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

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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

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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)