Chapter 06 Triangles Exercise-02
EXERCISE 6.2
1. In Fig. 6.17, (i) and (ii), $DE \| BC$. Find $EC$ in (i) and $AD$ in (ii).
Fig. 6.17
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Solution
(i)
Let $EC=x cm$
It is given that $D E \| B C$.
By using basic proportionality theorem, we obtain
$ \begin{aligned} & \dfrac{AD}{DB}=\dfrac{AE}{EC} \\ & \dfrac{1.5}{3}=\dfrac{1}{x} \\ & x=\dfrac{3 \times 1}{1.5} \\ & x=2 \\ & \therefore EC=2 cm \end{aligned} $
(ii)
Let $AD=x cm$
It is given that DE || BC.
By using basic proportionality theorem, we obtain
$\dfrac{AD}{DB}=\dfrac{AE}{EC}$
$\dfrac{x}{7.2}=\dfrac{1.8}{5.4}$
$x=\dfrac{1.8 \times 7.2}{5.4}$
$x=2.4$
$\therefore AD=2.4 cm$
2. $E$ and $F$ are points on the sides $P Q$ and $P R$ respectively of a $\triangle PQR$. For each of the following cases, state whether EF $\|$ QR :
(i) $PE=3.9 ~cm, EQ=3 ~cm, PF=3.6 ~cm$ and $FR=2.4 ~cm$
(ii) $PE=4 cm, QE=4.5 cm, PF=8 cm$ and $RF=9 cm$
(iii) $PQ=1.28 cm, PR=2.56 cm, PE=0.18 cm$ and $PF=0.36 cm$
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Solution
(i)
Given that, $PE=3.9 cm, EQ=3 cm, PF=3.6 cm, FR=2.4 cm$ $\dfrac{PE}{EQ}=\dfrac{3.9}{3}=1.3$ $\dfrac{PF}{FR}=\dfrac{3.6}{2.4}=1.5$
Hence, $\dfrac{PE}{EQ} \neq \dfrac{PF}{FR}$
Therefore, EF is not parallel to QR.
(ii)
$P E=4 cm, Q E=4.5 cm, P F=8 cm, R F=9 cm$
$\dfrac{PE}{EQ}=\dfrac{4}{4.5}=\dfrac{8}{9}$
$\dfrac{PF}{FR}=\dfrac{8}{9}$
Hence, $\dfrac{PE}{EQ}=\dfrac{PF}{FR}$
Therefore, $EF$ is parallel to QR.
(iii)
$P Q=1.28 cm, P R=2.56 cm, P E=0.18 cm, P F=0.36 cm$
$\dfrac{PE}{PQ}=\dfrac{0.18}{1.28}=\dfrac{18}{128}=\dfrac{9}{64}$
$\dfrac{PF}{PR}=\dfrac{0.36}{2.56}=\dfrac{9}{64}$
Hence, $\dfrac{P E}{P Q}=\dfrac{P F}{P R}$
Therefore, EF is parallel to QR.
3. In Fig. 6.18, if $L M \| C B$ and $L N \| C D$, prove that $\dfrac{AM}{AB}=\dfrac{AN}{AD} .$
Fig. 6.18
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Solution
In the given figure, $L M \| C B$
By using basic proportionality theorem, we obtain $\dfrac{AM}{AB}=\dfrac{AL}{AC}$
Similarly, $LN \| CD$
$\therefore \dfrac{AN}{AD}=\dfrac{AL}{AC}$
From $(i)$ and $(i i)$, we obtain
$\dfrac{AM}{AB}=\dfrac{AN}{AD}$
4. In Fig. 6.19, $DE \| AC$ and $DF \| AE$. Prove that $\dfrac{BF}{FE}=\dfrac{BE}{EC}$.
Fig. 6.19
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Solution
In $\triangle A B C, D E \| A C$
$\therefore \dfrac{B D}{D A}=\dfrac{B E}{E C} \quad$ (Basic Proportionality Theorem)
In $\triangle BAE, DF \| AE$
$\therefore \dfrac{BD}{DA}=\dfrac{BF}{FE} \quad$ (Basic Proportionality Theorem)
(ii)
From $(i)$ and $(i i)$, we obtain
$\dfrac{BE}{EC}=\dfrac{BF}{FE}$
5. In Fig. 6.20, DE $\| OQ$ and $DF \| OR$. Show that EF $\| QR$.
Fig. 6.20
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Solution
In $\triangle POQ, DE \| OQ$
$\therefore \dfrac{P E}{E Q}=\dfrac{P D}{D O} \quad$ (Basic proportionality theorem)
(i)
In $\triangle POR, DF \| OR$
$\therefore \dfrac{PF}{FR}=\dfrac{PD}{DO} \quad$ (Basic proportionality theorem)
From $(i)$ and $(i i)$, we obtain
$\dfrac{PE}{EQ}=\dfrac{PF}{FR}$
$\therefore EF \| QR \quad$ (Converse of basic proportionality theorem)
6. In Fig. 6.21, A, B and $C$ are points on $OP, OQ$ and $OR$ respectively such that $AB \| PQ$ and $AC \| PR$. Show that $BC \| QR$.
Fig. 6.21
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Solution
In $\triangle POQ, AB \| PQ$
$\therefore \dfrac{OA}{AP}=\dfrac{OB}{BQ} \quad$ (Basic proportionality theorem)
(i)
In $\triangle POR, AC \| PR$
$\therefore \dfrac{OA}{AP}=\dfrac{OC}{CR} \quad$ (By basic proportionality theorem) (ii)
From $(i)$ and $(i i)$, we obtain
$\dfrac{OB}{BQ}=\dfrac{OC}{CR}$
$\therefore BC \| QR \quad$ (By the converse of basic proportionality theorem)
7. Using Theorem 6.1 , prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).
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Solution
Consider the given figure in which $P Q$ is a line segment drawn through the mid-point $P$ of line $A B$, such that $PQ \| BC$
By using basic proportionality theorem, we obtain
$\dfrac{AQ}{QC}=\dfrac{AP}{PB}$
$\dfrac{AQ}{QC}=\dfrac{1}{1} \quad$ (P is the mid-point of $AB . \therefore AP=PB$ )
$\Rightarrow AQ=QC$
Or, $Q$ is the mid-point of $AC$.
8. Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).
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Solution
Consider the given figure in which $P Q$ is a line segment joining the mid-points $P$ and $Q$ of line $A B$ and $A C$ respectively.
i.e., $A P=P B$ and $A Q=Q C$
It can be observed that $\dfrac{AP}{PB}=\dfrac{1}{1}$
and $\dfrac{AQ}{QC}=\dfrac{1}{1}$
$\therefore \dfrac{AP}{PB}=\dfrac{AQ}{QC}$
Hence, by using basic proportionality theorem, we obtain
$PQ \| BC$
9. $A B C D$ is a trapezium in which $A B \| D C$ and its diagonals intersect each other at the point $O$. Show that $\dfrac{AO}{BO}=\dfrac{CO}{DO}$.
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Solution
Draw a line $EF$ through point $O$, such that
$EF \| CD$
In $\triangle ADC, EO \| CD$
By using basic proportionality theorem, we obtain
$\dfrac{AE}{ED}=\dfrac{AO}{OC}$
In $\triangle ABD$
$OE \| AB$
So, by using basic proportionality theorem, we obtain
$\dfrac{ED}{AE}=\dfrac{OD}{BO}$
$\Rightarrow \dfrac{AE}{ED}=\dfrac{BO}{OD}$
From equations (1) and (2), we obtain $\dfrac{AO}{OC}=\dfrac{BO}{OD}$
$\Rightarrow \dfrac{AO}{BO}=\dfrac{OC}{OD}$
10. The diagonals of a quadrilateral $A B C D$ intersect each other at the point $O$ such that $\dfrac{AO}{BO}=\dfrac{CO}{DO} \cdot$ Show that $ABCD$ is a trapezium.
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Solution
Let us consider the following figure for the given question.
Draw a line $O E \| A B$
In $\triangle A B D, O E \| A B$
By using basic proportionality theorem, we obtain
$\dfrac{AE}{ED}=\dfrac{BO}{OD}$
However, it is given that $\dfrac{AO}{OC}=\dfrac{OB}{OD}$
From equations (1) and (2), we obtain
$\dfrac{AE}{ED}=\dfrac{AO}{OC}$
$\Rightarrow EO$ || DC [By the converse of basic proportionality theorem]
$\Rightarrow AB|| OE|| DC$
$\Rightarrow AB \| CD$
$\therefore A B C D$ is a trapezium.
