Triangles Test

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Triangles Test - 25

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Weekly Quiz Competition

Question 1
1 / -0

If the corresponding sides of two triangles are proportional, then the two triangles are similar by which test

A
SSS test

B
SAS test

C
AAA test

D
ASA test

Solution

In $$\triangle ABC$$ and $$\triangle DEF$$, if the corresponding sides are proportional

$$\Rightarrow \dfrac{AB}{DE}=\dfrac{BC}{EF}=\dfrac{CA}{FD}$$

$$\Rightarrow$$ The triangles are similar by SSS rule
Question 2
1 / -0

In $$\triangle ABC$$ and $$\triangle PQR$$, $$\angle A=\angle R, \angle B=\angle Q, AB=28, BC=10, AC=24, PQ=5$$ and $$QR= 14$$. Then the length of the side $$PR$$ is :

A
$$10$$

B
$$12$$

C
$$20$$

D
$$26$$

Solution

Considering the triangles $$\Delta ABC$$ and $$\Delta PQR$$

$$\angle A = \angle R $$ ....Given

$$\angle B = \angle Q $$. Given
If two of their angles are equal, then the third angle must also be equal, because angles of a triangle always add to make 180
$$\Rightarrow \angle C = \angle P $$
So, by $$ AAA$$ similarity

$$\Delta ABC $$ $$\sim $$ $$\Delta RQP$$
the lengths of the corresponding sides of two similar triangles are proportional.

$$\Rightarrow \dfrac{AB}{QR} = \dfrac{AC}{PR} =\dfrac{BC}{PQ}$$
$$\dfrac{AC}{PR} =\dfrac{BC}{PQ}$$

$$\Rightarrow \dfrac{24}{PR} = \dfrac{10}{5} $$
$$\Rightarrow PR = 12$$
Question 3
1 / -0

If ratio of heights of two similar triangles is $$4:9$$, then ratio between their areas is?

A
$$2:3$$

B
$$3:2$$

C
$$81:16$$

D
$$16:81$$

Solution

Altitudes of similar triangles are in ratio $$4:9$$
Hence, area of these triangles $$=$$ square of the ratio of their heights or altitudes
$$=(4:9)^2$$
$$=16:81$$
Option $$(4)$$.
Question 4
1 / -0

State the criterion using which the above triangles are similar.

A
AAA

B
AAS

C
SAS

D
SSS

Solution

Let us first write the sides in proportions
$$\dfrac { AC }{ DF } =\dfrac { AB }{ DE } =\dfrac { CB }{ FE } \\ \Rightarrow \dfrac { 2 }{ 1 } =\dfrac { 4 }{ 2 } =\dfrac { 6 }{ 3 } \\ \Rightarrow 2=2=2\\ \therefore \triangle ACB\sim \triangle DFE$$
We get all sides in proportion so the given triangles are similar by $$SSS$$ criterion .
Option $$D$$ is correct
Question 5
1 / -0

In the given $$\Delta$$ $$ABC$$, if $$AB$$ $$=$$ $$AC$$ and $$BD$$ $$=$$ $$DC$$, then $$\angle$$ $$ADC$$ $$=$$ ___________.

A
$$60^o$$

B
$$120^o$$

C
$$90^o$$

D
$$45^o$$

Solution

Considering the triangles $$\Delta ABD$$ and $$\Delta ACD$$
$$AB=AC$$ ....Given
$$BD=DC$$ .....Given
$$AD=AD$$ ....Common
So, by $$SSS$$ congruency
$$\Delta ABD $$ $$\cong $$ $$\Delta ACD$$
Corresponding sides and angles of congruent triangles are also equal
$$\Rightarrow \angle ADC = \angle ADB $$
We know that $$BDC$$ is a straight line
$$\angle ADC+ \angle ADB = 180^{\circ}$$

$$\Rightarrow 2\angle ADC = 180^{\circ}$$
$$\Rightarrow \angle ADC = 90^{\circ}$$
Question 6
1 / -0

If all three angles in one triangle are the same as the corresponding angles in another triangle, then the triangles are similar by which test ?

A
SSS

B
SAS

C
AAA

D
AAS

Solution

It all the angles in triangle are same as the corresponding angle in the other triangle , then the triangles are similar by $$AAA$$ similarity criteria.
So option $$C$$ is correct.
Question 7
1 / -0

In $$\triangle ADE, \angle D=90^{o}, BC=6$$ and $$AB=10=BD$$. Then, find $$x$$.

A
$$12$$

B
$$13$$

C
$$14$$

D
$$15$$

Solution

given that $$\angle D = 90^{\circ} $$ $$ BC = 6 ,AB=10=BD$$
Considering the triangles $$\Delta ADE$$ and $$\Delta ABC$$
$$\angle D = \angle B = 90^{\circ} $$ ....GIVEN in figure
And they are corresponding angles and equal
$$\Rightarrow $$ BC is parallel to DE
$$\angle E = \angle C $$....corresponding angles of parallel lines BC and DE.
$$\angle A = \angle A $$ ..... common angle
So, by $$ AAA$$ similarity
$$\Delta ABC $$ $$\sim $$ $$\Delta ADE$$
the lengths of the corresponding sides of two similar triangles are proportional.
$$\Rightarrow \dfrac{AB}{AD} = \dfrac{AC}{AE} =\dfrac{BC}{DE}$$
$$\Rightarrow \dfrac{10}{20} =\dfrac{6}{x}$$
On solving above we get
$$x=12$$
Question 8
1 / -0

In $$\triangle ADE, \angle D=90^{o}$$, find $$x$$.

A
$$2$$

B
$$2.1$$

C
$$2.2$$

D
$$2.3$$

Solution

Considering the triangles $$\Delta ADE$$ and $$\Delta ABC$$
$$\angle D = \angle B = 90^{\circ} $$ ....GIVEN
And they are corresponding angles and equal
$$\Rightarrow $$ BC is parallel to DE
$$\angle E = \angle C $$....corresponding angles of parallel lines BC and DE.
$$\angle A = \angle A $$ ..... common angle
So, by $$ AAA$$ similarity
$$\Delta ABC $$ $$\sim $$ $$\Delta ADE$$
the lengths of the corresponding sides of two similar triangles are proportional.
$$\Rightarrow \dfrac{AB}{AD} = \dfrac{AC}{AE} =\dfrac{BC}{DE}$$
$$\Rightarrow \dfrac{10}{11} =\dfrac{2}{x}$$
On solving above we get
$$x=2.2$$
Question 9
1 / -0

The area of two similar triangles are $$200$$ and $$128$$, then the ratio of their corresponding altitude is __________

A
$$25:16$$

B
$$5:4$$

C
$$4:5$$

D
$$16:25$$

Solution

Since we know that ratio of areas of two similar triangles is equal to the square of the ratio of their altitude
therefore
Ratio of their altitude $$=\sqrt {\dfrac{{200}}{{128}}} $$

$$ = \sqrt {\dfrac{{100}}{{64}}} $$

$$ = \dfrac{{10}}{8}$$

$$ = \dfrac{5}{4}$$
$$ = 5:4$$
Question 10
1 / -0

If $$\triangle ABC\sim \triangle DEF$$ and $$AB:DE=3:4$$, then the ratio of area of triangles taken in order is

A
$$\dfrac{9}{16}$$

B
$$\dfrac{16}{9}$$

C
$$\dfrac{15}{9}$$

D
$$\dfrac{9}{15}$$

Solution

In similar triangles,
Ratio of areas of triangle = Square of ratio of corresponding sides
Ratio of areas of triangle=$${\dfrac {3^2}{4^2}}$$

=$$\dfrac{9}{16}$$

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Re-Attempt Weekly Quiz Competition

Triangles Test - 25

Result

Triangles Test - 25

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SHARING IS CARING

Question 1

SSS test

SAS test

AAA test

ASA test

Solution

Question 2

$$10$$

$$12$$

$$20$$

$$26$$

Solution

Question 3

$$2:3$$

$$3:2$$

$$81:16$$

$$16:81$$

Solution

Question 4

AAA

AAS

SAS

SSS

Solution

Question 5

$$60^o$$

$$120^o$$

$$90^o$$

$$45^o$$

Solution

Question 6

SSS

SAS

AAA

AAS

Solution

Question 7

$$12$$

$$13$$

$$14$$

$$15$$

Solution

Question 8

$$2$$

$$2.1$$

$$2.2$$

$$2.3$$

Solution

Question 9

$$25:16$$

$$5:4$$

$$4:5$$

$$16:25$$

Solution

Question 10

$$\dfrac{9}{16}$$

$$\dfrac{16}{9}$$

$$\dfrac{15}{9}$$

$$\dfrac{9}{15}$$

Solution

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