SChand CLASS 9 Chapter 11 Rectilinear Figures TEST

  TEST

 Question 1

Sol : 

The given figure is trapezium if. A pair of two opposite side are parallel (b) option (b) is right 

 Question 2

Sol: If The given Figure is of a parallelogram. so Their opposite sides are equal and parallel

$\begin{aligned} \Rightarrow & x=3 \\ & 2 x+1=3 x-2 \Rightarrow 3 x-2 x=1+2 \\ \Rightarrow & x=3 \\ & \text { so } x=3 \end{aligned}$

option(e) is right 

 Question 3

Sol: In figure , ABCD is a rectangle whose diagonals bisect each other At O.

$\triangle A B D$ is a rignt triangle as $\angle A=90^{\circ}$. 

option (b) is rigut

 Question 4

(i) In figure ABCD is a parallelogram 

So Their opposite side are equal and parallel and opposite angle are equal 

$x+6=5 x-8 \Rightarrow 5 x-x=6+8$

$\Rightarrow 4 x=14 \Rightarrow x=\frac{14}{4}=3.5$

$14 y+6 y=180$

$20 y=180^{\circ} \Rightarrow y=\frac{180^{\circ}}{20^{\circ}}=9$

$6 y=a \Rightarrow 6 \times 9=a \Rightarrow 9=54^{\circ}$

$b=14 y=14 \times 9=126^{\circ}$

So $a=54, b=126^{\circ}$

(ii) P Q R S is a rectangle

(IMAGE TO BE ADDED)

opposite sides are equal and parallel 

Diagonals bisect Each other.

$c=330^{\circ}$

Similarly a=e 

e$=<1$

and $\angle L+33^{\circ}=90^{\circ}$

$\Rightarrow \angle L=90^{\circ}-33^{\circ}=57^{\circ}$

So $\begin{aligned} e=\angle 1 &=57^{\circ} \text { and } a=e=57^{\circ} \\ \text { if } O P &=O Q \\ \text { so } c &=\angle 2=33^{\circ} \\ \text { so } d &=180^{\circ}-c-\angle 2=180^{\circ}-30^{\circ}-35^{\circ} \\ &=180-66^{\circ}=114^{\circ} \\ & \text { and } b+d=180^{\circ} \\ \Rightarrow b &=180^{\circ}-d=180^{\circ}-114^{\circ}=66^{\circ} \end{aligned}$

Hence 

$a=57^{\circ}, b=66^{\circ}, c=33^{\circ}, d=194^{\circ}, e=57^{\circ}$

(iii) MNPQ is a rhombus in which all sides are equal and diagonals bisect each other at 90 and diagonals bisects opposite angles 

$b=53^{\circ}$

$a=d$

$e=53^{\circ}$

but $e+d=90^{\circ}$

$53^{\circ}+d=90^{\circ} \Rightarrow d=90^{\circ}-53^{\circ}=37^{\circ}$

$a=d=37^{\circ}$

$L c=90^{\circ}$

$a=37^{\circ}, b=53^{\circ}, \angle c=90^{\circ}, d=37^{\circ}, e=53^{\circ}$

 Question 5

Sol: In the figure ABCD is a kite in which BC = CD , AB = AD and diagonals AC and BC intersect at right angle at O and AC bisects the opposite angles 

$\angle O B C=58^{\circ}, \angle D A B=50^{\circ}$

(IMAGE TO BE ADDED)

(i) In $\triangle B C D \cdot B C=C O$

so $\angle O B C=\angle O D C=58^{\circ}$

So $\angle B C U=180^{\circ}-\angle O B C-\angle O D C$

$=180^{\circ}-58^{\circ}-58^{\circ}=180^{\circ}-116^{\circ}=64^{\circ}$

(ii) $\angle D A O=\frac{1}{2} \angle B A D=\frac{1}{2} \times 50^{\circ}=25^{\circ}$

(iii) $\angle O D A=90^{\circ}-\angle O A O=90^{\circ}-25^{\circ}=65^{\circ}$

(iv) $\angle A D C=\angle O O A+\angle O D C$

$=65^{\circ}+58^{\circ}=123^{\circ}$

 Question 6

Sol: In $A B C, P C$ and $Q R$ are mid segments $A B=B C$

So $P Q \| B C$ and $P Q=\frac{1}{2} \quad B C=B R$.............(i)

But AB =BC and PR Their mid points 

So PB = BR ...........(ii)

And QR||AB and QR =$\frac{1}{2} A B=P B$..........(ii)

and $Q R \| A B$ and $A R=\frac{1}{2} A B=P B$..........(iii)

From (i), (ii) and (iii)

$P B=B R=Q R=P Q$

SQ BPQR is a rhombus or a square 

But $\angle B \neq 90^{\circ}$

So $B P Q R$ is a rhombus.