4 Mark Question - Maths STD 9 Questions - Vidyadip

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4 Mark Question

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18 questions · self-marked practice — reveal the answer and mark yourself.

Question 14 Marks

In the given figure, l || m and a transversal t cuts them, If $\angle7=80^\circ,$ find the measure of each of the remaining marked angles.

Answer

Given, $\angle7=80^\circ$Now, $\angle7+\angle8=180 ^\circ$ (linear pair)
$\Rightarrow80^\circ+\angle8=180^\circ$
$\Rightarrow\angle8 =100^\circ$
$\angle7=\angle5$ (vertically opposite angles)
$\Rightarrow\angle5=80^\circ$
Also, $\angle6=\angle8$ (vertically opposite angles)
$\Rightarrow\angle6=100^\circ$
Line l || line m and line t is a transversal.
$\Rightarrow\angle1=\angle5=80^\circ$ (corresponding angles)
$\angle2=\angle6=100^\circ$ (corresponding angles)
$\angle3=\angle7=80^\circ$ (corresponding angles)
$\angle4=\angle8=100^\circ$ (corresponding angles)

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Question 24 Marks

In the given figure, l || m and a transversal t cuts them, If $\angle1=120^\circ,$ find the measure of each of the remaining marked angles.

Answer

Given, $\angle1=120^\circ$ Now, $\angle1+\angle2=180 ^\circ$ (linear pair)$\Rightarrow120^\circ+\angle2=180^\circ$
$\Rightarrow\angle2 =60^\circ$
$\angle1=\angle3$ (vertically opposite angles)
$\Rightarrow\angle3=120^\circ$
Also, $\angle2=\angle4$ (vertically opposite angles)$\Rightarrow\angle4=60^\circ$
Line l || line m and line t is a transversal.$\Rightarrow\angle5=\angle1=120^\circ$ (corresponding angles)
$\angle6=\angle2=60^\circ$ (corresponding angles)
$\angle7=\angle3=120^\circ$ (corresponding angles)
$\angle8=\angle4=60^\circ$ (corresponding angles)

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Question 34 Marks

In the given figure, BA || ED and BC || EF. Show that $\angle\text{ABC}=\angle\text{DEF}.$

Answer

Construction: Produce DE to meet BC at Z.

Now, AB || DZ and BC is the transversal.$\Rightarrow\angle\text{ABC}=\angle\text{DZC}$ (corresponding angles) ….(i)
Also, EF || BC and DZ is the transversal.$\Rightarrow\angle\text{DZC}=\angle\text{DEF}$ (corresponding angles) ….(ii)
From (i) and (ii), we have$\angle\text{ABC}=\angle\text{DEF}$

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Question 44 Marks

In the figure given below, AB || CD. Find the value of x in each case.

Answer

Through E draw EG || CD. Now since EG || CD and ED is a transversal.

So, $\angle\text{GED}=\angle\text{EDC}=65^\circ$ [Alternate interior angles] Since EG || CD and AB || CD, EG || AB and EB is transversal. So, $\angle\text{BEG}=\angle\text{ABE}=35^\circ$ [Alternate interior angles] So, $\angle\text{DEB}=\text{x}^\circ$$\Rightarrow\angle\text{BEG}+\angle\text{GED}=35^\circ+65^\circ=100^\circ.$
Hence, x = 100.

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Question 54 Marks

In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O. Find the value of x. Also, find $\angle\text{AOD},\angle\text{COE}$ and $\angle\text{AOE}.$

Answer

We know that if two lines intersect, then the vertically-opposite angles are equal.$\therefore\angle\text{DOF}=\angle\text{COE}=5\text{x}^\circ$
$\angle\text{AOD}=\angle\text{BOC}=2\text{x}^\circ\text{ and}$
$\angle\text{AOE}=\angle\text{BOF}=3\text{x}^\circ$
Since, AOB is a straight line, we have:$\angle\text{AOE}+\angle\text{COE}+\angle\text{BOC}=180^\circ$
⇒ 3x + 5x + 2x = 180º ⇒ 10x = 180º ⇒ x = 18º Therefore,$\angle\text{AOD}=2\times18^\circ=36^\circ$
$\angle\text{COE}=5\times18^\circ=90^\circ$
$\angle\text{AOE}=3\times18^\circ=54^\circ$

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Question 64 Marks

Prove that the bisectors of two adjacent supplementary angles include a right angle.

Answer

Let AOB denote a straight line and let$\angle\text{AOC}$ and $\angle\text{BOC}$

be the supplementary angles. Then, we have:$\angle\text{AOE}=\angle\text{COE}=\frac{1}{2}\text{x}^\circ\text{ and}$
$\angle\text{BOF}=\angle\text{FOC}=\frac{1}{2}(180-\text{x})^\circ$
Therefore,$\angle\text{COE}+\angle\text{FOC}=\frac{1}{2}\text{x}+\frac{1}{2}(180^\circ-\text{x})$
$=\frac{1}{2}(\text{x}+180^\circ-\text{x})$
$=\frac{1}{2}(180^\circ)$
$=90^\circ$

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Question 74 Marks

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

Answer

Since$ AB\ ||\ CD$ and AC is a transversal.
So, $\angle\text{BAC}+\angle\text{ACD}=180^\circ$ [sum of consecutive interior angles is $180^{\circ}$]
$\Rightarrow\angle\text{ACD}=180^\circ-\angle\text{BAC}$
$=180^\circ-75^\circ=105^\circ$
$\angle\text{ECF}=\angle\text{ACD}$ [Vertically opposite angles]
$\Rightarrow\angle\text{ECF}=105^\circ$
Now in $\triangle\text{CEF},$$\angle\text{ECF}+\angle\text{CEF}+\angle\text{EFC}=180^\circ$
$\Rightarrow105^\circ+\text{x}^\circ+30^\circ=180^\circ$
$\Rightarrow\text{x}=180-30-105=45$
Hence, $x = 45.$

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Question 84 Marks

In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O, forming angles as shown. Find the values of x, y, z and t.

Answer

We know that if two lines intersect, then the vertically opposite angles are equal.$\therefore\angle\text{BOD}=\angle\text{AOC}=90^\circ$
Hence, t = 90º Also,$\angle\text{DOF}=\angle\text{COE}=50^\circ$
Hence, z = 50º Since, AOB is a straight line, we have:$\angle\text{AOC}+\angle\text{COE}+\angle\text{BOE}=180^\circ$
⇒ 90 + 50 + y = 180º ⇒ 140 + y = 180º ⇒ y = 40º Also,$\angle\text{BOE}=\angle\text{AOF}=40^\circ$
Hence, x = 40º$\therefore$ x = 40º, y = 40º, z = 50º and t = 90º

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Question 94 Marks

In the given figure, AB || CD. Find the value of x, y and z.

Answer

AB || CD and EF is transversal.$\Rightarrow\angle\text{AEF}=\angle\text{EFG}$ (alternate angles)
Given, $\angle\text{AEF}=75^\circ$$\Rightarrow\angle\text{EFG}=\text{y}=75^\circ$
Now, $\angle\text{EFC}+\angle\text{EFG}=180^\circ$ (linear pair)$\Rightarrow\text{x}+\text{y}=180^\circ$
$\Rightarrow\text{x}+75^\circ=180^\circ$
$\Rightarrow\text{x}=105^\circ$
$\angle\text{EGD}=\angle\text{EFG}+\angle\text{FEG}$ (Exterior angle property)
$\Rightarrow125^\circ=\text{y}+\text{z}$
$\Rightarrow125^\circ=75^\circ+\text{z}$
$\Rightarrow\text{z}=50^\circ$
Thus, x = 105°, y = 75° and z = 50°.

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Question 104 Marks

In the figure given below, AB || CD. Find the value of x in each case.

Answer

Through O draw OF || CD.

Now since OF || CD and OD is transversal.$\angle\text{CDO}+\angle\text{FOD}=180^\circ$
[sum of consecutive interior angles is $180^\circ$]
$\Rightarrow25^\circ+\angle\text{FOD}=180^\circ$
$\Rightarrow\angle\text{FOD}=180^\circ-25^\circ=155^\circ$
As OF || CD and AB || CD [Given] Thus, OF || AB and OB is a transversal.
So, $\angle\text{ABO}+\angle\text{FOB}=180^\circ$ [sum of consecutive interior angles is $180^\circ$]
$\Rightarrow55^\circ+\angle\text{FOB}=180^\circ $
$\Rightarrow\angle\text{FOB}=180^\circ-55^\circ=125^\circ$
Now, $\text{x}^\circ=\angle\text{FOB}+\angle\text{FOD}=125^\circ+155^\circ=280^\circ.$
Hence, x = 280.

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Question 114 Marks

In the given figure, AB || CD Find the value of x, y and z.

Answer

Since AB || CD and BC is a transversal. So, $\angle\text{ABC}=\angle\text{BCD}$$\Rightarrow\text{x}=35$
Also, AB || CD and AD is a transversal. So, $\angle\text{BAD}=\angle\text{ADC}$$\Rightarrow\text{x}=75$
In $\triangle\text{ABO},$ we have,$\angle\text{ABO}+\angle\text{BAO}+\angle\text{BOA}=180^\circ$
$\Rightarrow\text{x}^\circ+75^\circ+\text{y}^\circ=180^\circ$
$\Rightarrow35+75+\text{y}=180$
$\Rightarrow\text{y}=180-110=70$
$\therefore\text{x}=35,\text{y}=70$ and $\text{z}=75.$

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Question 124 Marks

In the given figure, AB || CD. Prove that $\angle\text{BAE}-\angle\text{ECD}=\angle\text{AEC}.$

Answer

Draw $\text{EF}||\text{AB}||\text{CD}$ through E. Now, $\text{EF}||\text{AB}$ and AE is the transversal. Then, $\angle\text{BAE}+\angle\text{AEF}=180^\circ$ [Angles on the same side of a transversal line are supplementary] Again, $\text{EF}||\text{CD}$ and CE is the transversal.Then,
$\angle\text{DCE}+\angle\text{CEF}=180^\circ$ [Angles on the same side of a transversal line are supplementary]$\Rightarrow\angle\text{DCE}+(\angle\text{AEC}+\angle\text{AEF})=180^\circ$
$\Rightarrow\angle\text{DCE}+\angle\text{AEC}+180^\circ-\angle\text{BAE}=180^\circ$
$\Rightarrow\angle\text{BAE}-\angle\text{DCE}=\angle\text{AEC}$

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Question 134 Marks

In the given figure, BA || ED and BC || EF. Show that $\angle\text{ABC}+\angle\text{DEF}=180^\circ.$

Answer

Construction: Produce ED to meet BC at Z.

Now, AB || EZ and BC is the transversal.$\Rightarrow\angle\text{ABZ}+\angle\text{EZB}=180^\circ$ (interior angles)
$\Rightarrow\angle\text{ABC}+\angle\text{EZB}=180^\circ\ .....(\text{i)}$
Also, EF || BC and EZ is the transversal.$\Rightarrow\angle\text{BZE}=\angle\text{ZEF}$ (alternate angles)
$\Rightarrow\text{BZE}=\angle\text{DEF}\ .....(\text{ii)}$
From (i) and (ii), we have$\angle\text{ABC}+\angle\text{DEF}=180^\circ$

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Question 144 Marks

Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other.

Answer

Let the two parallel lines be m and n. Let p ⊥ m.$\Rightarrow\angle1=90^\circ$
Let q ⊥ n.$\Rightarrow\angle2=90^\circ$
Now, m || n and p is a transversal.$\Rightarrow\angle1=\angle3$ (corresponding angles)
$\Rightarrow\angle3=90^\circ$
$\Rightarrow\angle3=\angle2$ (each 90°)
But, these are corresponding angles, when transversal n cuts lines p and q.$\therefore$ p || q.
Hence, two lines which are perpendicular to two parallel lines, are parallel to each other.

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Question 154 Marks

In the given figure, l || m and a transversal t cuts them, If $\angle1:\angle2=2:3,$ find the measure of each of the remaining marked angles.

Answer

Given, $\angle1:\angle2=2:3$ Now, $\angle1+\angle2=180^\circ$ (linear pair)$\Rightarrow2\text{x}+3\text{x}=180^\circ$
$\Rightarrow5\text{x}=180^\circ$
$\text{x}=36^\circ$
$\Rightarrow\angle1=2\text{x}=72^\circ$ and $\angle2=3\text{x}=108^\circ$
$\angle1=\angle3$ (vertically opposite angles)
$\Rightarrow\angle3=72^\circ$
Also, $\angle2=\angle4$ (vertically opposite angles)$\Rightarrow\angle4=108^\circ$
Line l || line m and line t is a transversal.$\Rightarrow\angle5=\angle1=72^\circ$ (corresponding angles)
$\angle6=\angle2=108^\circ$ (corresponding angles)
$\angle4=\angle3=72^\circ$ (corresponding angles)
$\angle8=\angle4=108^\circ$ (corresponding angles)

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Question 164 Marks

In the figure given below, AB || CD. Find the value of x in each case.

Answer

Through E, draw EF || CD.

Now since EF || CD and EC is transversal.$\angle\text{FEC}+\angle\text{ECD}=180^\circ$
[sum of consecutive interior angles is $180^\circ$]
$\Rightarrow\angle\text{FEC}+124^\circ=180^\circ$
$\Rightarrow\angle\text{FEC}=180^\circ-124^\circ=56^\circ$
Since EF || CD and AB || CD So, EF || AB and AE is a trasveral.
So, $\angle\text{BAE}+\angle\text{FEA}=180^\circ$ [sum of consecutive interior angles is $180^\circ$]
$\therefore116^\circ+\angle\text{FEA}=180^\circ$
$\Rightarrow\angle\text{FEA}=180^\circ-116^\circ=64^\circ$
Thus, $\text{x}^\circ=\angle\text{FEA}+\angle\text{FEC}$$=64^\circ+56^\circ=120^\circ.$

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Question 174 Marks

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

Answer

Since AB || CD and BC is a transversal. So, $\angle\text{ABC}=\angle\text{BCD}$ [atternate interior angles]$\Rightarrow70^\circ=\text{x}+\angle\text{ECD}\ ....(\text{i)}$
Now, CD || EF and CE is transversal. So, $\angle\text{ECD}+\angle\text{CEF}=180^\circ$ [sum of consecutive interior angles is 180º]$\therefore\angle\text{}\text{ECD}+130^\circ=180^\circ$
$\Rightarrow\angle\text{ECD}=180^\circ-130^\circ=50^\circ$
Putting $\angle\text{ECD}=50^\circ $ in (i) we get,$70^\circ=\text{x}^\circ+50^\circ$
$\Rightarrow\text{x}=70-50=20$

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Question 184 Marks

In the given figure, AB || CD. Prove that P + q - r = 180.

Answer

Through F, draw KH || AB || CD

Now, KF || CD and FG is a transversal.$\Rightarrow\angle\text{KFG}=\angle\text{FGD}=\text{r}^\circ (\text{i})$
[alternate angles] Again AE || KF, and EF is a transversal. So, $\angle\text{AEF}+\angle\text{KFE}=180^\circ$$\Rightarrow\angle\text{KFE}=180^\circ-\text{p}^\circ(\text{ii)}$
Adding (i) and (ii) we get,$\angle\text{KFG}+\angle\text{KFE}=180-\text{p}+\text{r}$
$\Rightarrow\angle\text{EFG}=180-\text{p}+\text{r}$
$\Rightarrow\text{q}=180-\text{p}+\text{r}$
$\text{i}.\text{e}.,\ \text{p}+\text{q}-\text{r}=180$

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