Triangles Test

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Triangles Test - 35

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Weekly Quiz Competition

Question 1
1 / -0

Which Similarity criteria will be used to prove: $$\displaystyle \Delta BCD\sim \Delta BAD$$?

A
AAA similarity criterion

B
SAS similarity criterion

C
SSS similarity criterion

D
All of the above

Solution

We can see from the given figure that, in $$\displaystyle \Delta BCD$$ and $$\displaystyle \Delta BAD$$
$$AD=DC$$
$$BD=BD$$
and $$\angle BDA=\angle BDC$$

So, the two triangles are similar by $$SAS$$ similarity criterion.

Hence, $$SAS$$ similarity criterion is used to prove $$\Delta BCD \sim \Delta BAD$$.
Question 2
1 / -0

Both triangles are similar by $$SAS$$ similarity rule. Find the ratio by which they are similar.

A
$$1 : 3$$

B
$$2 : 3$$

C
$$1 : 4$$

D
$$3 : 2$$

Solution

By $$SAS$$ Rule, $$\triangle ABC \cong \triangle PQR$$
Two sides and one angle are proportional
$$AB\sim PQ$$ and $$BC\sim QR$$
$$\Rightarrow \cfrac {AB}{PQ}=\cfrac {12}{36}$$ and $$\cfrac {BC}{QR}=\cfrac {15}{45}$$
$$\Rightarrow \cfrac {AB}{PQ}=\cfrac {BC}{QR}=\cfrac 13$$
$$\therefore$$ The ratio in which they are similar $$=1:3$$
Question 3
1 / -0

If $$\triangle PQR$$ and $$\triangle STU$$ are similar, find value of $$x.$$

A
$$24$$

B
$$12$$

C
$$16$$

D
$$32$$

Solution

Since, both the triangles are similar, the sides of the triangles are proportional:
$$\dfrac{PR}{SU} =\dfrac{PQ}{ST} = \dfrac {RQ}{UT}$$

$$\dfrac{10}{x} = \dfrac{5}{20}$$

$$x= \dfrac {8 \times 20}{5} = 32$$
Question 4
1 / -0

In $$\triangle ABC, \angle A = 90^o$$ & $$\angle B = 30^o$$. In $$\triangle DEF, \angle E = 30^o$$ & $$\angle F = 60^o$$.
Are triangles similar & if yes, name the similarlty criteria?

A
$$AAS$$

B
$$AAA$$

C
$$SSS$$

D
$$SAS$$

Solution

In $$\Delta DEF$$ by angle sum property,
$$\angle D = 180^{o} - 30^{o} - 60^{o}$$
$$\angle D= 90^{o}$$
$$\angle A= \angle D$$
$$\angle B= \angle E$$
By using angle sum property for $$\Delta ABC, $$
$$\angle C = 60^{o}$$
Also, $$\angle C = \angle F$$
Triangles are similar by $$AAA$$ similarity.
$$ABC \sim DEF$$
Question 5
1 / -0

Determine by choosing the best option for the similarly of triangles theorem. $$DE || BC$$. Both $$\triangle ADE$$ & $$ \triangle ABC$$ are similar by:

A
SAS

B
SSS

C
AAA

D
AAS

Solution

In $$\triangle ADE$$ and $$\triangle ABC$$
$$\angle D = \angle B $$          ....[alternate angles since $$DE || BC$$]
And, $$\angle E = \angle C$$      ....[alternate angles since $$DE || BC$$]
$$\angle A$$ is common.
$$\triangle ADE \sim \triangle ABC$$           ...AAA test of similarity
Question 6
1 / -0

$$\displaystyle DE\parallel QR$$, $$D, E, F $$ are midpoints of sides $$PQ, PR$$ and $$QR$$. By which rule are triangles formed equal?

A
SAS congruency Theorem

B
SSS congruency Theorem

C
AAA congruency Theorem

D
AA congruency Theorem

Solution

Given : $$D, E, F$$ are midpoints of sides $$PQ, PR$$ and $$QR$$
In $$\displaystyle \Delta ABC$$, D is the mid point of $$PQ$$.
$$E$$ is the mid point of $$PR$$.
$$\displaystyle DE\parallel QR$$, and $$\displaystyle DE=\frac { 1 }{ 2 } QR$$ ( By Mid point Theorem)
$$\displaystyle DE\parallel QF,DE=QF\left[ QF=\frac { 1 }{ 2 } QR \right] $$
$$\displaystyle \therefore \quad DEFQ$$ is a $$\displaystyle \parallel gm$$. .( opposite sides are equal)
In $$\displaystyle \Delta DEF$$ and $$\displaystyle \Delta FDQ$$
$$\displaystyle DE=QF,DQ=EF,DF=DF$$
$$\displaystyle \therefore \Delta DEF\cong \Delta FDQ$$ ( SSS congruency Theorem)
Question 7
1 / -0

Find the value of $$x$$ when $$DE \parallel AB.$$

A
$$8$$

B
$$4$$

C
$$16$$

D
none

Solution

BASIC PROPORTIONALITY THEOREM (BPT) : If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then the other two sides are divided in the same ratio. That is also known as Thales theorem.

$$DE \parallel AB,$$
by basic proportionality theorem,

$$ \dfrac{CD}{DA} = \dfrac{CE}{EB},$$

$$ \dfrac{2}{8} = \dfrac{4}{x}$$

$$x = \dfrac{4 \times 8}{2}$$

$$x = 16$$
Question 8
1 / -0

Diagonals of trapezium $$ABCD$$ intersect at $$O$$ and $$AB \parallel DC$$.
If $$AB = 3CD$$. Find ratio of areas of $$\triangle AOB $$ & $$\triangle COD$$

A
$$1 : 9$$

B
$$9 : 1$$

C
$$3 : 1$$

D
$$1 : 3$$

Solution

In $$\triangle AOB $$ & $$COD$$
$$\angle AOB = \angle COD$$      ...Given
$$\angle OAB = \angle OCD$$      ...Given
$$\angle OBA = \angle ODC$$      ...Given
$$\triangle AOB \sim \triangle COD$$       ...by AAA test of similarity
By theorem on ratio of areas of similar triangles,
$$\dfrac{A (\triangle AOB)}{A(\triangle COD)} = \left ( \dfrac{AB}{CD} \right)^2 $$
$$\dfrac{A (\triangle AOB)}{A(\triangle COD)} = \left ( \dfrac{3}{1} \right)^2 = \dfrac 91$$
Question 9
1 / -0

Converse of Pythagoras theorem states that in a triangle, if the square of larger side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle. What are we proving here? We prove $$\displaystyle \angle B=$$........ .

A
$$\displaystyle { 130 }^{ o }$$

B
$$\displaystyle { 90 }^{ o }$$

C
$$\displaystyle { 170 }^{ o }$$

D
$$\displaystyle { 110 }^{ o }$$

Solution

Given, a $$\triangle ABC$$ with
$$BC=a, AB=b$$ and $$AC=c$$
$$\Rightarrow a^2+b^2=c^2$$
Now construct a right angled triangle $$DEF$$ with $$EF=BC=a$$
$$AB=DE=b$$
Let $$DF=d$$ and $$\angle DEF=90^o$$
Since $$\triangle DEF$$ is a right angled triangle, we can use Pythagoras Theorem,
$$\therefore d^2=a^2+b^2$$
$$\therefore c=d$$
i.e, $$AC=DF$$
There by construction, by SSS test,
$$\triangle ABC \cong \triangle DEF$$
$$\therefore \triangle ABC$$ is a right angled triangle with $$\angle B=90^o$$
Question 10
1 / -0

The triangles $$\triangle ABC$$ & $$\triangle EBD$$ are similar by $$SAS$$ test. Find $$BD.$$

A
$$8$$

B
$$9$$

C
$$6$$

D
$$12$$

Solution

In $$\triangle ABC$$ & $$\triangle EBD$$ ($$SAS$$ similarly condition)

$$\dfrac{BC}{BD} = \dfrac{AB}{BE} = \dfrac{9}{BD} = \dfrac{27}{24}$$

$$BD = \dfrac{9 \times 24}{27} = BD = 8$$