SChand CLASS 9 Chapter 8 Triangles TEST

TEST


Question 1

(i)Among the given criterion 
SSA is not criterion other are criterion 

(ii)From the given figure, AC=FE
AB=FD
DC=DE
So $\triangle A B C \cong \triangle F D E$

Question 2

Sol:
(i) When x =18 , then 
$4 x-11=4 \times 18-11=72-11$
$=61 \Rightarrow R S=U T=61$
and $2 x=2 \times 18=36^{\circ}$

$\Rightarrow \angle S R T=\angle R T U=36^{\circ}$
$R T=R T \quad$ (common)
Hence $\triangle R S T \cong \triangle T U R$

(ii)In  the figure,
ABCD is a kite a which AB=DB
$\Rightarrow 3 x-9=3 \Rightarrow 3 x=3+9=12$
$\Rightarrow x=\frac{12}{3}=4$
$A C=D C$
$\Rightarrow 5=2 \times 4-3 \Rightarrow 5=8-3=5$
$B C=B C \quad$ (common)
So $\triangle A B C \cong \triangle D B C$

Question 3

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Sol; In $\triangle A B C$
$D$ is point on $B C$ and $A D$ bisects $\angle B A C$

i.e. $\angle B A D=\angle C A D$
In $\triangle A D C$, then
Ext, $\angle A D B>\angle C A D$
So In $\triangle A B D$
$B A>B D$

Question 4

(IMAGE TO BE ADDED)

Sol: Two sides of $\triangle A B C$
Let BC =6cm and AB=2.6cm
Now BC-AB =6-2.6= 3.4cm

So third side must be greater than 3.4cm
If sum of any two sides is greater than third side 
So 3.2cm can be the third side (d)

Question 5

(IMAGE TO BE ADDED)
Sol: In $\triangle A B C, \angle B=30^{\circ}, \angle C=80^{\circ}$
Then $\angle A=180^{\circ}-\left(30^{\circ}+80^{\circ}\right)$
$=180^{\circ}-110^{\circ}=70^{\circ}$
Then $A B>B C>A C$   (c)

Question 6

Sol: In $\triangle A B C$ ,two sides are 4cm and 10cm
Let AB = 4CM and BC = 10cm

'a' is third side

If Sum of any two sides >third side
$10<4+a \Rightarrow 10-4<9 \Rightarrow 6>9$

and $a<4+10$
$\Rightarrow a<14$

So 6<a<14 (d)






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