SChand CLASS 9 Chapter 9 Mid Point and Intercept Theorems TEST

TEST

Question 1

Sol: 

In the given $\triangle A B C$

AD=DB and BE=EC

So D and E are mid point of AB and BC respectively 

So $D E \| A C$ and $D E=\frac{1}{2} A C$

(a) DE is mid segment of $\triangle A B C$

(b) DE is parallel to AC

(C) AD=BD

(d) DE is half of AC

(e) Twice of $E C=B C$

Question 2

Sol: In the given figure,

If $Y L=L X, Y M=M Z$ and $X N=N Z$

So L,M and N are the mid points of side XY, YZ

and XZ respectively

So $L M=\frac{1}{2} \times 2, M N=\frac{1}{2} x y$

XY =10cm LM =6cm, $\angle m N Z$=$25^{\circ}$

(a) $\mathrm{Nm}=\frac{1}{2} \times y=\frac{1}{2} \times 10 \mathrm{~cm}=5 \mathrm{~cm}$

(b)$x z=2 \times L m=2 \times 6=12 \mathrm{~cm}$

(c) $\quad N Z=\frac{1}{2} \times \times 2=\frac{1}{2} \times 12=6 \mathrm{~cm}$

(d) $\angle L M N=25^{\circ}$

(If LM||XZ an  MN is a transversal)

(e) $\angle Y \times z=\angle m N z=25^{\circ}$

(f)$\angle X L M=\angle X N M$ (opposite angle of a ||gm)

$=180^{\circ}-25^{\circ}=155^{\circ}$

Question 3

(IMAGE TO BE ADDED)

SOL: In $\triangle A B C$

$\mathrm{Lm} \| \mathrm{AC}$ AND $\angle m=\frac{1}{2} A C$

$\Rightarrow 32=\frac{1}{2} \times 4 n \Rightarrow 2 n=32$

$\Rightarrow n=\frac{32}{2}=16 \mathrm{~cm}$

So $n=16 \mathrm{~cm}$

Question 4

(IMAGE TO BE ADDED)

Sol: AE =EC and BD =DC 

So $D E \| A B$ and $D E=\frac{1}{2} A B$

$\Rightarrow 6 n=4 n+12$

$\Rightarrow 6 n-4 n=+12$

$\Rightarrow 2 n=12$

$n=\frac{12}{2}=6$

Question 5

Sol: In the given figure

CG,EH and FJ are mid segment of $\triangle A B D, \triangle G C D$ and $\triangle G H E$ Respectively

$A B=33 \mathrm{~cm}, \angle A B C=57^{\circ}$

(a)

$\begin{aligned}&C G=\frac{1}{2} A B \\&=\frac{1}{2} \times 33=16.5 \mathrm{~cm}\end{aligned}$

(b) 

$\begin{aligned} E H &=\frac{1}{2} D C \\ &=\frac{1}{2} \times \frac{1}{2} D B \\ &=\frac{1}{4} D B=\frac{1}{4} \times 44=11 \mathrm{~cm} \end{aligned}$

(C) 

$\begin{aligned} F J &=\frac{1}{2} G H=\frac{1}{2} \times \frac{1}{2} G C \\ &=\frac{1}{4} G C=\frac{1}{4} \times 16.5 \mathrm{~cm}=4.125 \mathrm{~cm} \end{aligned}$

(d) $\quad M \angle D C G=\angle D B A$ 

=$57^{\circ}$

(e) $\begin{aligned} M \angle G H E &=\angle G C D \\ &=37^{\circ} \end{aligned}$

(f) 

$\begin{aligned} M \angle F J H &=180^{\circ}-\angle G H E \\ &=180^{\circ}-57^{\circ}=123^{\circ} \end{aligned}$