Triangles Test

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Triangles Test - 36

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Weekly Quiz Competition

Question 1
1 / -0

If the sides of two similar triangles are in the ratio $$1:7$$, find the ratio of their areas.

A
$$7:1$$

B
$$1:7$$

C
$$1:49$$

D
$$1:14$$

Solution

We know that the relation between area of two similar triangle:
If two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides.
Given, sides of two similar triangles are in the ratio $$1:7$$.
So, the ratio of their areas $$= 1:49$$.
Question 2
1 / -0

XZ and AB are parallel then, using proportionality theorem, find the value of $$YZ$$.

A
$$4$$

B
$$16$$

C
$$20$$

D
$$24$$

Solution

The lines $$XZ$$ and $$AB$$ are parallel.
Therefore, by the basic proportionality theorem,
$$\Rightarrow \dfrac{YB}{ZB} = \dfrac{YA}{XA}$$ $$\Rightarrow \dfrac{YB}{16} = \dfrac{2}{8}$$
$$\Rightarrow 8YB = 32$$
$$\therefore YB = 4$$
Therefore, $$YZ = 4 + 16 = 20$$
Question 3
1 / -0

In given figure $$DE \parallel BC$$, find value of $$x.$$

A
$$2$$

B
$$3$$

C
$$4$$

D
$$1$$

Solution

By basic proportionality theorem,
$$DE \parallel BC$$ ...Given
$$\Rightarrow \dfrac {AD}{DB} = \dfrac {AE}{EC} \Rightarrow \dfrac {x}{x - 2} = \dfrac {x + 2}{x - 1}$$

$$\Rightarrow x = 4$$
Question 4
1 / -0

Identify the SSS postulate, based on which the given pair of triangle can be said similar?

A
figure 2

B
figure 1

C
figure 4

D
figure 3
Question 5
1 / -0

Find the value of $$x$$.

A
$$x =$$ $$\dfrac{1}{3}$$

B
$$x =$$ $$\dfrac{4}{3}$$

C
$$x =$$ $$\dfrac{2}{3}$$

D
$$x =$$ $$\dfrac{5}{3}$$

Solution

The lines $$BA$$ and $$OP$$ are parallel.
Therefore, by the basic proportionality theorem,
$$\Rightarrow \dfrac{CO}{BO} = \dfrac{CP}{PA}$$ $$\Rightarrow \dfrac{4}{x} = \dfrac{3}{1}$$
$$\Rightarrow 3x = 4$$ $$\Rightarrow x =$$ $$\dfrac{4}{3}$$
Question 6
1 / -0

In given figure, if $$ST\parallel QR$$. Find $$PS.$$

A
$$\dfrac {9}{2} cm$$

B
$$\dfrac {7}{2} cm$$

C
$$\dfrac {3}{2} cm$$

D
$$2\ cm$$

Solution

$$\because ST \parallel QR$$
By basic proportionality theorem,
$$\dfrac {PS}{QS} = \dfrac {PT}{RT}$$

$$\dfrac {PS}{3} = \dfrac {3}{2}$$

$$PS = \dfrac {9}{2} \ cm$$
Question 7
1 / -0

Find the ratio of $$BO:CO$$.

A
$$3:10$$

B
$$10:3$$

C
$$3:1$$

D
$$1:3$$

Solution

The lines $$AB$$ and $$OT$$ are parallel.
Therefore, by the base proportionality theorem,
$$\Rightarrow \dfrac{CO}{BO} = \dfrac{CT}{AT}$$ $$\Rightarrow \dfrac{CO}{BO} = \dfrac{12}{40}$$
$$\Rightarrow \dfrac{CO}{BO} = \dfrac{3}{10}$$
So, the ratio of $$BO:CO = 10:3$$ (reciprocal of $$CO:BO = BO:CO$$)
Question 8
1 / -0

In $$\triangle DEF$$, Line PQ $$\parallel $$ side EF. Find DP.
$$DQ=1.8$$ cm, $$QF = 5.4$$ cm, $$EP = 7.2$$ cm.

A
$$2.4$$ cm

B
$$4.2$$ cm

C
$$3.5$$ cm

D
$$1.8$$ cm

Solution

In $$\triangle DEF$$, $$PQ \parallel EF$$
By Basic proportionality theorem,
$$\therefore \dfrac {DP}{PE} = \dfrac {DQ}{QF}$$
$$\therefore \dfrac {DP}{7.2} = \dfrac {1.8}{5.4}$$
$$\therefore DP = \dfrac {1.8 \times 7.2}{5.4}$$
$$\therefore DP = 2.4$$ cm.
Question 9
1 / -0

In $$\triangle ABC, DE\parallel BC$$ and $$\dfrac {AD}{DB} = \dfrac {3}{5}$$. If $$AC = 5.6$$. Find $$AE.$$

A
$$2.1\ cm$$

B
$$4.2\ cm$$

C
$$7\ cm$$

D
$$3\ cm$$

Solution

$$DE\parallel BC$$
By basic proportionality theorem,
$$\dfrac {AD}{DB} = \dfrac {AE}{EC} \Rightarrow \dfrac {AD}{DB} = \dfrac {AE}{AC - AE}$$
$$\Rightarrow \dfrac {3}{5} = \dfrac {AE}{5.6 - AE}$$
$$\Rightarrow AE = 2.1\ cm$$
Question 10
1 / -0

In $$\triangle ABC,\, AB = 6\sqrt{3} cm, \,AC = 12 cm$$ and $$BC = 6 cm$$. The $$\angle B$$ is :

A
$$120^o$$

B
$$60^o$$

C
$$90^o$$

D
$$45^o$$

Solution

Given:
In $$\triangle ABC$$
$$AB=6\sqrt3\,cm$$
$$AC=12\,cm$$
$$BC=6\,cm$$
Solution:
$$AC^2=144$$
$$AB^2=108$$
$$BC^2=36$$
Since,
$$AC^2=AB^2+BC^2$$
$$\therefore$$ By Converse of Pythagoras Theorem,
$$\triangle ABC$$ is an Right Angle Triangle at B.
$$\therefore\,\angle \,B=90^o$$