Exercise 4.5 - Chapter 4 Geometry 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $4.5$
Question $1 .$
Construct the $\Delta \mathrm{LMN}$ such that $\mathrm{LM}=7.5 \mathrm{~cm}, \mathrm{MN}=5 \mathrm{~cm}$ and $\mathrm{LN}=8 \mathrm{~cm}$. Locate its centroid.
Solution:
In $\triangle \mathrm{LMN}$
$\mathrm{LM}=7.5 \mathrm{~cm}$,
$\mathrm{MN}=5 \mathrm{~cm}$,
$\mathrm{LN}=8 \mathrm{~cm}$

Construction :
Step 1 : Draw $\Delta \mathrm{LMN}$ with $\mathrm{LNM}=8 \mathrm{~cm}, \mathrm{MN}=5 \mathrm{~cm}, \mathrm{LM}=7.5 \mathrm{~cm}$
Step 2 : Construct perpendicular bisectors for any two sides (LN and MN) to find the mid points of LM and MN.
Step $3:$ Draw the medians LD, ME. Let them meet at $G$.
Step 4: $\mathrm{G}$ is the centroid of the triangle LMN.
 

Question $2 .$
Draw and locate the centroid of the triangle $\mathrm{ABC}$ where right angle at $\mathrm{A}, \mathrm{AB}=4 \mathrm{~cm}$ and $\mathrm{AC}=3 \mathrm{~cm}$.
Solution:
In $\triangle \mathrm{ABC}$,
$\mathrm{AB}=4 \mathrm{~cm}$,
$\mathrm{AC}=3 \mathrm{~cm}$,
$\angle \mathrm{A}=90^{\circ}$

Construction:
Step 1: Draw $\triangle \mathrm{ABC}$ with $\mathrm{AB}=4 \mathrm{~cm}, \mathrm{AC}=3 \mathrm{~cm}, \angle \mathrm{A}=90^{\circ}$
Step 2 : Draw perpendicular bisectors of any two sides ( $\mathrm{AB}$ and $\mathrm{AC}$ ) to find the mid points of $\mathrm{AB}$ and $\mathrm{AC}$.
Step 3: Draw the medians $C D$ and BE. Let them meet at $G$.
Step $4: G$ is the centroid of the given triangle.
 

Question $3 .$
Draw the $\triangle \mathrm{ABC}$, where $\mathrm{AB}=6 \mathrm{~cm}, \angle \mathrm{B}=110^{\circ}$ and $\mathrm{AC}=9 \mathrm{~cm}$ and construct the centroid.
Solution:
In $\triangle \mathrm{ABC}$,
$\mathrm{AB}=6 \mathrm{~cm}$,
$\angle B=110^{\circ}$
$\mathrm{AC}=9 \mathrm{~cm}$

Construction:
Step 1: Draw $\triangle \mathrm{ABC}$ with $\mathrm{AB}=6 \mathrm{~cm}, \angle \mathrm{B}=110^{\circ}, \mathrm{AC}=9 \mathrm{~cm}$
Step 2 : Draw perpendicular bisectors of any two sides ( $\mathrm{BC}$ and $\mathrm{AB})$ to find the mid points of $\mathrm{BC}$ and $\mathrm{AB}$.
Step 3 : Construct medians $\mathrm{AD}$ and CE. Let them meet at $\mathrm{G}$.
Step $4: G$ is the centroid of the given $\triangle \mathrm{ABC}$.
 

Question $4 .$
Construct the $\triangle P Q R$ such that $P Q=5 \mathrm{~cm}, P R=6 \mathrm{~cm}$ and $\angle Q P R=60^{\circ}$ and locate its centroid.
Solution:
In $\triangle \mathrm{PQR}, \mathrm{PQ}=5 \mathrm{~cm}, \mathrm{PR}=6 \mathrm{~cm}, \angle \mathrm{QPR}=60^{\circ}$

Construction :
Step 1: Draw $\triangle \mathrm{PQR}$ with the given measurement
Step 2 : Draw perpendicular bisectors of any two sides (PQ and QR) to find the mid points of $P Q$ and $Q R$.
Step 3 : Draw medians PD and RE. Let them meet at $G$.
Step $4: G$ is the centroid of the given $\triangle \mathrm{PQR}$.
 

Question $5 .$
Draw $\triangle \mathrm{PQR}$ with sides $\mathrm{PQ}=7 \mathrm{~cm}, \mathrm{QR}=8 \mathrm{~cm}$ and $P R=5 \mathrm{~cm}$ and construct its Orthocentre.
Solution:
$\triangle \mathrm{PQR}$ with sides $\mathrm{PQ}=7 \mathrm{~cm}$,
$\mathrm{QR}=8 \mathrm{~cm}$,
$\mathrm{PR}=5 \mathrm{~cm} .$

Construction:
(i) Draw the $\triangle \mathrm{PQR}$ with the given measurements.
(ii) Construct altitudes from any two vertices ( $R$ and $P$ ) to their opposite sides (PQ and $Q R$ ) respectively.
(iii) The point of intersection of the altitude $\mathrm{H}$ is the Orthocentre at the given $\triangle \mathrm{PQR}$.
 

Question 6.
Draw an equilateral triangle of sides $6.5 \mathrm{~cm}$ and locate its Orthocentre.
Solution:
Equilateral triangle at sides $6.5 \mathrm{~cm}$.

Construction :
(i) Draw the $\triangle \mathrm{ABC}$ with the given measurements.
(ii) Construct altitudes from any two vertices $A$ and $B$, to their opposite sides $B C$ and $A C$ respectively.
(iii) The point intersection of the altitude $\mathrm{H}$ is the orthocentre of the given $\triangle \mathrm{ABC}$.
 

Question 7.
Draw $\triangle \mathrm{ABC}$, where $\mathrm{AB}=6 \mathrm{~cm}, \angle \mathrm{B}=110^{\circ}$ and $\mathrm{BC}=5 \mathrm{~cm}$ and construct its Orthocentre.
Solution:
$\triangle \mathrm{ABC}$, where $\mathrm{AB}=6 \mathrm{~cm}$,
$\mathrm{B}=110^{\circ} \text { and }$
$\mathrm{BC}=5 \mathrm{~cm} \text {. }$

(i) Draw the $\triangle \mathrm{ABC}$ with the given measurements.
(ii) Construct altitudes from any two vertices (A and B), to their opposite sides (BC and $A C$ ) respectively.
(iii) The point of intersection the altitude $\mathrm{H}$ is the orthocentre of the given $\triangle \mathrm{ABC}$.
 

Question 8.
Draw and locate the Orthocentre of a right triangle $\mathrm{PQR}$ where $\mathrm{PQ}=4.5 \mathrm{~cm}, \mathrm{QR}=6 \mathrm{~cm}$ and $\mathrm{PR}=7.5 \mathrm{~cm}$.
Solution:
Right triangle $\mathrm{PQR}$ where $\mathrm{PQ}=4.5 \mathrm{~cm}, \mathrm{QR}=6 \mathrm{~cm}$ and $\mathrm{PR}=7.5 \mathrm{~cm}$.

Construction:
(i) Draw the $\triangle \mathrm{PQR}$ with the given measurements.
(ii) Construct altitudes from any two vertices $R$ and $Q$, to their opposite sides $P Q$ and $P R$ respectively.
(iii) The point of intersection of the altitude $H$ is the orthocentre of the given $\triangle P Q R$.