5 Marks Questions

Question 15 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 $\angle A O C+\angle B O E=70^{\circ}$ and $\angle B O D=40^{\circ}$
We need to find $\angle B O E$ and reflex $\angle C O E$
From the given figure, we can conclude that $\angle A O E$ and $\angle B O E$ form a linear pair.
We know that sum of the angles of a linear pair is $180^{\circ}$
$\therefore \angle AOE + \angle BOE = 180^\circ $
$ \because \angle AOE = \angle AOC + \angle COE$
$ \therefore \angle AOC + \angle COE + \angle BOE = 180^\circ $
$ \therefore \angle AOC + \angle BOE + \angle COE = 180^\circ $
$ \Rightarrow 70^\circ + \angle COE = 180^\circ $
$ \Rightarrow \angle COE = 180^\circ - 70^\circ $
$= 110^\circ $
Reflex $\angle COE = 360^\circ - \angle COE$
$= 360^\circ - 110^\circ $
$= 250^\circ $
$ \angle AOC = \angle BOD$ (Vertically opposite angles), or
$\angle BOD + \angle BOE = 70$
But, we are given that $\angle BOD = 40^\circ .$
$40^\circ + \angle BOE = 70^\circ $
$ \angle BOE = 70^\circ - 40^\circ $
$= 30^\circ $.
Therefore, we can conclude that Reflex $\angle COE =250^{\circ}$ and $\angle BOE =30^{\circ}$

View full question & answer→

Question 25 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.

View full question & answer→

Question 35 Marks

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

Answer

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

From the equations $(i)$ and $(ii),$
$\Rightarrow \angle \ce{AOC + \angle AOD=\angle AOD + \angle BOD}$
$\ce{\angle AOC=\angle BOD}$
$(ii)$ Ray $OD$ stands on line $AB$.
$ \therefore \ce{\angle AOD + \angle BOD}=180^{\circ}[$ linear pair axiom $] \ldots (iii)$
Ray $OB$ stands on line $CD.$
$ \therefore \ce{\angle DOB + \angle BOC=180^{\circ}}$
From equations $(iii)$ and $(iv)$
$\Rightarrow \ce{\angle AOD + \angle BOD=\angle DOB+\angle BOC}$
$\ce{\angle AOD=\angle BOC}$
Hence proved.

View full question & answer→

MATHS 5 Marks Questions

Test yourself on this topic