5 Marks Questions

Question 15 Marks

Prove: Two distinct lines cannot have more than one point in common.

Answer

Proof : In the statement above, it is given that 'two lines intersect each other'. So, let $\mathrm{AB}$ and $\mathrm{CD}$ be two lines intersecting at $\mathrm{O}$ as shown in Fig. 6.8. They lead to two pairs of vertically opposite angles, namely,
(i) $\angle \mathrm{AOC}$ and $\angle \mathrm{BOD}$ (ii) $\angle \mathrm{AOD}$ and $\angle \mathrm{BOC}$.
We need to prove that $\angle \mathrm{AOC}=\angle \mathrm{BOD}$ and $\angle \mathrm{AOD}=\angle \mathrm{BOC}$.
Now, ray $\mathrm{OA}$ stands on line $\mathrm{CD}$.
Therefore, $\angle \mathrm{AOC}+\angle \mathrm{AOD}=180^{\circ}$ $\quad$ (Linear pair axiom) (1)
Can we write $\angle \mathrm{AOD}+\angle \mathrm{BOD}=180^{\circ}$ ? Yes! (Why?) $\quad$ (2)
From (1) and (2), we can write
$
\angle \mathrm{AOC}+\angle \mathrm{AOD}=\angle \mathrm{AOD}+\angle \mathrm{BOD}
$
This implies that $\angle \mathrm{AOC}=\angle \mathrm{BOD} \quad$ (Refer Section 5.2, Axiom 3)
Similarly, it can be proved that $\angle \mathrm{AOD}=\angle \mathrm{BOC}$
Now, let us do some examples based on Linear Pair Axiom and Theorem 6.1.

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Question 25 Marks

If two lines intersect, prove that the vertically opposite angles are equal.

Answer

Given, two lines $A B$ and $C D$ intersect at point $O$.
To prove
$(i) \angle AOC =\angle BOD$
$(ii) \angle A O D=\angle BOC$
Proof:
$(i)$ Ray $OA$ stands on line $CD$.
$\therefore \angle A O C+\angle A O D=180^{\circ} \ [$linear pair axiom$] ...(i)$
Ray $O D$ stands on line $A B$.
$\therefore \angle A O D+\angle B O D=180^{\circ}\ [$linear pair exiom $] \ldots (ii)$

From the equations $(i)$ and $(ii),$
$ \Rightarrow \angle A O C+\angle A O D$
$=\angle AOD+\angle BOD$
$ \angle A O C=\angle BOD $
$(ii)$ Ray $O D$ stands on line $A B$.
$ \therefore \angle A O D+\angle B O D=180^{\circ}\ [$ linear pair axiom $] \ (iii) $
Ray $O B$ stands on line $CD.$
$ \therefore \angle DOB+\angle BOC=180^{\circ} $
From equations $(iii)$ and $(iv)$
$ \Rightarrow \angle AOD+\angle BOD$
$=\angle DOB+\angle BOC$
$\angle AOD=\angle BOC $
Hence proved.

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Question 35 Marks

In Figure lines AB and CD intersect at O. If $\angle AOC + \angle BOE = {70^ \circ }$ and $\angle BOD = {40^ \circ }$,find $\angle BOE$ and reflex $\angle COE$
Fig.6.1.png

Answer

We are given that $\angle$AOC + $\angle$BOE = 70° and $\angle$BOD = 40°
We need to find $\angle$BOE and reflex $\angle$COE
From the given figure, we can conclude that $\angle$AOE and $\angle$BOE form a linear pair.
We know that sum of the angles of a linear pair is 180°
$\therefore$ $\angle$AOE + $\angle$BOE = 180°
$\because$ $\angle$AOE = $\angle$AOC + $\angle$ COE
$\therefore$ $\angle$AOC + $\angle$COE + $\angle$BOE = 180°
$\therefore$ $\angle$AOC + $\angle$BOE + $\angle$COE = 180°
$\Rightarrow$ 70° + $\angle$COE = 180°
$ \Rightarrow$ $\angle$COE = 180° - 70°
= 110°
Reflex $\angle$COE = 360° - $\angle$COE
= 360° - 110°
= 250°
$\angle$AOC = $\angle$BOD (Vertically opposite angles), or
$\angle$BOD + $\angle$BOE = 70
But, we are given that $\angle$BOD = 40°.
40° + $\angle$BOE = 70°
$\angle$BOE = 70° - 40°
= 30°.
Therefore, we can conclude that Reflex $\angle$COE = 250° and $\angle$BOE = 30°

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