Exercise 5.1 - Chapter 5 Geometry 8th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

EX $5.1$ : Chapter 5 Geometry Class 8th std Maths Guide Samacheer Kalvi Solutions
Question 1.
Fill in the blanks with the correct term from the given list.
(in proportion, similar, corresponding, congruent, shape, area, equal)
(i) Corresponding sides of similar triangles are ___
Answer:
in proportion
(ii) Similar triangles have the same ___  but not necessarily the same size.
Answer:
Shape
(iii) In any triangle ____ sides are opposite to equal angles.
Answer:
equal
(iv) The symbol is used to represent ___ triangles.
Answer:
congruent
(v) The symbol $\sim$ is used to represent ____ triangles.
Answer:
similar

Question $2 .$
In the figure, $\angle \mathrm{CIP} \equiv \angle \mathrm{COP}$ and $\angle \mathrm{HIP} \equiv \angle \mathrm{HOP}$. Prove that $\mathrm{IP} \equiv \mathrm{OP}$.

Question $3 .$
In the given figure, $\mathrm{AC} \equiv \mathrm{AD}$ and $\angle \mathrm{CBD} \equiv \angle \mathrm{DEC}$. Prove that $\triangle \mathrm{BCF} \equiv \Delta \mathrm{EDF}$.

Answer:

Question $4 .$
In the given figure, $\triangle \mathrm{BCD}$ is isosceles with base $\mathrm{BD}$ and $\angle \mathrm{BAE} \equiv \angle \mathrm{DEA}$. Prove that $\mathrm{AB} \equiv \mathrm{ED}$.

Question $5 .$
In the given figure, $\mathrm{D}$ is the midpoint of $\mathrm{OE}$ and $\angle \mathrm{CDE}=90^{\circ}$. Prove that $\triangle \mathrm{ODC} \equiv$ $\triangle \mathrm{EDC}$

Question $6 .$
Is $\triangle \mathrm{PRQ} \equiv \triangle \mathrm{QSP}$ ? Why?

Answer:
In $\triangle \mathrm{PRQ}$ and $\triangle \mathrm{PSQ}$ $\angle \mathrm{PRQ}=\angle \mathrm{PSQ}=90^{\circ}$ given
$\mathrm{PR}=\mathrm{QS}=3 \mathrm{~cm}$ given
$\mathrm{PQ}=\mathrm{PQ}=5 \mathrm{~cm}$ common
It satisfies RHS criteria
$\therefore \triangle \mathrm{PRQ}$ congruent to $\triangle \mathrm{QSP}$.

Question $7 .$
From the given figure, prove that $\triangle \mathrm{ABC} \sim \triangle \mathrm{EDF}$

Answer:
From the $\triangle \mathrm{ABC}, \mathrm{AB}=\mathrm{AC}$
It is an isosceles triangle
Angles opposite to equal sides are equal
$\begin{aligned}
&\therefore \angle \mathrm{B}=\angle \mathrm{C}=65^{\circ} \\
&\therefore \angle \mathrm{B}+\angle \mathrm{C}=65^{\circ}+65^{\circ} \\
&=130^{\circ}
\end{aligned}$
We know that sum of three angles is a triangle $=180^{\circ}$
$\begin{aligned}
&\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ} \\
&\angle \mathrm{A}+130^{\circ}=180^{\circ} \\
&\angle \mathrm{A}=180^{\circ}-130^{\circ} \\
&\angle \mathrm{A}=50^{\circ} \\
&\text { From } \triangle \mathrm{EDF}, \angle \mathrm{E}=50^{\circ} \\
&\therefore \text { Sum of Remaining angles }=180^{\circ}-50^{\circ}=130^{\circ} \\
&\mathrm{DE}=\mathrm{FD} \\
&\therefore \angle \mathrm{D}=\angle \mathrm{F} \\
&\text { From } \triangle \mathrm{ABC} \text { and } \triangle \mathrm{EDF} \therefore \triangle \mathrm{D}=\frac{130}{2}=65^{\circ} \\
&\angle \mathrm{A}=\angle \mathrm{E}=50^{\circ} \\
&\angle \mathrm{B}=\angle \mathrm{D}=65^{\circ} \\
&\angle \mathrm{C}=\angle \mathrm{F}=65^{\circ} \\
&\therefore \mathrm{By} \mathrm{AAA} \text { criteria } \triangle \mathrm{EDF} \sim \triangle \mathrm{ABC}
\end{aligned}$

Question 8 .
In the given figure $\mathrm{YH} \| \mathrm{TE}$. Prove that $\triangle \mathrm{WHY} \sim \triangle \mathrm{WET}$ and also find HE and TE.

Also $\triangle \mathrm{WHY} \sim \triangle \mathrm{WET}$
$\therefore$ Corresponding sides are proportionated
$\begin{aligned}
&\frac{\mathrm{WH}}{\mathrm{WE}}=\frac{\mathrm{HY}}{\mathrm{ET}}=\frac{\mathrm{WY}}{\mathrm{WT}} \\
&\frac{6}{6+\mathrm{HE}}=\frac{4}{\mathrm{ET}}=\frac{4}{16} \\
&\frac{6}{6+\mathrm{HE}}=\frac{4}{16} \\
&\Rightarrow 6+\mathrm{HE}=\frac{6}{4} \times 16 \\
&\Rightarrow 6+\mathrm{HE}=24 \\
&\therefore \mathrm{HE}=24-6 \\
&\mathrm{HE}=18 \\
&\mathrm{Again} \frac{4}{\mathrm{ET}}=\frac{4}{16} \\
&\mathrm{ET}=\frac{4}{4} \\
&\mathrm{ET}=16
\end{aligned}$

Question $9 .$
In the given figure, if $\triangle \mathrm{EAT} \sim \triangle \mathrm{BUN}$, find the measure of all angles.

Answer:
Given $\triangle \mathrm{EAT}=\triangle \mathrm{BUN}$
$\therefore$ Corresponding angles are equal
$\therefore \angle \mathrm{E}=\angle \mathrm{B} \ldots . .$ (1)
$\angle A=\angle U \ldots . .(2)$
$\angle T=\angle N \ldots . . .(3)$
$\angle \mathrm{E}=\mathrm{x}^{\circ}$
$\angle A=2 x^{\circ}$
Sum of three angles of a triangle $=180^{\circ}$
In $\triangle \mathrm{EAT}, \mathrm{x}+2 \mathrm{x}+\angle \mathrm{T}=180^{\circ}$
$\angle T=180^{\circ}-\left(x^{9}+2 x^{9}\right)$
$\angle T=180^{\circ}-3 x^{\circ}$
Also in $\triangle \mathrm{BUN}$
$\begin{aligned}
&(x+40)^{\circ}+x^{\circ}+\angle U=180^{\circ} \\
&x+40^{\circ}+x+\angle U=180^{\circ} \\
&2 x^{\circ}+40^{\circ}+\angle U=180^{\circ} \\
&\angle U=180^{\circ}-2 x-40^{\circ}=140^{\circ}-2 x^{\circ}
\end{aligned}$
Now by (2)
$\begin{aligned}
&\angle \mathrm{A}=\angle \mathrm{U} \\
&2 \mathrm{x}=140^{\circ}-2 \mathrm{x}^{\circ} \\
&2 \mathrm{x}+2 \mathrm{x}=140^{\circ} \\
&4 \mathrm{x}=140^{\circ} \\
&\mathrm{x}=\frac{140}{4}=35^{\circ} \\
&\angle \mathrm{A}=2 \mathrm{x}^{\circ}=2 \times 35^{\circ}=70^{\circ}
\end{aligned}$
$\begin{aligned}
&\angle \mathrm{N}=\mathrm{x}+40^{\circ}=35^{\circ}+40^{\circ}=75^{\circ} \\
&\therefore \angle \mathrm{T}=\angle \mathrm{N}=75^{\circ} \\
&\angle \mathrm{E}=\angle 8=35^{\circ} \\
&\angle \mathrm{A}=\angle \mathrm{U}=70^{\circ}
\end{aligned}$

Question $10 .$
In the given figure, UB $\| \mathrm{AT}$ and $\mathrm{CU}=\mathrm{CB}$ Prove that $\triangle \mathrm{CUB}-\triangle \mathrm{CAT}$ and hence $\triangle \mathrm{CAT}$ is isosceles.

Answer:

Objective Type Questions
Qucstion $11 .$
Two similar triangles will always have angles
(A) acute
(B) obtuse
(C) right
(D) matching
Answer:
(D) matching
 

Question $12 .$
If in triangles $\mathrm{PQR}$ and $\mathrm{XYZ}, \frac{\mathrm{PQ}}{\mathrm{XY}}=\frac{\mathrm{QR}}{2 \mathrm{X}}$ then they will be similar if
(A) $\angle Q=\angle Y$
(B) $\angle P=\angle X$
(C) $\angle \mathrm{Q}=\angle \mathrm{X}$
(D) $\angle \mathrm{P}=\angle Z$
Ans:
(C) $\angle Q=\angle X$
 

Question $13 .$
A flag pole $15 \mathrm{~m}$ high casts a shadow of $3 \mathrm{~m}$ at 10 a.m. The shadow cast by a building at the same time is $18.6 \mathrm{~m}$. The height of the building is
(A) $90 \mathrm{~m}$
(B) $91 \mathrm{~m}$
(C) $92 \mathrm{~m}$
(D) $93 \mathrm{~m}$
Answer:
(D) $93 \mathrm{~m}$
 

Question $14 .$
If $\triangle \mathrm{ABC}-\triangle \mathrm{PQR}$ in which $\angle \mathrm{A}=53^{\circ}$ and $\angle \mathrm{Q}=77^{\circ}$, then $\angle \mathrm{R}$ is
(A) $50^{\circ}$
(B) $60^{\circ}$
(C) $70^{\circ}$
(D) $80^{p}$
Answer:
(A) $50^{\circ}$

Question $15 .$
In the figure, which of the following statements is true?

(A) $\mathrm{AB}=\mathrm{BD}$
(B) $\mathrm{BD}<\mathrm{CD}$
(C) $\mathrm{AC}=\mathrm{CD}$
(D) $\mathrm{BC}=\mathrm{CD}$
Answer:
(C) $\mathrm{AC}=\mathrm{CD}$