Exercise 3.3 (Revised) - Chapter 3 - Understanding Quadrilaterals - Ncert Solutions class 8 - Maths

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NCERT Solutions for Class 8 Maths Chapter 3 - Understanding Quadrilaterals

Ex 3.3 Question 1.

Given a parallelogram ABCD. Complete each statement along with the definition or property used.

(i) $\mathrm{AD}=$ $\qquad$
(ii) $\angle \mathrm{DCB}=$ $\qquad$
(iii) $\mathrm{OC}=$ $\qquad$
(iv) $m \angle \mathrm{DAB}+m \angle \mathrm{CDA}=$ $\qquad$
Answer.

(i) $\mathrm{AD}=\mathrm{BC}$
[Since opposite sides of a parallelogram are equal]
(ii) $\angle \mathrm{DCB}=\angle \mathrm{DAB}$
[Since opposite angles of a parallelogram are equal]
(iii) $\mathrm{OC}=\mathrm{OA}$
[Since diagonals of a parallelogram bisect each other]
(iv) $m \angle \mathrm{DAB}+m \angle \mathrm{CDA}=180^{\circ}$
[Adjacent angles in a parallelogram are supplementary]
Ex 3.3 Question 2.

Consider the following parallelograms. Find the values of the unknowns $x, y, z$.

Note: For getting correct answer, read $3^{\circ}=30^{\circ}$ in figure (iii)
Answer.

(i) $\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ}$

[Adjacent angles in a parallelogram are supplementary]

$
\begin{aligned}
& \Rightarrow 100^{\circ}+x=180^{\circ} \\
& \Rightarrow x=180^{\circ}-100^{\circ}=80^{\circ}
\end{aligned}
$

And $z=x=80^{\circ}$
[Since opposite angles of a parallelogram are equal]
Also $y=100^{\circ}$
[Since opposite angles of a parallelogram are equal]
(ii) $x+50^{\circ}=180^{\circ}$
[Adjacent angles in a $\| \mathrm{gm}$ are supplementary]

$
\begin{aligned}
& \Rightarrow x=180^{\circ}-50^{\circ}=130^{\circ} \\
& \Rightarrow z=x=130^{\circ}
\end{aligned}
$
[Corresponding angles]
$
\Rightarrow \mathrm{y}=\mathrm{x}=130 \text { degrees }
$
[Since opposite angles of a parallelogram are equal]
(iii) $x=90^{\circ}$
[Vertically opposite angles]

$
\Rightarrow y+x+30^{\circ}=180^{\circ}
$
[Angle sum property of a triangle]
$
\begin{aligned}
& \Rightarrow y+90^{\circ}+30^{\circ}=180^{\circ} \\
& \Rightarrow y+120^{\circ}=180^{\circ} \\
& \Rightarrow y=180^{\circ}-120^{\circ}=60^{\circ} \\
& \Rightarrow z=y=60^{\circ}
\end{aligned}
$
[Alternate angles]
(iv) $z=80^{\circ}$
[Corresponding angles]
$
\Rightarrow x+80^{\circ}=180^{\circ}
$
[Adjacent angles in a $\|$ gm are supplementary]

$
\Rightarrow x=180^{\circ}-80^{\circ}=100^{\circ}
$

And $y=80^{\circ}$
[Opposite angles are equal in a $\| \mathrm{gm}$ ]

(v) $y=112^{\circ}$
[Opposite angles are equal in a $\| \mathrm{gm}$ ]

$
\Rightarrow 40^{\circ}+y+x=180^{\circ}
$
[Angle sum property of a triangle]
$
\begin{aligned}
& \Rightarrow 40^{\circ}+112^{\circ}+x=180^{\circ} \Rightarrow 152^{\circ}+x=180^{\circ} \\
& \Rightarrow x=180^{\circ}-152^{\circ}=28^{\circ}
\end{aligned}
$

And $z=x=28^{\circ}$
[Alternate angles]
Ex 3.3 Question 3.

Can a quadrilateral $\mathrm{ABCD}$ be a parallelogram, if:
(i) $\angle \mathrm{D}+\angle \mathrm{B}=180^{\circ}$ ?
(ii) $\mathrm{AB}=\mathrm{DC}=8 \mathrm{~cm}, \mathrm{AD}=4 \mathrm{~cm}$ and $\mathrm{BC}=4.4 \mathrm{~cm}$ ?
(iii) $\angle \mathrm{A}=70^{\circ}$ and $\angle \mathrm{C}=65^{\circ}$ ?

Answer.

(i) $\angle \mathrm{D}+\angle \mathrm{B}=180^{\circ}$
It can be, but here, it needs to be a square or a rectangle.

(ii) No, in this case, because one pair of opposite sides are equal and another pair of opposite sides are unequal. So, it is not a parallelogram.

(iii) No. $\angle \mathrm{A} \neq \angle \mathrm{c}$.

Since opposite angles are equal in parallelogram and here opposite angles are not equal in quadrilateral $A B C D$. Therefore it is not a parallelogram.

Ex 3.3 Question 4.

Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measures.

Answer.

ABCD is a quadrilateral in which angles $\angle \mathrm{A}=\angle \mathrm{C}=110^{\circ}$.
Therefore, it could be a kite.

Ex 3.3 Question 5.

The measure of two adjacent angles of a parallelogram are in the ratio $3: 2$. Find the measure of each of the angles of the parallelogram.

Answer.

Let two adjacent angles be $3 x$ and $2 x$.

Since the adjacent angles in a parallelogram are supplementary.
$
\begin{aligned}
& \therefore 3 x+2 x=180^{\circ} \\
& \Rightarrow 5 x=180^{\circ}
\end{aligned}
$

$
\Rightarrow x=\frac{180^{\circ}}{5}=36^{\circ}
$
$\therefore$ One angle $=3 x=3 \times 36^{\circ}=108^{\circ}$
And Another angle $=2 x=2 \times 36^{\circ}=72^{\circ}$
Ex 3.3 Question 6.

Two adjacent angles of a parallelogram have equal measure. Find the measure of the angles of the parallelogram.

Answer.

Let each adjacent angle be $x$.
Since the adjacent angles in a parallelogram are supplementary.
$
\begin{aligned}
& \therefore x+x=180^{\circ} \\
& \Rightarrow 2 x=180^{\circ} \\
& \Rightarrow x=\frac{180^{\circ}}{2}=90^{\circ}
\end{aligned}
$

Hence, each adjacent angle is $90^{\circ}$.

Ex 3.3 Question 7.

The adjacent figure HOPW is a parallelogram. Find the angle measures $x=y$ and $z$. State the properties you use to find them.

$
\angle \mathrm{HOP}+70^{\circ}=180^{\circ}
$

Answer.

Here $\angle \mathrm{HOP}=180^{\circ}-70^{\circ}=110^{\circ}$
[Angles of linear pair]
And $\angle \mathrm{E}=\angle \mathrm{HOP}$

[Opposite angles of a $\| \mathrm{gm}$ are equal]
$
\begin{aligned}
& \Rightarrow x=110^{\circ} \\
& \angle \mathrm{PHE}=\angle \mathrm{HPO}
\end{aligned}
$
[Alternate angles]
$
\therefore y=40^{\circ}
$

Now $\angle \mathrm{EHO}=\angle \mathrm{O}=70^{\circ}$
[Corresponding angles]
$
\begin{aligned}
& \Rightarrow 40^{\circ}+z=70^{\circ} \\
& \Rightarrow z=70^{\circ}-40^{\circ}=30^{\circ}
\end{aligned}
$

Hence, $x=110^{\circ}, y=40^{\circ}$ and $z=30^{\circ}$
Ex 3.3 Question 8.

The following figures GUNS and RUNS are parallelograms. Find $X$ and $y_{\text {- (Lengths are }}$ in $\mathrm{cm}$ )

Answer.

(i) In parallelogram GUNS,
$
\mathrm{GS}=\mathrm{UN}
$
[Opposite sides of parallelogram are equal]
$
\Rightarrow 3 x=18
$

$
\Rightarrow x=\frac{18}{3}=6 \mathrm{~cm}
$

Also GU $=\mathrm{SN}$
[Opposite sides of parallelogram are equal]
$
\begin{aligned}
& \Rightarrow 3 y-1=26 \\
& \Rightarrow 3 y=26+1 \\
& \Rightarrow 3 y=27 \\
& \Rightarrow y=\frac{27}{3}=9 \mathrm{~cm}
\end{aligned}
$

Hence, $\boldsymbol{X}=6 \mathrm{~cm}$ and $\boldsymbol{X}=9 \mathrm{~cm}$.
(ii) In parallelogram RUNS,
$
y+7=20
$
[Diagonals of $\| \mathrm{gm}$ bisects each other]
$
\Rightarrow y=20-7=13 \mathrm{~cm}
$

$
\begin{aligned}
& \text { And } x+y=16 \\
& \Rightarrow x+13=16 \\
& \Rightarrow x=16-13 \\
& \Rightarrow x=3 \mathrm{~cm}
\end{aligned}
$

Hence, $x=3 \mathrm{~cm}$ and $y=13 \mathrm{~cm}$.
Ex 3.3 Question 9.

In the figure, both RISK and CLUE are parallelograms. Find the value of $x$.

Answer.

In parallelogram RISK,
$
\angle \mathrm{RIS}=\angle \mathrm{K}=120^{\circ}
$
[Opposite angles of a ||gm are equal]
$
\begin{aligned}
& \angle m+120^{\circ}=180^{\circ} \text { [Linear pair] } \\
& \Rightarrow \angle m=180^{\circ}-120^{\circ}=60^{\circ}
\end{aligned}
$

And $\angle \mathrm{ECI}=\angle \mathrm{L}=70^{\circ}$
[Corresponding angles]
$
\Rightarrow m+n+\angle \mathrm{ECI}=180^{\circ}
$
[Angle sum property of a triangle]
$
\begin{aligned}
& \Rightarrow 60^{\circ}+n+70^{\circ}=180^{\circ} \\
& \Rightarrow 130^{\circ}+n=180^{\circ} \\
& \Rightarrow n=180^{\circ}-130^{\circ}=50^{\circ}
\end{aligned}
$

Also $x=n=50^{\circ}$
[Vertically opposite angles]
Ex 3.3 Question 10.

Explain how this figure is a trapezium. Which of its two sides are parallel?

Answer.

Here, $\angle \mathrm{M}+\angle \mathrm{L}=100^{\circ}+80^{\circ}=180^{\circ}$
[Sum of interior opposite angles is $180^{\circ}$ ]
$\therefore \mathrm{NM}$ and $\mathrm{KL}$ are parallel.
Hence, KLMN is a trapezium.
Ex 3.3 Question 11.

Find $m \angle \mathrm{c}$ in figure, if $\overline{\mathrm{AB}} \| \overline{\mathrm{DC}}$

Answer.

Here, $\angle \mathrm{B}+\angle \mathrm{c}=180^{\circ}$
$
\begin{aligned}
& {[\because \overline{\mathrm{AB}} \| \overline{\mathrm{DC}}]} \\
& \therefore 120^{\circ}+m \angle \mathrm{C}=180^{\circ} \\
& \Rightarrow m \angle \mathrm{C}=180^{\circ}-120^{\circ}=60^{\circ}
\end{aligned}
$
Ex 3.3 Question 12.

Find the measure of $\angle \mathrm{P}$ and $\angle \mathrm{S}$ if $\overline{\mathrm{SP}} \| \overline{\mathrm{RQ}}$ in given figure.
(If you find $m \angle \mathbf{R}$ is there more than one method to find $m \angle \mathrm{P}$ )

Answer.

Here, $\angle \mathrm{P}+\angle \mathrm{Q}=180^{\circ}$
[Sum of co-interior angles is $180^{\circ}$ ]
$
\Rightarrow \angle \mathrm{P}+130^{\circ}=180^{\circ}
$

$
\begin{aligned}
& \Rightarrow \angle \mathrm{P}=180^{\circ}-130^{\circ} \\
& \Rightarrow \angle \mathrm{P}=50^{\circ} \\
& \because \angle \mathrm{R}=90^{\circ} \text { [Given] } \\
& \therefore \angle \mathrm{s}+90^{\circ}=180^{\circ} \\
& \Rightarrow \angle \mathrm{s}=180^{\circ}-90^{\circ} \\
& \Rightarrow \angle \mathrm{s}=90^{\circ}
\end{aligned}
$

Yes, one more method is there to find $\angle \mathrm{P}$.
$
\angle \mathrm{s}+\angle \mathrm{R}+\angle \mathrm{Q}+\angle \mathrm{p}=360^{\circ}
$
[Angle sum property of quadrilateral]
$
\begin{aligned}
& \Rightarrow 90^{\circ}+90^{\circ}+130^{\circ}+\angle \mathrm{P}=360^{\circ} \\
& \Rightarrow 310^{\circ}+\angle \mathrm{P}=360^{\circ} \\
& \Rightarrow \angle \mathrm{P}=360^{\circ}-310^{\circ} \\
& \Rightarrow \angle \mathrm{P}=50^{\circ}
\end{aligned}
$