Exercise 5.4 - Chapter 5 Geometry 8th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

$\mathbf{E x} 5.4$
I. Construct the following quadrilaterals with the given measurements and also find their area.
Question $1 .$
$\mathrm{ABCD}, \mathrm{AB}=5 \mathrm{~cm}, \mathrm{BC}=4.5 \mathrm{~cm}, \mathrm{CD}=3.8 \mathrm{~cm}, \mathrm{DA}=4.4 \mathrm{~cm}$ and $\mathrm{AC}=6.2 \mathrm{~cm}$.
Answer:
Given $\mathrm{ABCD}, \mathrm{AB}=5 \mathrm{~cm}, \mathrm{BC}=4.5 \mathrm{~cm}, \mathrm{CD}=3.8 \mathrm{~cm}, \mathrm{DA}=4.4 \mathrm{~cm}$ and $\mathrm{AC}=6.2$
$\mathrm{cm}$

Steps:
- Draw a line segment $\mathrm{AB}=5 \mathrm{~cm}$
- With $\mathrm{A}$ and $\mathrm{B}$ as centers drawn ares of radii $6.2 \mathrm{~cm}$ and $4.5 \mathrm{~cm}$ respectively and let them cut at $C$.
- Joined $A C$ and $B C$.
- With $\mathrm{A}$ and $\mathrm{C}$ as centrers drawn arcs of radii $4.4 \mathrm{~cm}$ and $3.8 \mathrm{~cm}$ respectively and let them at D.
- Joined AD and CD.
- $\mathrm{ABCD}$ is the required quadrilateral.
Calculation of Area:
Area of the quadrilateral $\mathrm{ABCD}=\frac{1}{2} \times \mathrm{d} \times\left(\mathrm{h}_{1} 1+\mathrm{h}_{2}\right)$ sq. units $=\frac{1}{2} \times 6.2 \times(2.6+3.6) \mathrm{cm}^{2}=3.1 \times 6.2=19.22 \mathrm{~cm}^{2}$

Question $2 .$

PLAY, PL $=7 \mathrm{~cm}, \mathrm{LA}=6 \mathrm{~cm}, \mathrm{AY}=6 \mathrm{~cm}, \mathrm{PA}=8 \mathrm{~cm}$ and $\mathrm{LY}=7 \mathrm{~cm}$
Answer:
Given PLAY, $\mathrm{PL}=7 \mathrm{~cm}, \mathrm{LA}=6 \mathrm{~cm}, \mathrm{AY}=6 \mathrm{~cm}, \mathrm{PA}=8 \mathrm{~cm}$ and $\mathrm{LY}=7 \mathrm{~cm}$

Steps:
- Drawn a line segment $\mathrm{PL}=7 \mathrm{~cm}$
- With $\mathrm{P}$ and $\mathrm{L}$ as centers, drawn arcs of radii $8 \mathrm{~cm}$ and $6 \mathrm{~cm}$ respectively, let them cut at $\mathrm{A}$.
- Joined PA and LA.
- With $L$ and $A$ as centers, drawn arcs of radii $7 \mathrm{~cm}$ and $6 \mathrm{~cm}$ respectively and let them cut at $Y$.
- Joined LY, PY and AY.
- PLAY is the required quadrilateral.
Calculation of Area:
Area of the quadrilateral PLAY $=\frac{1}{2} \times \mathrm{d} \times\left(\mathrm{h}_{1}+\mathrm{h}_{2}\right)$ sq. units $=\frac{1}{2} \times 8 \times(5.1+1.4)$ $\mathrm{cm}^{2}$
$\frac{1}{2} \times 8 \times 6.5 \mathrm{~cm}^{2}=26 \mathrm{~cm}^{2}$
Area of the quadrilateral $=26 \mathrm{~cm}^{2}$

Question $3 .$
$\mathrm{PQRS}, \mathrm{PQ}=\mathrm{QR}=3.5 \mathrm{~cm}, \mathrm{RS}=5.2 \mathrm{~cm}, \mathrm{SP}=5.3 \mathrm{~cm}$ and $\angle \mathrm{Q}=120^{\circ}$.
Answer:
Given $\mathrm{PQ}=\mathrm{QR}=3.5 \mathrm{~cm}, \mathrm{RS}=5.2 \mathrm{~cm}, \mathrm{SP}=5.3 \mathrm{~cm}$ and $\angle \mathrm{Q}=120^{\circ}$

Steps:
- Draw a line segment $P Q=3.5 \mathrm{~cm}$
- Made $\angle Q=120^{\circ}$. Drawn the ray $Q X$.
- With $Q$ as centre drawn an arc of radius $3.5 \mathrm{~cm}$. Let it cut the ray QX at $R$.
- With $\mathrm{R}$ and $\mathrm{P}$ as centres drawn arcs of radii $5.2 \mathrm{~cm}$ and $5.5 \mathrm{~cm}$ respectively and let them cut at $S$.
- Joined PS and RS.
- PQRS is the required quadrilateral.
Calculation of Area:
Area of the quadrilateral $P Q R S=\frac{1}{2} \times d \times\left(h_{1}+h_{2}\right)$ sq. units
$\begin{aligned}
&=\frac{1}{2} \times 6 \times(4.3+17) \mathrm{cm}^{2} \\
&=3 \times 6 \mathrm{~cm}^{2} \\
&=18 \mathrm{~cm}^{2}
\end{aligned}$
Area of the quadrilateral $\mathrm{PQRS}=18 \mathrm{~cm}^{2}$

Question $4 .$
MIND, MI $=3.6 \mathrm{~cm}, \mathrm{ND}=4 \mathrm{~cm}, \mathrm{MD}=4 \mathrm{~cm}, \angle \mathrm{M}=50^{\circ}$ and $\angle \mathrm{D}=100^{\circ}$.
Answer:
Given $\mathrm{MI}=3.6 \mathrm{~cm}, \mathrm{ND}=4 \mathrm{~cm}, \mathrm{MD}=4 \mathrm{~cm}, \angle \mathrm{M}=50^{\circ}$ and $\angle \mathrm{D}=100^{\circ}$

Steps:
- Drawn a line segment $\mathrm{MI}=3.6 \mathrm{~cm}$
- At M on MI made an angle $\angle \mathrm{IMX}=500$
- Drawn an are with center $M$ and radius $4 \mathrm{~cm}$ let it cut MX it D
- At D on DM made an angle $\angle \mathrm{MDY}=100^{\circ}$
- With I as center drawn an arc of radius $4 \mathrm{~cm}$, let it cut DY at $N$.
- Joined DN and IN.
- MIND is the required quadrilateral.
Calculation of Area:
Area of the quadrilateral MIND $=\frac{1}{2} \times \mathrm{d} \times\left(\mathrm{h}_{1}+\mathrm{h}_{2}\right)$ sq. units $=\frac{1}{2} \times 3.2 \times(2.7+33)$ $\mathrm{cm}^{2}$
$=\frac{1}{2} \times 3.2 \times 6 \mathrm{~cm}^{2}=9.6 \mathrm{~cm}^{2}$
Area of the quadrilateral $=9.6 \mathrm{~cm}^{2}$

Question $5 .$
AGRI, $\mathrm{AG}=4.5 \mathrm{~cm}, \mathrm{GR}=3.8 \mathrm{~cm}, \angle \mathrm{A}=60^{\circ}, \angle \mathrm{G}=110^{\circ}$ and $\angle \mathrm{R}=90^{\circ}$.
Answer:
$\mathrm{AG}=4.5 \mathrm{~cm}, \mathrm{GR}=3.8 \mathrm{~cm}, \angle \mathrm{A}=60^{\circ}, \angle \mathrm{G}=110^{\circ}$ and $\angle \mathrm{R}=90^{\circ}$.

Steps:
- Draw a line segment $\mathrm{AG}=4.5 \mathrm{~cm}$
- At $\mathrm{G}$ on $\mathrm{AG}$ made $\angle \mathrm{AGX}=110^{\circ}$
- With $G$ as centre drawn an arc of radius $3.8 \mathrm{~cm}$ let it cut $G X$ at $R$.
- At R on GR made $\angle G R Z=90^{\circ}$
- At A on $\mathrm{AG}$ made $\angle \mathrm{GAY}=90^{\circ}$
- AY and RZ meet at I.
- AGRI is the required quadrilateral.
Calculation of Area:
Area of the quadrilateral $\mathrm{AGRI}=\frac{1}{2} \times \mathrm{d} \times\left(\mathrm{h}_{1}+\mathrm{h}_{2}\right)$ sq. units $=\times 6.8 \times(2.9+2.4)$ $\mathrm{cm}^{2}$
$=\frac{1}{2} \times 6.8 \times 5.3 \times \mathrm{cm}^{2}$
Area of the quadrilateral $=18.02 \mathrm{~cm}^{2}$

II. Construct the following trapeziums with the given measures and also find their area.
Question $1 .$
AIMS with $\overline{\mathrm{AI}} \| \overline{\mathrm{SM}}, \mathrm{AI}=6 \mathrm{~cm}, \mathrm{IM}=5 \mathrm{~cm}, \mathrm{AM}=9 \mathrm{~cm}$ and $\mathrm{MS}=6.5 \mathrm{~cm}$.
Answer:
Given $\mathrm{AI}=6 \mathrm{~cm}, \mathrm{IM}=5 \mathrm{~cm}$
$\mathrm{AM}=9 \mathrm{~cm}$, and $\overline{\mathrm{AI}} \| \overline{\mathrm{SM}}$ $\mathrm{MS}=6.5 \mathrm{~cm}$
$M S=6.5 \mathrm{~cm}$

Construction:
Steps:
- Draw a line segment $\mathrm{AI}=6 \mathrm{~cm}$.
- With A and I as centres, draw arcs of radii $9 \mathrm{~cm}$ and $5 \mathrm{~cm}$ respectively and let them cut at $M$
- Join AM and IM.
- Draw MX parallel to $\mathrm{AI}$
- With M as centre, draw an arc of radius $6.5 \mathrm{~cm}$ cutting MX at S.
- Join AS AIMS is the required trapezium.
Calculation of Area
Area of the trapezium AIMS $=\frac{1}{2} \times h \times(a+b)$ sq.units
$\begin{aligned}
&=\frac{1}{2} \times 4.6 \times(6+6.5) \\
&=\frac{1}{2} \times 4.6 \times 12.5 \\
&=28.75 \mathrm{Sq} . \mathrm{cm}
\end{aligned}$

Question $2 .$
CUTE with $\overline{\mathrm{CD}} \| \overline{\mathrm{ET}}, \mathrm{CU}=7 \mathrm{~cm}, \angle \mathrm{UCE}=80^{\circ} \mathrm{CE}=6 \mathrm{~cm}$ and $\mathrm{TE}=5 \mathrm{~cm}$.
Answer:
Given: In the trapezium CUTE,
$\begin{aligned}
&\mathrm{CU}=7 \mathrm{~cm}, \angle \mathrm{UCE}=80^{\circ}, \\
&\mathrm{CE}=6 \mathrm{~cm}, \mathrm{TE}=5 \mathrm{~cm} \text { and } \overline{\mathrm{CD}} \| \overline{\mathrm{ET}}
\end{aligned}$

Construction:
Steps:
- Draw a line segment $\mathrm{CU}=7 \mathrm{~cm}$.
- Construct an angle $\angle \mathrm{UCE}=80^{\circ}$ at $\mathrm{C}$
- With C as centre, draw an arc of radius $6 \mathrm{~cm}$ cutting $C Y$ at $E$
- Draw EX parallel to CU
- With $\mathrm{E}$ as centre, draw an arc $7 \mathrm{~cm}$ of radius $5 \mathrm{~cm}$ cutting $E X$ at $T$
- Join UT. CUTE is the required trapezium.
Calculation of area:
Area of the trapezium CUTE $=\frac{1}{2} \times h \times(a+b)$ sq. units
$=\frac{1}{2} \times 5.9 \times(7+5)$ sq. units
$=35.4$ sq.cm
 

Question $3 .$
ARMY with $\overline{\mathrm{AR}} \| \overline{\mathrm{YM}}, \mathrm{AR}=7 \mathrm{~cm}, \mathrm{RM}=6.5 \mathrm{~cm} \angle \mathrm{RAY}=100^{\circ}$ and $\angle \mathrm{ARM}=60^{\circ}$
Answer:
Given: In the trapezium ARMY
$\mathrm{AR}=7 \mathrm{~cm}, \mathrm{RM}=6.5 \mathrm{~cm}$,
$\angle \mathrm{RAY}=100^{\circ}$ and $\mathrm{ARM} 60^{\circ}, \overline{\mathrm{AR}} \| \overline{\mathrm{YM}}$

Steps:
- Draw a line segment $\mathrm{AR}=7 \mathrm{~cm}$.
- Construct an angle $\angle \mathrm{RAX}=100^{\circ}$ at $\mathrm{A}$
- Construct an angle $\angle \mathrm{ARN}=60^{\circ}$ at $\mathrm{R}$
- With $R$ as centre, draw an arc of radius $6.5 \mathrm{~cm}$ cutting $R N$ at $M$
- Draw MY parallel to AR
- ARMY is the required trapezium.
Calcualtion of Area:
Area of the trapezium ARMY $=\frac{1}{2} \times \mathrm{h} \times(\mathrm{a}+\mathrm{b})$ sq. units
$\begin{aligned}
&=\frac{1}{2} \times 5.6 \times(7+4.8) \text { sq. units } \\
&=\frac{1}{2} \times 5.6 \times 1.18 \\
&=33.04 \text { sq.cm }
\end{aligned}$

Question $4 .$
CITY with $\overline{\mathrm{CI}} \| \overline{\mathrm{YT}}, \mathrm{CI}=7 \mathrm{~cm}, \mathrm{IT}=5.5 \mathrm{~cm}, \mathrm{TY}=4 \mathrm{~cm}$ and $\mathrm{YC}=6 \mathrm{~cm}$.
Answer:
Given: In the trapezium CITY,
$\begin{aligned}
&\mathrm{CI}=7 \mathrm{~cm}, \mathrm{IT}=5.5 \mathrm{~cm}, \mathrm{TY}=4 \mathrm{~cm} \\
&\mathrm{YC}=6 \mathrm{~cm}, \text { and } \overline{\mathrm{CI}} \| \overline{\mathrm{YT}}
\end{aligned}$

Construction:
Steps:
- Draw a line segment $\mathrm{CI}=7 \mathrm{~cm}$.
- Mark a point D on CI such that $C D=4 \mathrm{~cm}$
- With D and I as centres, draw arcs of radii $6 \mathrm{~cm}$ and $5.5 \mathrm{~cm}$ respectively. Let them cut at T. Join DT and IT.
- With $\mathrm{C}$ as centre, draw an arc of radius $6 \mathrm{~cm}$.
- Draw TY parallel to $\mathrm{Cl}$. Let the line cut the previous arc at Y.
- Join CY. CITY is the required trapezium.
Construction of area:
Area of the trapezium CITY $=\frac{1}{2} \times h \times(a+b)$ sq. units
$\begin{aligned}
&=\frac{1}{2} \times 5.5 \times(7+4) \text { sq.units } \\
&=\frac{1}{2} \times 5.5 \times 11 \\
&=30.25 \mathrm{sq} . \mathrm{cm}
\end{aligned}$