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MATHS 3 Marks Question

Understanding Quadrilaterals · Gujarat Board (GSEB) STD 8 (GSEB - English Medium). 21 questions.

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Question 13 Marks
After showing $\text{m} \angle \text{R = m} \angle \text{N}=70^{\circ},$ can you find $\text{m} \angle \text{I}$ and $\text{m} \angle \text{G}$ by any other method?
Answer
Yes, we can find $\text{m} \angle \text{I}$ and $\text{m} \angle \text{G}$ by any other method. Given that in a parallelogram RING,
Image
$m \angle R=70^{\circ}$
Here, $R G \| I N$ and $R I$ is a transversal.
We know that if two parallel lines are cut by a transversal, then each pair of interior angles on the same side of the transversal are supplementary.
So, $m \angle R+m \angle I=180^{\circ} \Rightarrow 70^{\circ}+m \angle I=180^{\circ}$
$\Rightarrow \quad m \angle I=180^{\circ}-70^{\circ}=110^{\circ}$
Similarly, $R I \| G N$ and $R G$ is a transversal.
$\therefore \quad m \angle R+m \angle G=180^{\circ} \quad$ [interior angles]
$\Rightarrow \quad 70^{\circ}+m \angle G=180^{\circ}$
$\Rightarrow \quad m \angle G=180^{\circ}-70^{\circ}$
$\Rightarrow \quad m \angle G=110^{\circ}$
Hence, $\quad m \angle I=m \angle G=110^{\circ}$
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Question 23 Marks
Find $x$, in the following figures.
Image
Answer
(i) We know that the sum of the exterior angles of any polygon is $360^{\circ}$.
$\begin{array}{lr}\therefore & x+125^{\circ}+125^{\circ}=360^{\circ} \\ \Rightarrow & x+250^{\circ}=360^{\circ} \\ \Rightarrow & x=360^{\circ}-250^{\circ}=110^{\circ}\end{array}$
Hence, the value of $x$ is $110^{\circ}$.
(ii) Let given, polygon be $A B C D E$. It is clear that $D M$ is a straight line.
Image
$90^{\circ}+\angle 1=180^{\circ}$ [by linear pair angle]
$\Rightarrow \quad \angle 1=180^{\circ}-90^{\circ}=90^{\circ}$
Now, the sum of the exterior angles of any polygon is $360^{\circ}$.
$x+90^{\circ}+60^{\circ}+90^{\circ}+70^{\circ}=360^{\circ}$
$\Rightarrow \quad x+310^{\circ}=360^{\circ}$
$\Rightarrow \quad x=360^{\circ}-310^{\circ}=50^{\circ}$
Hence, the value of $x$ is $50^{\circ}$.
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Question 33 Marks
.Find the value of $\angle P$ and $\angle S$ if $\overline{S P} \| \overline{R Q}$ In the following figure. If you find $m \angle R$, Is there more than one method to find $m \angle P$?
Image
Answer
Given, $P Q R S$ is a trapezium, in which $\angle Q=130^{\circ}, \angle R=90^{\circ}$ and $\overline{S P} \| \overline{R Q}$.
Since, $\overline{S P} \| \overline{R Q}$ and $P Q$ is a transversal, so
$\angle P+\angle Q=180^{\circ}\quad$ $[\because$ interior angles on the same side]
$\Rightarrow \quad \angle P+130^{\circ}=180^{\circ}$
$\Rightarrow \quad \angle P=180^{\circ}-130^{\circ}=50^{\circ}$
Similarly, $\quad \angle S+\angle R=180^{\circ} \quad[\because S R$ is a transversal$]$
$\Rightarrow \quad \angle S+90^{\circ}=180^{\circ}$
$\Rightarrow \quad \angle S=180^{\circ}-90^{\circ}=90^{\circ}$
Hence, $\angle P=50^{\circ}$ and $\angle S=90^{\circ}$.
We may find $\angle P$ by one more method which is given below. We know that the sum of all the angles of a quadrilateral is $360^{\circ}$.
$\therefore \quad \angle P+\angle Q+\angle R+\angle S=360^{\circ}$
$\Rightarrow \quad \angle P+130^{\circ}+90^{\circ}+90^{\circ}=360^{\circ}$
$\Rightarrow \quad \angle P+310^{\circ}=360^{\circ}$
$\Rightarrow \quad \angle P=360^{\circ}-310^{\circ}=50^{\circ}$
Hence, the measure of $\angle P$ is $50^{\circ}$.
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Question 43 Marks
The adjacent figure HOPE is a parallelogram. Find the angle $x, y$ and $z$. State the properties you use to find them.
Image
Answer
Given, $H O P E$ is a parallelogram, in which $\angle E H P=40^{\circ}$ and exterior angle $\angle O=70^{\circ}$.
So, interior $\angle P O H=180^{\circ}-70^{\circ}=110^{\circ}$ [by linear pair]
In a parallelogram, opposite angles are equal.
$\therefore \quad x=\angle P O H=110^{\circ}$
Now, $\quad H E \| P O$ [opposite sides of a parallelogram] and $H P$ is a transversal.
So, $\quad \angle E H P=\angle H P O \quad$ [alternate interior angles]
$\therefore \quad \angle H P O=y=40^{\circ}$
In $\triangle H P O$, we have
$\angle H P O+\angle O H P+\angle P O H=180^{\circ}\quad$ $\left[\because\right.$ sum of three angles of a triangle is $\left.180^{\circ}\right]$
$\Rightarrow \quad 40^{\circ}+z+110^{\circ}=180^{\circ}$
$\Rightarrow \quad z+150^{\circ}=180^{\circ}$
$\Rightarrow \quad z=180^{\circ}-150^{\circ}=30^{\circ}$
Hence, the values of $x, y$ and $z$ are $110^{\circ}, 40^{\circ}$ and $30^{\circ}$, respectively and we use the following properties :
(i) The angles in a linear pair are supplementary.
(ii) Opposite sides of a parallelogram are parallel.
(iii) If two parallel lines are cut by a transversal, then each pair of alternate interior angles are equal.
(iv) Sum of three angles of a triangle is $180^{\circ}$.
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Question 53 Marks
The measures of two adjacent angles of a parallelogram are in the ratio $3: 2$. Find each angle of the parallelogram.
Answer
Let two adjacent angles $A$ and $B$ of a parallelogram $A B C D$ be $3 x$ and $2 x$, respectively.
Image
We know that the sum of two adjacent angles of a parallelogram is $180^{\circ}$.
$\therefore \quad \angle A+\angle B=180^{\circ}$
$\Rightarrow \quad 3 x+2 x=180^{\circ}$
$\Rightarrow \quad 5 x=180^{\circ}$
$\Rightarrow \quad x=\frac{180^{\circ}}{5}=36^{\circ}$
Therefore, $\quad \angle A=3 x=3 \times 36^{\circ}=108^{\circ}$
and $\quad \angle B=2 x=2 \times 36^{\circ}=72^{\circ} $
Also, we know that opposite angles of a parallelogram are equal.
$\therefore \quad \angle C=\angle A=108^{\circ}$
and $\angle D=\angle B=72^{\circ}$
Hence, the angles of a parallelogram are $\angle A=108^{\circ}$, $\angle B=72^{\circ}, \angle C=108^{\circ}$ and $\angle D=72^{\circ}$, respectively.
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Question 63 Marks
Can a quadrilateral $A B C D$ be a parallelogram if
(i) $\angle D+\angle B=180^{\circ}?$
(ii) $A B=D C=8 cm, A D=4 cm$ and $B C=4.4 cm?$
(iii) $\angle A=70^{\circ}$ and $\angle C=65^{\circ}?$
Answer
(i) In a quadrilateral $A B C D$,
$\angle D+\angle B=180^{\circ}=$ Sum of two adjacent angles
So, the quadrilateral may be parallelogram but not always.
(ii) In a quadrilateral $A B C D$,
$A B=D C=8 cm, A D=4 cm$ and $B C=4.4 cm$
Here, opposite sides $A D$ and $B C$ are not equal but in a parallelogram opposite sides are of equal length.
So, quadrilateral $A B C D$ cannot be a parallelogram.
(iii) In a quadrilateral $A B C D$,
$\angle A=70^{\circ}$ and $\angle C=65^{\circ}$
Here, $\angle A \neq \angle C$
But in a parallelogram, opposite angles are equal. So, quadrilateral $A B C D$ cannot be a parallelogram.
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Question 73 Marks
What is a regular polygon? State the name of a regular polygon of
(i) 3 sides $\quad$ (ii) 4 sides$\quad$(iii) 6 sides
Answer
A polygon is called a regular polygon if it is both equiangular and equilateral i.e. its all interior angles are equal and its all sides are equal.
(i) The name of a regular polygon of 3 sides is an equilateral triangle.
(ii) The name of a regular polygon of 4 sides is a square.
(iii) The name of a regular polygon of 6 sides is a regular hexagon.
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Question 83 Marks
In a parallelogram WISH, find $\angle S W H, \angle O S H$ and $\angle S H O$.
Image
Answer
We have, a parallelogram WISH.
Here, $\quad W H \| I S$ and $I W \| H S$
$\therefore \quad \angle W I H=\angle I H S=35^{\circ} \quad$ [alternate angles]
Similarly, $\angle H I S=\angle I H W=25^{\circ} \quad$ [alternate angles]
Now, using angle sum property of a triangle in $\triangle W O H$, we get
$\angle W O H+\angle O W H+\angle O H W=180^{\circ}$
$\Rightarrow \quad 110^{\circ}+\angle O W H+25^{\circ}=180^{\circ}\quad$ $[\because \angle O H W=\angle I H W] $
$\Rightarrow \angle O W H=180^{\circ}-\left(110^{\circ}+25^{\circ}\right)$
$=180^{\circ}-135^{\circ}=45^{\circ}=\angle S W H$
In $\triangle O H S$, we have
$\angle H O S=180^{\circ}-110^{\circ}=70^{\circ}$
$\Rightarrow \angle O H S=35^{\circ}\quad$ [proved]
$\therefore \quad 70^{\circ}+35^{\circ}+\angle O S H=180^{\circ}$
$\Rightarrow \quad \angle O S H=180^{\circ}-105^{\circ}=75^{\circ}$
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Question 93 Marks
In the given figure, ABCD and BDCE are parallelograms with common base $D C$. If $B C \perp B D$, then find $\angle B E C$.
Image
Answer
We have, parallelograms $A B C D$ and $B D C E$ and $B C \perp B D$
$\therefore \quad \angle A B C=180^{\circ}-30^{\circ}=150^{\circ}\quad$
[since, $\angle D A B$ and $\angle A B C$ are adjacent angles of parallelogram $A B C D$ ]
$\therefore \quad \angle C B E=180^{\circ}-\angle A B C=180^{\circ}-150^{\circ}=30^{\circ}$
Also, it is given that
$\angle D B C=90^{\circ}$
$\therefore \quad \angle D B E=90^{\circ}+30^{\circ}=120^{\circ}$
But $\quad \angle D B E=\angle D C E=120^{\circ}$
$\therefore \quad \angle B E C=180^{\circ}-120^{\circ}=60^{\circ}\quad$
[since, $\angle B E C$ and $\angle D C E$ are adjacent angles of parallelogram DCEB]
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Question 103 Marks
A playground in a town is in the form of a kite. The perimeter is 106 m . If one of its sides is 23 m, what are the lengths of other three sides?
Answer
In the kite $A B C D$, the sides $A B$ and $A D$ are equal.
Also, the sides $B C$ and $C D$ are equal.
Image
Let $A B=23 m$, then
$A B+B C+C D+D A=106 m$
$\Rightarrow 23+B C+C D+23=106\quad$ $[\because A B=A D]$
$\Rightarrow \quad B C+C D=106-46=60$
$\Rightarrow \quad B C+B C=60 m[\because C D=B C]$
$\Rightarrow \quad 2 B C=60 m$
$\Rightarrow \quad B C=60 \times \frac{1}{2}=30 m=C D$
Thus, $A B=23 m, A D=23 m, B C=30 m$ and $C D=30 m$
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Question 113 Marks
In the following figure, $F D\|B C\| A E$ and $A C \| E D$. Find the value of $x$.
Image
Answer
Produced $D F$ such that it intersects $A B$ at $G$.
Image
In $\triangle A B C, \angle C=180^{\circ}-\left(52^{\circ}+64^{\circ}\right)=180^{\circ}-116^{\circ}=64^{\circ}$
Now, we see that, $D G \| B C$ and $D G \| A E$
$\therefore \quad \angle A C B=\angle A F G\quad$ $[\because F G \| B C$ and $F C$ is a transversal, so corresponding angles]
$\Rightarrow \quad 64^{\circ}=\angle A F G$
Also, GFD is a straight line.
$\therefore \quad \angle G F A+\angle A F D=180^{\circ}$
$\Rightarrow \quad 64^{\circ}+\angle A F D=180^{\circ}$
$\Rightarrow \quad \angle A F D=180^{\circ}-64^{\circ}=116^{\circ}$
Also, $\quad F D \| A E$ and $A F \| E D$
So, $A E D F$ is a parallelogram.
$\therefore \quad \angle A F D=\angle A E D\quad$ [$\because$ opposite angles in a parallelogram are equal]
or $\quad\angle A E D=116^{\circ}=x$
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Question 123 Marks
Quadrilateral EFGH is a rectangle, in which J is the point of intersection of the diagonals. Find the value of $x$ if $J F=(8 x+4)$ and $E G=(24 x-8).$
Image
Answer
Since, EFGH is a rectangle.
$\therefore \quad H F=2 J F=2 \times(8 x+4)=16 x+8$
and $\quad E G=24 x-8\quad$ [given]
Thus, $\quad H F=E G\quad$ [$\because$ diagonals are equal in a rectangle]
$\therefore \quad 16 x+8=24 x-8$
$\Rightarrow 24 x-16 x=8+8$
$\Rightarrow \quad 8 x=16$
$\Rightarrow \quad x=16 \times \frac{1}{8}=2$
Thus, the value of $x$ is 2 .
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Question 133 Marks
ABCD is a rectangle with $\angle B A C=68^{\circ}$. Determine the value of $\angle D B C$.
Answer
We have, $A B C D$ is the rectangle and let $O$ be the Intereection ooint of $\triangle C$ and $B D$, where $\angle B A C=68^{\circ}$
Image
In a rectangle, diagonals are equal.
$\therefore \quad A C=B D \Rightarrow \frac{A C}{2}=\frac{B D}{2} \Rightarrow O C=B O\quad$ [ $\because$ in a rectangle, diagonals bisect each other]
$\Rightarrow \quad \angle C B O=\angle O C B=x\quad$ [$\because$ angles opposite to the equal sides in a triangle are equal]
Also, we have $A B \| C D$ and $A C$ is transversal.
$\therefore \quad \angle C A B=\angle A C D=68^{\circ} \quad$ [alternate angles]
and $\angle B C D=90^{\circ} \Rightarrow \angle D C O+\angle O C B=90^{\circ}$
$\Rightarrow \quad \angle D C A+\angle O C B=90^{\circ}$
$\Rightarrow \quad 68^{\circ}+x=90^{\circ} \quad\left[\because \angle D C A=68^{\circ}\right]$
$\Rightarrow \quad x=90^{\circ}-68^{\circ}=22^{\circ} \Rightarrow \angle O C B=22^{\circ}$
$\Rightarrow \quad \angle O B C=22^{\circ} \quad[\because \angle O C B=\angle O B C]$
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Question 143 Marks
If RENT is a rectangle. Its diagonals are meet at $O$. Find the value of $x$ if $O R=(5 x+4)$ and $O T=(3 x+12)$.
Image
Answer
Since, diagonals of rectangle bisect each other and
$O R=\frac{1}{2} R N \text { and } O T=\frac{1}{2} T E .$
Also, diagonals of a rectangle are equal.
$\therefore \quad R N=T E$
$\Rightarrow \quad \frac{1}{2} R N=\frac{1}{2} T E$
$\Rightarrow \quad O R=O T$
$\Rightarrow \quad 5 x+4=3 x+12 \Rightarrow 5 x-3 x=12-4=8$
$\Rightarrow \quad 2 x=8 \Rightarrow x=4$
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Question 153 Marks
In the following rectangle LEAP, find the value of $\angle E A L, \angle L A P$ and $\angle L O P$.
Image
Answer
We have, $L E A P$ is a rectangle and $\angle L O E=60^{\circ}$
Since, $\angle E O P=$ Straight angle $=180^{\circ}$
$\therefore \quad \angle E O L+\angle L O P=180^{\circ}$
$\Rightarrow \quad 60^{\circ}+\angle L O P=180^{\circ}$
$\Rightarrow \quad \angle L O P=180^{\circ}-60^{\circ}=120^{\circ}$
In a rectangle, we can say that
$L O=O P=O A=O E$
$\therefore \quad \angle O P A=\angle O A P=x($ say $)$
Now, in $\triangle O P A$, we have
$\angle A O P=60^{\circ}=\angle L O E\quad$ [ $\because$ vertically opposite angles]
$\therefore \quad \angle A O P+x+x=180^{\circ} \Rightarrow 60^{\circ}+2 x=180^{\circ}$
$\Rightarrow \quad 2 x=180^{\circ}-60^{\circ}=120^{\circ}$
$\Rightarrow \quad x=60^{\circ} \Rightarrow \angle O A P=60^{\circ}=\angle L A P$
Also, $\quad \angle E A P=90^{\circ}$
$\begin{aligned} \Rightarrow & & \angle E A L+\angle L A P & =90^{\circ} \\ \Rightarrow & & \angle E A L+\angle O A P & =90^{\circ} \\ \Rightarrow & & \angle E A L & =90^{\circ}-\angle O A P\end{aligned}$
$=90^{\circ}-60^{\circ}=30^{\circ}$
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Question 163 Marks
In a parallelogram PQRS, the bisectors of $\angle P$ and $\angle Q$ meet at $O$. Find $\angle P O Q$.
Answer
Since, $O P$ and $O Q$ are the bisectors of $\angle P$ and $\angle Q$ respectively (see the figure given below).
Image
So, $\quad \angle O P Q=\frac{1}{2} \angle P$
and $\quad\angle O Q P=\frac{1}{2} \angle Q$
In $\triangle P O Q$
$\angle O P Q+\angle P Q O+\angle P O Q=180^{\circ}\quad$ [angle sum property of a triangle]
$\Rightarrow \quad \frac{1}{2} \angle P+\angle P O Q+\frac{1}{2} \angle Q=180^{\circ}$
$\Rightarrow \quad \angle P O Q=180^{\circ}-\frac{1}{2}(\angle P+\angle Q)$
$=180^{\circ}-\frac{1}{2} \times 180^{\circ}=90^{\circ}$
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Question 173 Marks
The adjacent angles of a parallelogram are $(3 x-4)^{\circ}$ and $(2 x-1)^{\circ}$. Find all angles of the parallelogram.
Answer
We know that in a parallelogram, sum of any two adjacent angles is $180^{\circ}$.
Let $\quad\angle A=(3 x-4)^{\circ}$
and $\quad\angle B=(2 x-1)^{\circ}$
$\therefore(3 x-4)^{\circ}+(2 x-1)^{\circ}=180^{\circ}$
Image
$\Rightarrow \quad 5 x^{\circ}-5^{\circ}=180^{\circ}$
$\Rightarrow \quad 5 x^{\circ}=180^{\circ}+5^{\circ}$
$\Rightarrow \quad x^{\circ}=185^{\circ} \times \frac{1}{5^{\circ}}=37^{\circ}$
$\therefore \quad \angle A=3 \times 37^{\circ}-4=107^{\circ}$ and $\quad\angle B=2 \times 37^{\circ}-1^{\circ}=73^{\circ}$
Since, in a parallelogram, opposite angles are equal.
$\therefore \angle A=\angle C=107^{\circ}$ and $\angle B=\angle D=73^{\circ}$
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Question 183 Marks
PQRS is a trapezium such that $P Q \| P S, \angle P: \angle S=2: 1$, $\angle Q: \angle R=7: 5$. Find the angles of the trapezium.
Answer
We have, a trapezium $P Q R S$ such that $P Q \| R S$.
Let $\angle P$ and $\angle S$ be $2 x, x$ respectively and $\angle Q, \angle R$ be $7 y$ and $5 y$, respectively.
Image
But $\angle P+\angle S=180^{\circ}$ and $\angle Q+\angle R=180^{\circ}$
$\Rightarrow \quad 2 x+x=180^{\circ}$ and $7 y+5 y=180^{\circ}$
$\Rightarrow \quad 3 x=180^{\circ}$ and $12 y=180^{\circ}$
$\Rightarrow \quad x=180^{\circ} \times \frac{1}{3}$ and $y=180^{\circ} \times \frac{1}{12}$
$\Rightarrow \quad x=60^{\circ}$ and $y=15^{\circ}$
$\therefore \quad \angle P=2 \times 60^{\circ}=120^{\circ}$ and $\angle S=60^{\circ}$,
$\angle Q=7 \times 15^{\circ}=105^{\circ}$
and $\quad \angle R=5 \times 15^{\circ}=75^{\circ}$
Thus, angles of the trapezium $P Q R S$ are $120^{\circ}, 105^{\circ}, 75^{\circ}$ and $60^{\circ}$.
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Question 193 Marks
In a trapezium FARE, EP and RP are bisectors of $\angle E$ and $\angle R$, respectively. Find $\angle F A R$ and $\angle E F A$.
Image
Answer
We have, FARE is a trapezium, where $E R \| F A$ and $E P$ and $R P$ are bisectors of $\angle E$ and $\angle R$, respectively.
Thus, $\quad \angle P E F=\angle P E R$ and $\angle P R E=\angle P R A$
$\therefore \quad \angle P E F=25^{\circ}$ and $\angle P R A=30^{\circ}$
So, $\quad \angle E=25^{\circ}+25^{\circ}=50^{\circ}$
and $\quad \angle R=30^{\circ}+30^{\circ}=60^{\circ}$
Now, we know that in a trapezium,
$\angle E+\angle F=180^{\circ}$ and $\angle R+\angle A=180^{\circ}$
$\therefore \quad \angle F=180^{\circ}-50^{\circ}=130^{\circ}$
and $\quad \angle A=180^{\circ}-60^{\circ}=120^{\circ}$
or $\angle E F A=130^{\circ}$ and $\angle F A R=120^{\circ}$
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Question 203 Marks
In a trapezium $A B C D, M C$ and $M D$ are bisectors of $\angle C$ and $\angle D$, respectively. Find $\angle A B C$ and $\angle B A D$.
Image
Answer
Given, $M D$ and $M C$ are bisectors of $\angle B C D$ and $\angle A D C$, respectively.
So, $\angle M C B=40^{\circ}, \angle A D M=30^{\circ}$ and $A B \| C D$
In a trapezium,
$\angle B+\angle C=180^{\circ}$
$\Rightarrow \angle B+\angle D C M+\angle M C B=180^{\circ}$
$\Rightarrow \quad \angle B+40^{\circ}+40^{\circ}=180^{\circ}$
$\Rightarrow \quad \angle B=180^{\circ}-80^{\circ}=100^{\circ}$
$\therefore \quad \angle B=100^{\circ}$
Similarly, $\quad \angle A+\angle D=180^{\circ}$
$\Rightarrow \angle A+\angle M D A+\angle M D C=180^{\circ}$
$\Rightarrow \quad \angle A+30^{\circ}+30^{\circ}=180^{\circ}$
$\therefore \quad \angle A=180^{\circ}-60^{\circ}=120^{\circ}$
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Question 213 Marks
The sides of a pentagon are produced in order and the exterior angles so obtained are
$x^{\circ}$,$3x^{\circ}$, (4x + 10$^{\circ}$), (2x + 5$^{\circ}$) and $5x^{\circ}$, respectively.
Find the value of x and each exterior angle of a pentagon.
Answer
We know that the sum of exterior angles of a regular polygon is $360^{\circ}$.
Image
So, $x+3 x+(4 x+10)+(2 x+5)+5 x=360^{\circ}$
$\Rightarrow \quad 15 x+15^{\circ}=360^{\circ}$
$\Rightarrow \quad 15 x=360^{\circ}-15^{\circ}$
$\Rightarrow \quad 15 x=345^{\circ}$
$\Rightarrow \quad x=\frac{345^{\circ}}{15}=23^{\circ}$
$\therefore \quad 3 x=3 \times 23^{\circ}=69^{\circ}$,
$\begin{aligned} 4 x+10 & =4 \times 23^{\circ}+10=92+10=102^{\circ}, \\ 2 x+5 & =2 \times 23^{\circ}+5=46+5=51^{\circ} \\ 5 x & =5 \times 23^{\circ}=115^{\circ} \\ x & =23^{\circ}\end{aligned}$
So, exterior angles are $23^{\circ}, 69^{\circ}, 102^{\circ}, 51^{\circ}$ and $115^{\circ}$.
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