ROQ = 180^o ....(iv)
Now, on adding Eqs.(i) and (iv), we get
$\angle$MOP + $\angle$POQ + $\angle$MOS + $\angle$SOR + $\angle$ROQ = 180^o + 180^o
$\Rightarrow$ $\angle$MOP + $\angle$MOS + $\angle$POQ + $\angle$SOR + $\angle$ROQ = 360^o ....(iv)
But $\angle$MOP + $\angle$MOS = $\angle$POS
Then, from Eq.(v), we get
$\angle$POS + $\angle$POQ + $\angle$SOR + $\angle$ROQ = 360^o
Hence proved.

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