5 Marks Questions

Question 15 Marks

If two straight lines intersect in such a way that one of the angles formed measures $90^\circ$, show that each of the remaining angles measures $90^\circ $.

Answer

We know that if two lines intersect, then the vertically-opposite angles are equal.

$\angle\text{AOC}=90^\circ.$ Then, $\angle\text{AOC}=\angle\text{BOD}=90^\circ.$
And let $\angle\text{BOC}=\angle\text{AOD}=\text{x}$
Also, we know that the sum of all angles around a point is $360^\circ$.
$\therefore\angle\text{AOC}+\angle\text{BOD}+\angle\text{AOD}+\angle\text{BOC}=360^\circ$
$\Rightarrow90^\circ+90^\circ+\text{x}+\text{x}=360^\circ$
$\Rightarrow2\text{x}=180^\circ$
$\Rightarrow\text{x}=90^\circ$
Hence, $\angle\text{BOC}=\angle\text{AOD}=90^\circ$
$\therefore\angle\text{AOC}=\angle\text{BOD}=\angle\text{BOC}=\angle\text{AOD}=90^\circ$
Hence, the measure of each of the remaining angles is $90^\circ$ .

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Question 25 Marks

In the given figure, $AB\ ||\ CD$. Find the value of $x$.

Answer

Since $AB\ ||\ CD$ and $PQ$ a transversal.
So, $\angle\text{PEF}=\angle\text{EGH}$ [Corresponding angles]
$\Rightarrow\angle\text{EGH}=85^\circ$
$\angle\text{EGH}$ and $\angle\text{QGH}$ form a linear pair.
So, $\angle\text{EGH}+\angle\text{QGH}=180^\circ$
$\Rightarrow\angle\text{QGH}=180^\circ-85^\circ=95^\circ$
Similarly, $\angle\text{GHQ}+115^\circ=180^\circ$
$\Rightarrow\angle\text{GHQ}=180^\circ-115^\circ=65^\circ$ In $\triangle\text{GHQ},$
we have, $\Rightarrow\angle\text{GQH}+\angle\text{GHQ}+\angle\text{QGH}=180^\circ$
$\text{x}^\circ+65^\circ+95^\circ=180^\circ$
$\Rightarrow\text{x}=180-65-95=180-160$
$\therefore\text{x}=20$

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Question 35 Marks

In the given figure, $AB\ ||\ CD$ and $EF\ ||\ GH$. Find the value of $x, y, z$ and $t.$

Answer

$\angle\text{PRQ}=\text{x}^\circ=60^\circ$ [vertically opposite angles]
Since $EF || GH,$ and $RQ$ is a transversal.
So, $\angle\text{x}=\angle\text{y}$ [Alternate angles]
$\Rightarrow\text{y}=60 AB || CD$ and $PR$ is a transversal.
So, $\angle\text{PRD}=\angle\text{APR}$ [Alternate angles]
$\Rightarrow\angle\text{PRQ}+\angle\text{QRD}=\angle\text{APR}$
$[\text{since }\angle\text{PRD}=\angle\text{PRQ}+\angle\text{QRD]}$
$\Rightarrow\text{x}+\angle\text{QRD}=110^\circ$
$\Rightarrow\angle\text{QRD}=110^\circ-60^\circ=50^\circ$ In $\triangle\text{QRS},$
we have, $\angle\text{QRD}+\text{t}^\circ+\text{y}^\circ=180^\circ$
$\Rightarrow50+\text{t}+60=180$
$\Rightarrow\text{t}=180-110=70$ Since, $AB || CD$ and $GH$ is a transversal
So, $z^o = t^o = 70^o$ [Alternate angles]
$\therefore x = 60 , y = 60, z = 70$ and $t = 70$

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Question 45 Marks

In the given figure, the two lines $AB$ and $CD$ intersect at a point $O$ such that $\angle\text{BOC}=125^\circ.$ Find the values of $x, y$ and $z.$

Answer

Here, $\angle\text{AOC}$ and $\angle\text{BOC}$ from a liner pair.
$\therefore\angle\text{AOC}+\angle\text{BOC}=180^\circ$
$\Rightarrow\text{x}^\circ+125^\circ=180^\circ$
$\Rightarrow\text{x}^\circ=180^\circ-125^\circ=55^\circ$
Now, $\angle\text{AOD}=\angle\text{BOC}=125^\circ$ (Vertically opposite angles)
$\therefore\text{y}^\circ=125^\circ$
$\angle\text{BOD}=\angle\text{AOC}=55^\circ$ (Vertically opposite angles)
$\therefore\text{y}^\circ=55^\circ$ Thus, the respective values of $x, y$ and $z$ are $55, 125$ and $55.$

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Question 55 Marks

Two lines $AB$ and $CD$ intersect at a point $O,$ such that $\angle\text{BOC}+\angle\text{AOD}=280^\circ,$ as shown in the figure. Find all the four angles.

Answer

We know that if two lines intersect, then the vertically-opposite angles are equal.
Let $\angle\text{BOC}=\angle\text{AOD}=\text{x}^\circ$ Then, $\text{x}+\text{x}=280$
$\Rightarrow2\text{x}=280$
$\Rightarrow\text{x}=140^\circ$
$\therefore\angle\text{BOC}=\angle\text{AOD}=140^\circ$
Also, let $\angle\text{AOC}=\angle\text{BOD}=\text{y}^\circ$
We know that the sum of all angles around a point is $360^\circ .$
$\therefore\angle\text{AOC}+\angle\text{BOC}+\angle\text{BOD}+\angle\text{AOD}=360^\circ$
$\Rightarrow\text{y}+140+\text{y}+140=360^\circ$
$\Rightarrow2\text{y}=80^\circ$
$\text{y}=40^\circ$ Hence, $\angle\text{AOC}=\angle\text{BOD}=40^\circ$
$\therefore\angle\text{BOC}=\angle\text{AOD}=140^\circ$ and $\angle\text{AOC}=\angle\text{BOD}=40^\circ$

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Question 65 Marks

Two lines $AB$ and $CD$ intersect each other at a point $O$ such that $\angle\text{AOC}:\angle\text{AOD}=5:7.$ Find all the angles.

Answer

Let $\angle\text{AOC}=5\text{k}$ and $\angle\text{AOD}=7\text{k,}$ where $k$ is some constant.
Here, $\angle\text{AOC}$ and $\angle\text{AOD}$ form a linear pair.
$\therefore\angle\text{AOC}+\angle\text{AOD}=180^\circ$
$\Rightarrow5\text{k}+7\text{k}=180^\circ$
$\Rightarrow12\text{k}=180^\circ$
$\Rightarrow\text{k}=15^\circ$
$\therefore\angle\text{AOC}=5\text{k}=5\times15^\circ=75^\circ$
$\angle\text{AOD}=7\text{k}=7\times15^\circ=105^\circ$
Now, $\angle\text{BOD}=\angle\text{AOC}=75^\circ$ (Vertically opposite angles)
$\angle\text{BOC}=\angle\text{AOD}=105^\circ$ (Vertically opposite angles)

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Question 75 Marks

In the given figure, three lines $AB, CD$ and $EF$ intersect at a point $O$ such that $\angle\text{AOE}=35^\circ$ and $\angle\text{BOD}=40^\circ.$ Find the measure of $\angle\text{AOC},\angle\text{BOF},\angle\text{COF}$ and $\angle\text{DOE}.$

Answer

In the given figure,
$\angle\text{AOC}=\angle\text{BOD}=40^\circ$ (Vertically opposite angles)
$\angle\text{BOF}=\angle\text{AOE}=35^\circ$ (Vertically opposite angles)
Now, $\angle\text{EOC}$ and $\angle\text{COF}$ form a linear pair.
$\therefore\angle\text{EOC}+\angle\text{COF}=180^\circ$
$\Rightarrow(\angle\text{AOE}+\angle\text{AOC})+\angle\text{COF}=180^\circ$
$\Rightarrow35^\circ+406\circ+\angle\text{COF}=180^\circ$
$\Rightarrow75^\circ+\angle\text{COF}=180^\circ$
$\Rightarrow\angle\text{COF}=180^\circ-75^\circ=105^\circ$
Also, $\angle\text{DOE}=\angle\text{COF}=105^\circ$ (Vertically opposite angles)

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Question 85 Marks

In the given figure, $AB\ ||\ CD$. Find the value of $x$.

Answer

Through $C$ draw $FG || AE$

Now, since $CG\ ||\ BE$ and $CE$ is a transversal.
So, $\angle\text{GCE}=\angle\text{CEA}=20^\circ$ [Alternate angles]
$\therefore\angle\text{DCG}=130^\circ-\angle\text{GCE}$
$=130^\circ-20^\circ=110^\circ$
Also, we have $AB\ ||\ CD$ and $FG$ is a transversal.
So, $\angle\text{BFC}=\angle\text{DCG}=110^\circ$ [Corresponding angles]
As, $FG\ ||\ AE, AF$ is a transversal.
$\angle\text{BFG}=\angle\text{FAE}$ [Corresponding angles]
$\therefore\text{x}^\circ=\angle\text{FAE}=110^\circ.$
Hence, $x = 110$.

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Question 95 Marks

In the figure, $AB || PQ$. Find the value of $x$ and$ y.$

Answer

Since $AB || PQ$ and $EF$ is a transversal.
So, $\angle\text{CEB}=\angle\text{EFQ}$ [Corresponding angles]
$\Rightarrow\angle\text{EFQ}=75^\circ$
$\Rightarrow\angle\text{EFG}+\angle\text{GFQ}=75^\circ$
$25^\circ+\text{y}^\circ=75^\circ$
$\Rightarrow\text{y}=75-25=50$
Also, $\angle\text{BEF}+\angle\text{EFQ}=180^\circ$ [sum of consecutive interior angles is $180^o$]
$\angle\text{BEF}=180^\circ-\angle\text{EFQ}$
$=180^\circ-75^\circ$
$\angle\text{BEF}=105^\circ$
$\therefore\angle\text{FEG}+\angle\text{GEB}=\angle\text{BEF}=105^\circ$
$\Rightarrow\angle\text{FEG}=105^\circ-\angle\text{GEB}=105^\circ-20^\circ=85^\circ$ In $\triangle\text{EFG}$
we have, $\text{x}^\circ+25^\circ+\angle\text{FEG}=180^\circ$
$\Rightarrow\text{x}^\circ+25^\circ+85^\circ=180^\circ$
$\Rightarrow\text{x}^\circ+110^\circ=180^\circ$
$\Rightarrow\text{x}^\circ=180^\circ-110^\circ$
$\Rightarrow\text{x}^\circ=70^\circ$
Hence, $x = 70.$

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