Chapter 6: Triangles

Ans: Let us assume that \[\text{EC = x cm}\]

Given that $\,\text{DE  }\!\!|\!\!\text{  }\!\!|\!\!\text{  BC}$

But from basic proportionality theorem, we know that

$\dfrac{\text{AD}}{\text{DB}}$ $=$ $\dfrac{\text{AE}}{\text{EC}}$

$\dfrac{\text{1}\text{.5}}{\text{3}}$ $=$ $\dfrac{\text{1}}{\text{x}}$

\[\text{x = }\dfrac{\text{3 x 1}}{\text{1}\text{.5}}\]

\[x\text{ }=\text{ }2\]

\[\therefore \]\[\text{EC = 2 cm}\]

Ans: 

Let us assume that \[\text{AD = x cm}\]

Given that \[\text{DE  }\!\!|\!\!\text{  }\!\!|\!\!\text{  BC}\text{.}\]

But from basic proportionality theorem we know that

$\dfrac{\text{AD}}{\text{DB}}$ $\text{=}$ $\dfrac{\text{AE}}{\text{EC}}$

$ \dfrac{\text{x}}{\text{7}\text{.2}}\text{ = }\dfrac{\text{1}\text{.8}}{\text{5}\text{.4}} $

$ \text{x = }\dfrac{\text{1}\text{.8 x 7}\text{.2}}{\text{5}\text{.4}} $

$ \text{x = 2}\text{.4} $

\[\therefore \text{AD = 2}\text{.4}\]$\text{cm}$

Ans:

Given, \[\text{PE = 3}\text{.9 cm, EQ = 3 cm, PF = 3}\text{.6 cm}\],\[\text{FR = 2}\text{.4 cm}\]

$\dfrac{\text{PF}}{\text{EQ}}$ $\text{=}$$\dfrac{\text{3}\text{.9}}{\text{3}}$\[\text{ }\!\!~\!\!\text{ = 1}\text{.3}\]

$\dfrac{\text{PF}}{\text{FR}}$ \[\text{=}\] $\dfrac{\text{3}\text{.6}}{\text{2}\text{.4}}$ \[\text{= 1}\text{.5}\]

Hence, $\dfrac{\text{PE}}{\text{EQ}}$ \[\ne \] $\dfrac{\text{PF}}{\text{FR}}$

Therefore , \[\text{EF}\] is parallel to \[\text{QR}\].

(ii) In a $\text{ }\!\!\Delta\!\!\text{ PQR,}$ \[\text{E}\] and \[\text{F}\] are any two points on the sides \[\text{PQ}\] and \[\text{PR}\] respectively. State whether \[\text{EF  }\!\!|\!\!\text{  }\!\!|\!\!\text{  QR}\] for \[\text{PE = 4 cm, QE = 4}\text{.5 cm, PF = 8 cm}\] and \[\text{RF = 9 cm}\]

 Ans:

\[\text{PE = 4 cm,QE = 4}\text{.5 cm,PF = 8 cm,RF = 9 cm}\]

$\dfrac{\text{PE}}{\text{EQ}}\text{ = }\dfrac{\text{4}}{\text{4}\text{.5}}\text{ = }\dfrac{\text{8}}{\text{9}} $

$ \dfrac{\text{PF}}{\text{FR}}\text{ = }\dfrac{\text{8}}{\text{9}} $

Hence, $\dfrac{\text{PE}}{\text{EQ}}\text{ = }\dfrac{\text{PF}}{\text{FR}}$

Therefore, \[\text{EF}\] is parallel to \[\text{QR}\].

Ans:

\[\text{PQ = 1}\text{.28 cm,PR = 2}\text{.56 cm,PE = 0}\text{.18 cm,PF = 0}\text{.36 cm}\]

$\dfrac{\text{PE}}{\text{PQ}}\text{ = }\dfrac{\text{0}\text{.18}}{\text{1}\text{.28}}\text{ = }\dfrac{\text{18}}{\text{128}}\text{ = }\dfrac{\text{9}}{\text{64}} $

$\dfrac{\text{PF}}{\text{PR}}\text{ = }\dfrac{\text{0}\text{.36}}{\text{2}\text{.56}}\text{ = }\dfrac{\text{9}}{\text{64}} $

Hence, $\dfrac{\text{PE}}{\text{PQ}}\text{ = }\dfrac{\text{PF}}{\text{PR}}$

Therefore, \[\text{EF}\] is parallel to \[\text{QR}\].

Ans:

Given that in the figure, \[\text{LM  }\!\!|\!\!\text{  }\!\!|\!\!\text{  CB}\]

But from basic proportionality theorem, we know that

$\dfrac{\text{AM}}{\text{AB}}\text{ = }\dfrac{\text{AL}}{\text{AC}}\text{ }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{ (i)}$

Also, \[\text{LN  }\!\!|\!\!\text{  }\!\!|\!\!\text{  CD}\]

$\therefore \dfrac{\text{AN}}{\text{AD}}\text{ = }\dfrac{\text{AL}}{\text{AC}}\text{ }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{ (ii)}$

From (i) and (ii), we get

$\dfrac{\text{AM}}{\text{AB}}\text{ = }\dfrac{\text{AN}}{\text{AD}}$

Ans:

In

$\text{ }\!\!\Delta\!\!\text{ ABC,DE  }\!\!|\!\!\text{  }\!\!|\!\!\text{  AC}$ 

$\therefore \dfrac{\text{BD}}{\text{DA}}\text{ = }\dfrac{\text{BE}}{\text{EC}} $

(By Basic proportionality theorem)

$\text{In}$ 

$\text{ }\!\!\Delta\!\!\text{ BAE,DF  }\!\!|\!\!\text{  }\!\!|\!\!\text{  AE} $

$ \therefore \dfrac{\text{BD}}{\text{DA}}\text{ = }\dfrac{\text{BE}}{\text{FE}} $

By Basic proportionality theorem

From (i) and (ii),we get

$\dfrac{\text{BE}}{\text{EC}}\text{ = }\dfrac{\text{BF}}{\text{FE}}$

Ans:

$\text{In}$ 

$ \text{ }\!\!\Delta\!\!\text{ POQ,DE }\!\!|\!\!\text{  }\!\!|\!\!\text{ OQ} $

$ \therefore \dfrac{\text{PE}}{\text{EQ}}\text{=}\dfrac{\text{PD}}{\text{DO}} $              ……………………(i) By basic proportionality theorem$\text{In}$

$ \text{ }\!\!\Delta\!\!\text{ POR,DF  }\!\!|\!\!\text{  }\!\!|\!\!\text{  OR} $

$ \therefore \dfrac{\text{PF}}{\text{FR}}\text{=}\dfrac{\text{PD}}{\text{DO}}$

……………………(ii) By basic proportionality theorem

From (i) and (ii),we get

$\dfrac{\text{PE}}{\text{EQ}}\text{ = }\dfrac{\text{PF}}{\text{FR}} $

$ \therefore \text{EF  }\!\!|\!\!\text{  }\!\!|\!\!\text{  QR} $                             Converse of Basic proportionality theorem

Ans:

In

$\text{ }\!\!\Delta\!\!\text{ POQ,AB  }\!\!|\!\!\text{  }\!\!|\!\!\text{  PQ} $

$\therefore \dfrac{\text{OA}}{\text{OP}}\text{ = }\dfrac{\text{OB}}{\text{PQ}} $

$……………………(i) By basic proportionality theorem

$\text{In}$ 

$\text{ }\!\!\Delta\!\!\text{ POR,AC  }\!\!|\!\!\text{  }\!\!|\!\!\text{  PR} $

\[\therefore \dfrac{\text{OA}}{\text{OP}}\text{ = }\dfrac{\text{OC}}{\text{CR}}\]  ………………(ii) By basic proportionality theorem

From (i) and (ii),we get

$\dfrac{\text{OB}}{\text{BQ}}\text{ = }\dfrac{\text{OC}}{\text{CR}} $

$ \therefore \text{BC  }\!\!|\!\!\text{  }\!\!|\!\!\text{  CR} $

Converse of Basic proportionality theorem

Ans:

Let us assume in the given figure in which \[\text{PQ}\] is a line segment passing through the mid-point \[\text{P}\] of line \[\text{AB}\], such that \[\text{PQ  }\!\!|\!\!\text{  }\!\!|\!\!\text{  BC}\].

From basic proportionality theorem, we know that

$\dfrac{\text{AQ}}{\text{QC}}\text{ = }\dfrac{\text{AP}}{\text{PB}} $

$ \dfrac{\text{AQ}}{\text{QC}}\text{ = 1} $

As \[\text{P}\] is the midpoint of \[\text{AB}\] ,\[\text{AP  =  PB}\]

\[\Rightarrow \text{AQ = QC}\]

Or

\[\text{Q}\] is the midpoint of \[\text{AC}\]

Ans:

Let us assume that the given figure in which \[\text{PQ}\] is a line segment joined by the mid-points \[\text{P and Q}\] of lines \[\text{AB and AC}\] respectively. 

i.e., \[\text{AP  =  PB and AQ  =  QC}\]

Also it is clear that

$\dfrac{\text{AP}}{\text{PB}}\text{ = 1}$ and

$\dfrac{\text{AQ}}{\text{QC}}\text{ = 1} $

$ \therefore \dfrac{\text{AP}}{\text{PB}}\text{ = }\dfrac{\text{AQ}}{\text{QC}} $ 

Hence, using basic proportionality theorem, we get 

\[\text{PQ  }\!\!|\!\!\text{  }\!\!|\!\!\text{  BC}\]

Ans:

Draw a line  \[\text{EF}\] through point \[\text{O}\] , such that 

In \[\text{ }\!\!\Delta\!\!\text{ ADC}\], \[\text{EO  }\!\!|\!\!\text{  }\!\!|\!\!\text{  CD}\]

Using basic proportionality theorem, we get

$\dfrac{\text{AE}}{\text{ED}}\text{ = }\dfrac{\text{AO}}{\text{OC}}$____________________(i)

In \[\text{ }\!\!\Delta\!\!\text{ ABD}\]\[\text{, OE  }\!\!|\!\!\text{  }\!\!|\!\!\text{  AB}\] 

So, using basic proportionality theorem, we get

\[\frac{\text{AE}}{\text{ED}}\ =\ \frac{\text{BO}}{\text{DO}}\ \] ___________________(ii)

From equation (i) and (ii), we get 

$\frac{\text{AO}}{\text{CO}}\text{= }\frac{\text{BO}}{\text{DO}}$

$\therefore \ \frac{\text{AO}}{\text{BO}}\text{= }\frac{\text{CO}}{\text{DO}}$

Ans: 

Let us assume the following figure for the given question.

Draw a line \[\text{OE  }\!\!|\!\!\text{  }\!\!|\!\!\text{  AB}\]

In \[\text{;ABD, OE  }\!\!|\!\!\text{  }\!\!|\!\!\text{  AB}\]

Using basic proportionality theorem, we get 

$\dfrac{\text{AE}}{\text{ED}}\text{ = }\dfrac{\text{BO}}{\text{OD}}\text{ }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{ (i)}$

However, it is given that 

$\frac{\text{AO}}{\text{BO}}\text{ = }\frac{\text{CO}}{\text{DO}} $

$ \therefore \text{ }\frac{\text{AO}}{\text{CO}}\text{ = }\frac{\text{BO}}{\text{DO}}\ \text{ }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ (ii)} $

From equations (i) and (ii), we get 

$\dfrac{\text{AE}}{\text{ED}}\text{ = }\dfrac{\text{AO}}{\text{OC}} $

$ \Rightarrow \text{EO  }\!\!|\!\!\text{  }\!\!|\!\!\text{  DC} $

By the converse of basic proportionality theorem

$\Rightarrow \text{ AB }\left| \left| \text{ OE } \right| \right|\text{ DC}  $

$\Rightarrow \text{AB  }\!\!|\!\!\text{  }\!\!|\!\!\text{  CD} $

\[\therefore \text{ ABCD}\] is a trapezium.