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How to check for similarity in triangles? - class-IX

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_____ condition is not considered for the similarity of triangle.

  1. $SAS$

  2. $SSS$

  3. $AAA$

  4. $ASA$


Correct Option: A
Explanation:

$SAS$ is a congruency condition not a smililarity condition of triangle

So option $A$ is correct.

If  in $\Delta PQR$,M and N are  points on PQ and PR and $PQ=1.28 ,PR=2.56,PM=0.18,PN=0.36$ cm. 
then $MN||QR.$

  1. True

  2. False


Correct Option: A
State true or false:

In quadrilateral $ABCD$, its diagonals $AC$ and $BD$ intersect at point $O$, such that
$\displaystyle \dfrac{OC}{OA}=\dfrac{OD}{OB}=\dfrac{1}{3}$, then
$\triangle OAB \sim \triangle OCD$
  1. True

  2. False


Correct Option: A
Explanation:

In $\triangle$s, $OAB$ and $OCD$
$\dfrac{OC}{OA} = \dfrac{OD}{OB}$ (Given)
$\angle AOB = \angle COD$ (Vertically opposite angles)
Thus, $\triangle OAB \sim \triangle OCD$ (SAS rule)

In triangle ABC ; M is mid-point of AB, N mid-point of AC and D is any point in base BC. Then:

  1. MN bisects AD

  2. MN divides AD in the ratio 1:3

  3. MN divides AD in the ratio 1:2

  4. MN divides AD in the ratio 1:4


Correct Option: A
Explanation:

In $\triangle ABC$, $M$ is mid point of $AB$ and $N$ is mid point of $AC$
$D$ is any point of BC
Now, Join AD and MN such that they met at O
In $\triangle ABC$
M is mid point of AB and N is mid point point of AC
Hence, $MN \parallel BC$ and $MN = \frac{1}{2} BC$

Now, In $\triangle ABD$
$MO \parallel BC$ and M is mid point of AB
Thus, $O$ is mid point of AD
Hence, $MN$ bisects $AD$

In triangle $ABC$, angle $B$ is obtuse. $D$ and $E$ are mid-points of sides $AB$ and $BC$ respectively and $F$ is a point on side $AC$ such that $EF$ is parallel to $AB$. Then, $BEFD$ is a parallelogram. State True or False. 

  1. True

  2. False


Correct Option: A
Explanation:

Given: $D$ is mid point of $AB$ and $E$ is mid point of $BC$, $F$ is any point on $AC$ and $EF \parallel AB$

Now, in $\triangle ABC$,
E is mid point of BC and $EF \parallel AB$
By Mid point Theorem, $F$ is mid point of $AC$

Also, D is mid point of AB and F is mid point of AC
Hence, by mid point theorem, $DF \parallel BE$
Since, $DF \parallel BE$ and $EF \parallel AB or BD$
Hence, BEFD is parallelogram.

If in two triangles $DEF$ and $PQR$, $\angle D=\angle Q$ and $\angle R=\angle E$, then which of the following is not true?

  1. $\cfrac{EF}{PR}=\cfrac{DF}{PQ}$

  2. $\cfrac{DE}{PQ}=\cfrac{EF}{RP}$

  3. $\cfrac{DE}{QR}=\cfrac{DF}{PQ}$

  4. $\cfrac{EF}{RP}=\cfrac{DE}{QR}$


Correct Option: B

If in the triangles $ABC$ and $DEF$, angle $A$ is equal to angle $E$, both are equal to ${40}^{o}$, $AB:ED=AC:EF$ and angle $F$ is ${65}^{o}$, then angle $B$ is:

  1. ${35}^{o}$

  2. ${65}^{o}$

  3. ${75}^{o}$

  4. ${85}^{o}$


Correct Option: C

D is the mid point of the base BC of a triangle ABC. DM and DN are perpendiculars on AB and AC respectively. If $DM=DN$, the triangle  is

  1. Isosceles

  2. Equilateral

  3. Right angled

  4. Scalene


Correct Option: A
Explanation:

Given: D is mid point of BC. $DM \perp AB$ and $DN \perp AC$, $DM = DN$
Now, In $\triangle DMB$ and $\triangle DNC$,
$DM = DN$ (Given)
$\angle DMB = \angle DNC$ (Each $90^{\circ}$)
$BD = DC$ (D is mid point of BC)
Thus, $\triangle DMB \cong \triangle DNC$ (SAS rule)
Thus, $\angle B = \angle C$ (By cpct)
hence, $\triangle ABC$ is an Isosceles triangle.

If the medians of two equilateral triangles are in the ratio $3:2,$ then what is ratio of the sides$: ?$

  1. $1:1$

  2. $2:3$

  3. $3:2$

  4. $\sqrt{3}:\sqrt{2}$


Correct Option: C
Explanation:

Equilateral triangles are similar triangles.
In similar triangles, the ratio of their corresponding sides is the same as the ratio of their medians.
Hence, ratio of sides = $3: 2$

$\displaystyle \triangle ABC\sim \triangle PQR$ If ar(ABC)=2.25$\displaystyle m^{2}$ ar(PQR)=6.25$\displaystyle m^{2}$, PQ=0.5 m, then length of AB is

  1. $30 cm$

  2. $1.5 cm$

  3. $50 cm$

  4. $2 m$


Correct Option: A
Explanation:

The area of two similar triangles are proportional to the square of their corresponding sides

$\therefore \dfrac{Ar(ABC)}{ar(PQR)}=\dfrac{(AB)^2}{(PQ)^2}$
$\Rightarrow \dfrac{2.25}{6.25}=\dfrac{(AB)^2}{(.5)^2}$
$\Rightarrow AB^2=\dfrac{2.25\times .5}{6.25}$
$\Rightarrow AB^2=.09$
$\Rightarrow AB=.3m=30cm$

If $\displaystyle \triangle ABC\sim \triangle DEF$ BC=4 cm, EF=5 cm and ar $\displaystyle \left ( \triangle ABC \right )=80cm2$,the ar$\displaystyle \left ( \triangle DEF \right )$ is

  1. $\displaystyle 120cm^{2}$

  2. $\displaystyle 125cm^{2}$

  3. $\displaystyle 150cm^{2}$

  4. $\displaystyle 200cm^{2}$


Correct Option: B
Explanation:

If two triangles are equals than the ratio of their square is equal to the ratio of their corresponding sides.

$\therefore \dfrac{arc(\triangle ABC)}{arc(\triangle DEF)}=\dfrac{BC^2}{EF^2}$
$\Rightarrow \dfrac{80}{arc(\triangle DEF)}=\dfrac{16}{25}$
$\Rightarrow arc(\triangle DEF)=\dfrac{80\times 25}{16}=125 cm^2$


If the ratio of the corresponding sides of two similar triangles is 2:3 then the ratio of their corresponding altitude is

  1. 3 : 5

  2. 16 : 81

  3. 4 : 9

  4. 2 : 3


Correct Option: D
Explanation:

If two triangles are similar ,then the ratio of their corresponding sides and altitude are also same. Therefore the ratio of the altitude is 2:3.

If $\displaystyle \triangle ABC\cong \triangle RQP,\angle A=80^{\circ},\angle B=60^{\circ}$, then the value of $\displaystyle \angle P$ is

  1. $\displaystyle 60^{\circ}$

  2. $\displaystyle 50^{\circ}$

  3. $\displaystyle 40^{\circ}$

  4. $\displaystyle 80^{\circ}$


Correct Option: C
Explanation:

If two triangles are equal then their corresponding angle are also equal

$\therefore \angle A=\angle R,$$\angle B=\angle Q$ and$ \angle C=\angle P$
In $\triangle ABC$
$\angle A+\angle B+\angle C=180^{\circ}$
$80^{\circ}+60^{\circ}+\angle C=180^{\circ}$
$\angle C=180^{\circ}-140^{\circ}=40^{\circ}$
$\therefore \angle P=40^{\circ}$

If ABC and DEF are similar triangles such that $\displaystyle \angle A=47^{\circ}$ and $\displaystyle \angle B=83^{\circ}$ then $\displaystyle \angle F$ is

  1. $\displaystyle 60^{\circ}$

  2. $\displaystyle 70^{\circ}$

  3. $\displaystyle 50^{\circ}$

  4. $\displaystyle 100^{\circ}$


Correct Option: C
Explanation:

Since triangle ABC and DEF are similar 

$\therefore \angle A=\angle D,\angle B=\angle E   and  \angle  C=\angle F$
In $\triangle ABC$
$\angle A+\angle B+\angle C=180^\circ$
$47^\circ+83^\circ+\angle C=180^\circ$
$\angle C=180^\circ-130^\circ$
$\angle C=50^\circ$
$\therefore \angle F=50^\circ$

The perimeters of two similar triangles ABC and LMN are 60 cm and 48 cm respectively If LM=8 cm, the length of AB is

  1. $10\ cm$

  2. $8\ cm$

  3. $6\ cm$

  4. $4\ cm$


Correct Option: A
Explanation:

If two triangles are similar then the ratio of their perimeter is equal to the ratio of their corresponding sides.

$\therefore \dfrac{perimeter\ ABC}{perimeter\ LMN}=\dfrac{AB}{LM}$
$\Rightarrow \dfrac{60}{48}=\dfrac{AB}{8}$
$\Rightarrow AB=\dfrac{60\times 8}{48}=10 cm$

If $\displaystyle \triangle ABC$ and $\displaystyle \triangle PQR$ are similar triangles such that $\displaystyle \angle A=32^{\circ}$ and $\displaystyle \angle R=65^{\circ}$ then $\displaystyle \angle B$ is

  1. $\displaystyle 83^{\circ}$

  2. $\displaystyle 42^{\circ}$

  3. $\displaystyle 65^{\circ}$

  4. $\displaystyle 97^{\circ}$


Correct Option: A
Explanation:

Since triangle ABC anD PQR are similar 

$\therefore \angle A=\angle P,\angle B=\angle Q   and  \angle  C=\angle R=65^\circ$
In $\triangle ABC$
$\angle A+\angle B+\angle C=180^\circ$
$32^\circ+\angle B+65=180^\circ$
$\angle B=180^\circ-97^\circ$
$\angle B=83^\circ$

In $\displaystyle \triangle LMN,\triangle L=60^{\circ},\angle M=50^{\circ}$ If $\displaystyle \angle LMN\sim \triangle PQR$ then the value of $\displaystyle \angle R$ is

  1. $\displaystyle 40^{\circ}$

  2. $\displaystyle 60^{\circ}$

  3. $\displaystyle 70^{\circ}$

  4. $\displaystyle 110^{\circ}$


Correct Option: C
Explanation:

$\because \triangle LMN=\triangle PQR$ then

$\angle L=\angle P$,$\angle M=Q$,$\angle N=\angle R$
In $\triangle LMN$$\angle L=60^{\circ},\angle M=50^{\circ}$
$\angle L+\angle M+\angle N=180^{\circ}$
$60^{\circ}+50^{\circ}+\angle N=180^{\circ}$
$\angle N=180^{\circ}-110^{\circ}$
$\angle N=70^{\circ}$
$\therefore \angle R=70^{\circ}$

The area of two similar triangles ABC and PQR are 25 $\displaystyle cm^{2}$ and $\displaystyle 49cm^{2}$ If QR=9.8 cm then BC is

  1. 9.0 cm

  2. 7 cm

  3. 49 cm

  4. 41 cm


Correct Option: B
Explanation:

If two triangles are equals than the ratio of their square is equal to the ratio of their corresponding sides.

$\therefore \dfrac{arc(\triangle ABC)}{arc(\triangle PQR)}=\dfrac{BC^2}{QR^2}$
$\Rightarrow \dfrac{25}{49}=\dfrac{BC^2}{(9.8)^2}$
$\Rightarrow \dfrac{5}{7}=\dfrac{BC}{9.8}$
$\Rightarrow BC=\dfrac{9.8\times 5}{7}=7.0 cm^2$

If the ratio of the corresponding sides of the two similar triangles is 2 : 3 then the ratio of their corresponding attitudes is

  1. $2 : 3$

  2. $4 : 9$

  3. $16 : 81$

  4. none of these


Correct Option: A
Explanation:

If two triangles are same than their sides and corresponding attitudes are also same then the ratio of the corresponding altitude is 2:3. 

The perimeters of two similar triangles ABC and PQR are 60 cm and 48 cm respectively If PQ=8 cm length of AB is

  1. $10\ cm$

  2. $8\ cm$

  3. $6\ cm$

  4. $4\ cm$


Correct Option: A
Explanation:

$P _1=AB+BC+AC=60  cm$

$P _2=PQ+QR+RP=48  cm$
PQ=8 cm
$\dfrac{P _1}{P _2}=\dfrac{AB}{PQ}$
$\Rightarrow \dfrac{60}{48}=\dfrac{AB}{8}$
$\Rightarrow AB=\dfrac{60\times 8}{48}=10 cm$

SAS criterion is true when two sides  and the included angle is congruent with the when two sides  and the included angle of the other triangle are equal. The included angle means

  1. The side between two sides

  2. The angle not between two sides

  3. The line between two sides

  4. The angle between two sides


Correct Option: D
Explanation:

If two sides and the included angle of one triangle are congruent to the corresponding parts of another triangle, then the triangles are congruent.
Therefore, D is the correct answer.

If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar.

  1. AAA similarity criterion

  2. SAS similarity criterion

  3. SSS similarity criterion

  4. All of the above


Correct Option: A
Explanation:

If the corresponding angles are equal then the triangles are similar by $AAA$ similarity criteria.

Option $A$ is correct.

If $\Delta {ABC} \sim \Delta PQR, \angle{B} = \angle{Q}$ is said to be ________ similarity of postulate.

  1. SAS similarity postulate

  2. AAA similarity postulate

  3. SSS similarity postulate

  4. AAS similarity postulate


Correct Option: A
Explanation:

$\Delta {ABC} \sim \Delta PQR, \angle{B} = \angle{Q}$ is said to be SAS similarity of postulate.
Because, SAS Similarity Postulate states, "If an angle of one triangle is congruent to the corresponding angle of another triangle and the sides that include this angle are proportional, then the two triangles are similar."

When we construct a triangle similar to a given triangle as per given scale factor, we construct on the basis of ...........

  1. SSS Similarity

  2. AAA similarity

  3. Basic proportionality theorem

  4. $A$ and $C$ are correct


Correct Option: A
Explanation:

As we consider only sides, therefore, SSS similarity is used.
Option A is correct.  

Goldfish are sold at Rs.15 each. The rectangular coordinate graph showing the cost of 1 to 12 goldfish is:

  1. a straight line segment

  2. a set of horizontal parallel line segments

  3. a set of vertical parallel line segments

  4. a finite set of distinct points

  5. a straight line


Correct Option: D
Explanation:

$\angle MAB=\angle PAO\longrightarrow (1),\hspace{1mm} O\hspace{1mm} be\hspace{1mm} center\ \angle AMB=90°=\angle AOP\ (1)\Longrightarrow 90°-\angle MAB=90°-\angle PAO\ \angle MBA=\angle APO$

By AAA property,
$\triangle APO\sim \triangle ABM\ \cfrac { \bar { AP }  }{ \bar { AB }  } =\cfrac { \bar { AO }  }{ \bar { AM }  } \ \therefore \bar { AP } \cdot \bar { AM } =\bar { AO } \cdot \bar { AB } $

For $\triangle ABC$ and $\triangle PQR$, if $m\angle A=m\angle R $ and $m\angle C=m\angle Q$, then $ABC \longleftrightarrow $_________ is a similarity.

  1. $RQP$

  2. $PQR$

  3. $RPQ$

  4. $QPR$


Correct Option: C
Explanation:

For $\triangle ABC $ and $\triangle PQR$,
$m\angle A = m\angle R$
$m\angle C= m\angle Q$
$\therefore $ by AA criteria for similarity 

$ABC \longleftrightarrow RPQ $ is a similarity.

Say true or false.

If in two triangles, two angles of one triangle are respectively equal to the two angles of the other triangle, then the two triangles are similar.

  1. True

  2. False


Correct Option: A
Explanation:

If two angles of a triangle is equal to two angles of another triangle, then the third angle of both triangles will be equal.
$\therefore$By AAA Theorem of Similarity, the two triangles are similar.

In $\triangle DEF$ &$ \triangle PQR,\ m \angle R$  & _____, then both triangles are similar.

  1. $\dfrac{DE}{PQ}=\dfrac{EF}{QR}$

  2. $\dfrac{DE}{PQ}=\dfrac{DF}{PR}$

  3. $\dfrac{DE}{PR}=\dfrac{DF}{RQ}$

  4. $\dfrac{DE}{QR}=\dfrac{EF}{PR}$


Correct Option: A

$ABC$ and $BDE$ are two equilateral triangles such that $D$ is the mid point of $BC$. Ratio of the areas of triangle $ABC$ and $BDE$ is

  1. $2:1$

  2. $1:2$

  3. $4:1$

  4. $1:4$


Correct Option: C
Explanation:

$\triangle ABC \sim \triangle BDE$                            (both are equilateral triangles)


$\Rightarrow \triangle ABC : \triangle BDE = AB^2 : BD^2$

                                          $= AB^2 :  (\dfrac{1}{2} BC)^{2} $
                                          
                                          $ = AB^2 : \dfrac{1}{4} BC^2 $

                                          $= 4 : 1$           $(\because AB = BC)$
Hence proof.

In $ \triangle ABC, $ If $\angle ADE = \angle B,$ then  $ \Delta ADE ~ \Delta ABC$ are similar

  1. True

  2. False


Correct Option: A

In $\triangle A B C$, D is a point on AB such that $A D = \frac { 1 } { 4 } A B$ and E is a point on AC such that $A E = \frac { 1 } { 4 } A C$ then $D E = \frac { 1 } { 8 } B C$

  1. True

  2. False


Correct Option: B

State true or false:

In a trapezium ABCD, side AB is parallel to side DC; and the diagonals AC and BD intersect each other at point P, then
$\displaystyle \Delta APB$ is similar to $\displaystyle \Delta CPD.$

  1. True

  2. False


Correct Option: A
Explanation:

In $\triangle$ APB and $\triangle$ CPD,
$\angle APB = \angle CPD$ (Vertically opposite angles)
$\angle ABP = \angle CDP$ (Alternate angles of parallel sides AB and CD)
$\angle BAP = \angle DCP$ (Alternate angles of parallel sides AB and CD)
Hence, $\triangle APB \sim \triangle CPD$ (AAA rule)

State true or false:

In parallelogram $ ABCD $. $ E $ is the mid-point of $ AB $ and $ AP $ is parallel to $ EC $ which meets $ DC $ at point $ O $ and $ BC $ produced at $ P $. Hence 
$ O $ is mid-point of $ AP $.

  1. True

  2. False


Correct Option: A
Explanation:

In $\triangle$s, APB and ECB,

$\angle ABP = \angle EBC $ (Common angle)

$\angle PAB = \angle CEB$ (Corresponding angles of parallel lines)

$\angle APB = \angle ECB $ (Third angle of the triangle)

Thus $\triangle APB \sim \triangle ECB$ (AAA rule)

Hence, $\dfrac{AB}{EB} = \dfrac{BP}{BC}$ (Corresponding sides of similar triangles)

$2 = \dfrac{BP}{BC}$

$BP = 2 BC$

Now, in $\triangle$s $OPC$ and $APB,$

$\angle OPC = \angle APB$ (Common angle)

$\angle POC = \angle PAB$ (Corresponding angles of parallel lines)

$\angle PCO = \angle PBA$ (Third angle of a triangle)

$\triangle OPC \sim \triangle APB$ (AAA rule)

hence, $\dfrac{PC}{BP} = \dfrac{OP}{AP}$  (Corresponding sides)

$\dfrac{1}{2} = \dfrac{OP}{AP}$ 

$OP = \dfrac{1}{2} AP$

hence, $O$ is the midpoint of $AP$.

State true or false:

In parallelogram $ ABCD $. $ E $ is the mid-point of $ AB $ and $ AP $ is parallel to $ EC $ which meets $ DC $ at point $ O $ and $ BC $ produced at $ P $. Hence
$ BP= 2AD $


  1. True

  2. False


Correct Option: A
Explanation:

In $\triangle$s, APB and ECB,
$\angle ABP = \angle EBC $ (Common angle)
$\angle PAB = \angle CEB$ (Corresponding angles of parallel lines)
$\angle APB = \angle ECB $ (Third angle of the triangle)
Thus $\triangle APB \sim \triangle ECB$ 
Hence, $\frac{AB}{EB} = \frac{BP}{BC}$ (Corresponding sides of similar triangles)
$2 = \frac{BP}{BC}$
$BP = 2 BC$
$BP = 2 AD$  (BC = AD)

In quadrilateral ABCD, the diagonals AC and BD intersect each at point O. If $AO=2CO$ and $BO=2DO$; Then,

$\displaystyle \Delta AOB$ is similar to $\displaystyle \Delta COD$

  1. True

  2. False


Correct Option: B
Explanation:

Given: $AO = 2 CO$ or $\dfrac{AO}{CO} = 2$
Also given, $BO = 2 DO$ or $\dfrac{BO}{DO} = 2$
In $\triangle AOB$ and $\triangle COD$, we know 
$\angle AOB = \angle COD$
$\dfrac{AO}{CO} = \dfrac{BO}{DO}$
Thus, $\triangle AOB \sim \triangle COD$ (SAS rule)

$\angle BAC$ of triangle $ABC$ is obtuse and $AB=AC$. $P$ is a point in $BC$ such that $PC= 12$ cm. $ PQ $ and $PR$ are perpendiculars to sides $AB$ and $AC$ respectively. If $PQ= 15$ cm and $=9$ cm; find the length of $PB$.

  1. $20$

  2. $24$

  3. $36$

  4. $18$


Correct Option: A
Explanation:

Given: $AB = AC$, $PQ \perp AB$ and $PR \perp AC$
Since, $AB = AC$
$\angle ABC = \angle ACB$...(I) (Isosceles triangle property)

Now, In $\triangle PBQ$ and $\triangle PRC$
$\angle PBQ = \angle PCR$ (From I)
$\angle PQB = \angle PRC$ (Each $90^{\circ}$)
$\angle QPB = \angle RPC$ (Third angle)
Thus, $\triangle QPB \sim \triangle RPC$ (AAA rule)
Hence, $\dfrac{PQ}{PR} = \dfrac{PB}{PC}$
$\dfrac{15}{9} = \dfrac{PB}{12}$
$PB = \dfrac{15 \times 12}{9}$
$PB = 20$ cm

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