Maths 2 Marks Questions

1. $\triangle $ABC $ \sim $ $\triangle $AMP
2. $\frac{{CA}}{{PA}} = \frac{{BC}}{{MP}}$

1. In $\triangle $ABC $ \sim $ $\triangle $AMP
   $\angle$ABC = $\angle$AMP (1) ........ [Each equal to $90^o$]
   $\angle$BAC=$\angle$MAP (2).........[Common angle]
   In view of (1) and (2)
   $\triangle $ABC $ \sim $ $\triangle $AMP ..........AA similarity criterion
2. $\triangle $ABC $ \sim $ $\triangle $AMP.........Proved above in(i)
   $\therefore $ $\frac{{CA}}{{PA}} = \frac{{BC}}{{MP}}$.........Corresponding sides of two similar triangles are proportional.

In $ \Delta$'s ABE and CFB, we have

$ \angle$AEB = $ \angle$CBF [Alternate angles]
$ \angle$A = $ \angle$C [Opposite angles of a parallelogram]
Thus, by AA-criterion of similarity, we have,
$ \Delta A B E \sim \Delta C F B.$

Given: In the figure, $\triangle ABE \cong \triangle ACD$
To prove: $\triangle ADE \sim \triangle ABC$
Proof:
$\because \triangle ABE \cong \triangle ACD$........[Given]
$\therefore $ AB = AC........[CPCT
AE = AD ........(1)
Also, $\angle$ DAE= $\angle$ BAC.......[Common $\angle$ ].......(2)
In view of (1) and [SAS similarity criterion]

According to questions it is given that S and T are points on sides PR and QR of  $\triangle PQR$ such that $\angle P = \angle R T S$

To Prove $\triangle R P Q \sim \triangle R T S$
Proof In $\triangle RPQ$ and $\triangle RTS$, we have
$\angle P = \angle R T S$ (given)
$\angle R = \angle R$ (common)
$\therefore \quad \triangle R P Q \sim \triangle R T S$ [by AA-similarity].

From the triangle, $\frac{A B}{Q R}=\frac{B C}{P R}=\frac{A C}{P Q}=0.5$
Hence the corresponding sides are propotional. Thus the corresponding angles will be equal. The triangles ABC and QRP are similar i.e, $\vartriangle ABC \sim \vartriangle QRP$ by SSS similarity.

From the figure:
$\angle$A = $\angle$P = 60°
$\angle$B = $\angle$Q = 80°
$\angle$C = $\angle$R = 40°
Therefore, $\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$ [By AAA similarity]
Now corresponding sides of triangles will be propotional,
$\frac{A B}{PQ}=\frac{B C}{QR}=\frac{C A}{RP}$

In $\triangle OPQ,$$\because AB||PQ$
$\therefore \frac{{OA}}{{AP}} = \frac{{OB}}{{BQ}}$......(1)
By basic proportionality theorem
In $\triangle OPR,\,\,\,\because AV||PR$

$\therefore \frac{{OA}}{{AP}} = \frac{{OC}}{{CR}}$.......[By basic proportionality theorem]
From (1) and (2)
$\frac{{OB}}{{BQ}} = \frac{{OC}}{{OR}}$
$\therefore BC||QR$......[By converse basic proportionality theorem]

In $\triangle PQO$ $\because DE||OQ$
$\therefore \frac{{PD}}{{DO}} = \frac{{PE}}{{EQ}}$ ....... (1) [By basic proportionality theorem]
In $\triangle PRO\,\because DF||OR$
$\therefore \frac{{PD}}{{DO}} = \frac{{PF}}{{FR}}$....... (2) [By basic proportionality theorem]
from (1) and (2), $\frac{{PE}}{{EQ}} = \frac{{PF}}{{FR}}$
$\therefore $ $EF||QR$ ...... [By converse of basic proportionality theorem]

In $\triangle ABE,$ we have $DF||AE,$then
$\frac{{BD}}{{AD}} = \frac{{BF}}{{FE}}\,$ [By BPT] ...... (i)
In $\triangle ABC,$ we have $DE||AC,$ then
$\frac{{BD}}{{AD}} = \frac{{BE}}{{EC}}\,$ [By BPT] ...... (2)
From (i) and (2), We get
$\frac{{BF}}{{FE}} = \frac{{BE}}{{EC}}$

According to question it is given that In $\triangle ALM$, $L M \| C B$
$\therefore \quad \frac { A B } { A M } = \frac { A C } { A L }$

Therefore, by Thales' theorem
$\Rightarrow \frac { A M } { A B } = \frac { A L } { A C }$ ..........(i)
In $\triangle ALN$, $L N \| C D$
$\therefore \quad \frac { A C } { A L } = \frac { A D } { A N }$

Therefore by Thales theorem
$\Rightarrow \frac { A L } { A C } = \frac { A N } { A D }$ ...........(ii)
From (i) and (ii) we get
$\frac { A M } { A B } = \frac { A N } { A D }$

We have
PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm
EQ = PQ - PE = 1.28 - 0.18 = 1.10 cm
and FR = PR - PF = 2.56 - 0.36 = 2.20 cm
Now we can find
$\frac{{PE}}{{EQ}} = \frac{{0.18}}{{1.10}} = \frac{9}{{55}}$
$\frac{{PF}}{{PR}} = \frac{{0.36}}{{2.20}} = \frac{9}{{55}}$

$\therefore \frac{{PE}}{{EQ}} = \frac{{PF}}{{FR}}$
$\therefore EF ||QR$ (By converse of basic proportionality theorem)

From the given information, we can have
$\frac{{PE}}{{EQ}} = \frac{4}{{4.5}} = \frac{{40}}{{45}} = \frac{8}{9}.....(I)$
$\frac{{PF}}{{RF}} = \frac{8}{9}....(II)$
From (I) and (II), it is clear that $\frac{{PE}}{{QE}} = \frac{{PF}}{{RF}}$
Therefore, EF||QR (By converse of basic proportionality theorem)

We have
$\frac{{PE}}{{EQ}} = \frac{{3.9}}{3} = \frac{{1.3}}{1}.....(I)$
$\frac{{PF}}{{FR}} = \frac{{3.6}}{{2.4}} = \frac{3}{2} = \frac{{1.5}}{1}....(II)$
From (I) and (II),
we get
$\frac{{PE}}{{EQ}} \ne \frac{{PF}}{{FR}}$
Therefore, EF is not parallel to QR. (By converse of basic proportionality theorem)

In $\triangle ABC,$

1. ​​​​ $\because DE||BC$
   $\therefore \frac{{AD}}{{DB}} = \frac{{AE}}{{EC}}$ ..... By Basic Proportionality theorem​​​​​​​
   $\Rightarrow \frac{{1.5}}{3} = \frac{1}{{EC}}$
   $\Rightarrow EC = \frac{3}{{1.5}}$
   EC = 2cm
2. In​​​​​​​ $\vartriangle ABC,\because DE||BC$
   $\therefore \frac{{AD}}{{DB}} = \frac{{AE}}{{EC}}......$By Basic Proportionality theorem​​​​​​​
   $\therefore \frac{{AD}}{{DB}} = \frac{{1.8}}{{5.4}}$
   $\Rightarrow \frac{{7.2 \times 1.8}}{{5.4}}$
   $\Rightarrow$ AD = 2.4cm

From the given figure we have,
PQ || RS (Given)
So, $\angle$P = $\angle$S (Alternate angles)
and $\angle$Q = $\angle$R (Alternate angles)
Also, $\angle$POQ = $\angle$SOR (Vertically opposite angles)
Therefore, $\triangle POQ \sim \triangle SOR$ (AAA similarity criterion)

According to the question,we are given that,
$\frac { P S } { S Q } = \frac { P T } { T R }$

$\Rightarrow \quad S T \| Q R$ [By using the converse of Basic Proportionality Theorem]
$\Rightarrow \quad \angle P S T = \angle P Q R$ [Corresponding angles]
$\Rightarrow \quad \angle P R Q = \angle P Q R$ $[ \because \angle P S T = \angle P R Q ( \text { Given } ) ]$
$\Rightarrow$ PQ = PR [ $\because$ Sides opposite to equal angles are equal]
$\Rightarrow \quad \Delta$ PQR is isosceles.