### Introduction to similarity - class-IX

 Description: introduction to similarity Number of Questions: 73 Created by: Tags: similar triangles exploring geometrical figures similarity in geometrical shapes properties of parallel lines and their transversal triangles geometry maths pythagoras' theorem and similar shapes similarity congruence
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Two polygons of different number of side ...... be similar.

1. Cannot

2. Can

3. Both A and B

4. None

Correct Option: A
Explanation:

Ex: A triangle and a square are not similar figures.
Therefore, A  is the correct answer.

If the corresponding angles are equal then the two figures having samenumber of sides are said to be

1. Figures

2. Not similar

3. None of the above

4. Similar

Correct Option: D
Explanation:

The angles are of the same measure means the figures are similar.
Therefore, D is the correct answer.

If the same photograph is printed in different sizes , we say it is

1. Not similar

2. Similar

3. Common

4. None

Correct Option: B
Explanation:

Photograph being same it is printed in the same shape but in passport size, card size, and mini size. They are said to be similar.
Therefore, B is the correct answer.

Two quadrilaterals, a square and a rectangle are not similar as they ......... in shape as well as size.

1. Differ

2. Are same

3. Do not siffer

4. Angles also differ

Correct Option: A
Explanation:

When two quadrilaterals having corresponding angles equal but their corresponding sides are not equal, such figures are not similar.
Therefore, A is the correct answer.

Ratio of two corresponding sides of two similar triangles is $4:9$. Then ratio of their area is ___.

1. $\dfrac{16} {81}$

2. $\dfrac{34} {81}$

3. $\dfrac{81} {16}$

4. $None\ of\ these$

Correct Option: A
Explanation:

Ratio of areas of two similar triangles is equal to the squares of the ratio of their sides.

Ratio of sides $=\dfrac{4}{9}$
Ratio of areas $=\left( \dfrac { 4 }{ 9 } \right) ^{ 2 }=\dfrac { 16 }{ 81 }$

$\triangle PQR \sim \triangle XYZ, \dfrac{XY}{PQ}=\dfrac{3}{2}$ then $\dfrac{Area\ of\ \triangle PQR}{Area\ of\ \triangle XYZ}=$____.

1. $\dfrac{9}{4}$

2. $\dfrac{4}{9}$

3. $\dfrac{3}{2}$

4. $\dfrac{2}{3}$

Correct Option: B
Explanation:

$\dfrac{{XY}}{{PQ}} = \dfrac{3}{2}$

$\Rightarrow \dfrac{{PQ}}{{XY}} = \dfrac{2}{3}$

Now,  $\dfrac{{Area{\rm{ of }}\Delta {\rm{PQR}}}}{{Area{\rm{ of }}\Delta {\rm{XYZ}}}} = {\left( {\dfrac{2}{3}} \right)^2} = \dfrac{4}{9}$

ABCD is a tetrahedron and O is any point. If the lines joining O to the vertices meet the opposite at P, Q, R and S, then $\frac{OP}{AP}+\frac{OQ}{BQ}+\frac{OR}{CR}+\frac{OS}{DS}=2$.

1. True

2. False

Correct Option: B

It is given that $\Delta ABC \sim \Delta PQR$ with $\dfrac{BC}{QR} = \dfrac{1}{3}$. Then $\dfrac{ar (\Delta PQR)}{ar (\Delta ABC)}$ is equal to

1. $9$

2. $3$

3. $\dfrac{1}{3}$

4. $\dfrac{1}{9}$

Correct Option: A
Explanation:
If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.

Since, $\Delta ABC \sim \Delta PQR$

$\therefore \dfrac{ar (\Delta PQR)}{ar (\Delta ABC)} = \dfrac{PR^2}{AC^2} = \dfrac{QR^2}{BC^2} = \dfrac{9}{1} =9 \ \ \ .......... \left [ \therefore \dfrac{QR}{BC} = \dfrac{3}{1} \right ]$

$CM$ and $RN$ are respectively the medians of $\triangle {ABC}$ and $\triangle{PQR}$. If $\triangle {ABC}\sim \triangle{PQR}$, then
$\cfrac{CM}{RN}=\cfrac{AB}{PQ}$

1. True

2. False

Correct Option: A

In a square $ABCD$, the bisector of the angle $BAC$ cut $BD$ at $X$ and $BC$ at $Y$ then triangles $ACY, ABX$ are similar.

1. True

2. False

Correct Option: A

Assume that, $\Delta RST \sim \Delta XYZ$. Complete the following statement.

$\displaystyle \frac{RT}{XY} = \frac{- -}{YZ}, \frac{RS}{XY} = \frac{ST}{- -}, \frac{XY}{ - -} = \frac{YZ}{ST}$

1. ST, YZ, RT

2. ST, YZ, RS

3. YT, YS, RZ

4. ST, YZ, RZ

Correct Option: B
It is given that $\triangle FED\sim \triangle STU$. Is it true to say that $\cfrac{DE}{UT}=\cfrac{EF}{TS}$?
1. Yes

2. No

3. Cannot say

4. None

Correct Option: B
Explanation:

$\triangle FED \sim \triangle STU$
The corresponding sides of both the triangles are $F\leftrightarrow S$, $E\leftrightarrow T$, $D\leftrightarrow U$. With this correspondence,
$\cfrac{EF}{ST}=\cfrac{DE}{TU}$

Consider the following statements:
(1) If three sides of triangle are equal to three sides of another triangle, then the triangles are congruent.
(2) If three angles of a triangle are respectively equal to three angles of another triangle, then the two triangles are congruent.

Of these statements,

1. $(1)$ is correct and $(2)$ is false

2. Both $(1)$ and $(2)$ are false

3. Both $(1)$ and $(2)$ are correct

4. $(1)$ is false and $(2)$ is correct

Correct Option: A
Explanation:

If three sides of triangle are equal to other three sides of the triangle, then the two triangles are congruent by SSS rule.

If three angles of the triangle are equal to other three sides of the triangle, then the two triangles are similar by AAA rule but not congruent.

If in trianges $ABC$ and $DEF$, $\cfrac{AB}{DE}=\cfrac{BC}{FD}$, then they will be similar, when:

1. $\angle B=\angle E$

2. $\angle A=\angle D$

3. $\angle B=\angle D$

4. $\angle A=\angle F$

Correct Option: C
Explanation:

In $\triangle ABC$ and $\triangle DEF$,
$\dfrac{AB}{DE} = \dfrac{BC}{FD}$ (Given)

The angle between these sides are $\angle B$ and $\angle D$. Thus, If the containing angles are equal. The triangles will be similar..

In $\triangle PQR,$ $PQ=4$ cm, $QR=3$ cm, and $RP=3.5$ cm. $\triangle DEF$ is similar to $\triangle PQR.$ If $EF=9$ cm, then what is the perimeter of $\triangle DEF: ?$

1. $10.5$ cm

2. $21$ cm

3. $31.5$ cm

4. Cannot be determined as data is insufficient

Correct Option: C
Explanation:

$PQ = 4, QR = 3$ and $RP = 3.5$
Also, $EF = 9$
Now, perimeter of $\triangle PQR = PQ + QR + RP = 4 + 3 + 3.5 = 10.5$
Given, $\triangle DEF \sim \triangle PQR$
$\dfrac{EF}{QR} = \dfrac{Perimeter(\triangle DEF)}{Perimeter(\triangle PQR)}$
$\dfrac{9}{3} = \dfrac{Perimeter(\triangle DEF)}{10.5}$
Perimeter $(\triangle DEF) = 31.5$ cm

The perimeter of two similar triangles are $24$ cm and $16$ cm, respectively. If one side of the first triangle is $10$ cm, then the corresponding side of the second triangle is

1. $9$ cm

2. $\dfrac{20}3$ cm

3. $\dfrac{16}3$ cm

4. $5$ cm

Correct Option: B
Explanation:
In similar triangles, ratio of the sides is equal to the ratio of the perimeters.
Thus, $\dfrac{p _1}{p _2} = \dfrac{s _1}{s _2}$
$\dfrac{24}{16} = \dfrac{10}{s _2}$
$s _2 = \dfrac{20}{3}$
Thus, side of the other triangle is $\dfrac{20}{3}$ cm.

In a $\triangle ABC$, $BC=AB$ and $\angle B={ 80 }^{ 0 }$. Then $\angle A$ is equal to?

1. ${ 80 }^{ 0 }$

2. ${ 40 }^{ 0 }$

3. ${ 50 }^{ 0 }$

4. ${ 100 }^{ 0 }$

Correct Option: C
Explanation:

Given: $BC = AB$, $\angle B = 80^{\circ}$
Since, $BC = AB$
$\angle A = \angle C = x$ (Angles opposite to equal sides are equal)
Sum of angles of a triangle = 180
$\angle A + \angle B + \angle C = 180$
$x + 80 + x = 180$
$2x = 100$
$x = 50^{\circ}$
Thus, $\angle A = 50^{\circ}$

The area of two similar triangles $\displaystyle \Delta ABC$ and $\displaystyle \Delta DEF$ are 144 $\displaystyle cm^{2}$ and 81 $\displaystyle cm^{2}$ respectively If the longest side of larger $\displaystyle \Delta ABC$ be 36 cm then the longest side of the smaller triangle $\displaystyle \Delta DEF$ is

1. 20 cm

2. 26 cm

3. 27 cm

4. 30 cm

Correct Option: C
Explanation:

In similar triangle ABC & DBF
$\frac{AB}{De}=\frac{BC}{EF}=\frac{AC}{DF}=\frac{ratio ofArea of triangleABC}{ratio ofArea of triangleDEF}$
THEN $\frac{9}{12}=\frac{x}{36}$  (where x is longest side of the smaller triangle )
So x=27 cm

The perimeters of two similar triangles are $25\;cm$ and $15\;cm$ respectively. If one side of first triangle is $9\;cm$, then the corresponding side of the other triangle is

1. $6.2\;cm$

2. $3.4\;cm$

3. $5.4\;cm$

4. $8.4\;cm$

Correct Option: C
Explanation:

$The\quad perimeter\quad of\quad triangle\quad is\quad 25cm\quad and\quad 15cm.\ The\quad ratio\quad of\quad Perimeter\quad of\quad triangle\quad is\quad 25:15=5:3\ The\quad first\quad side\quad is\quad 9cm\quad ,let\quad the\quad other\quad side=x\ Hence,\quad \dfrac { 9 }{ x } =\dfrac { 5 }{ 3\ } \ \Rightarrow x=\dfrac { 3\times 9 }{ 5 } =\dfrac { 27 }{ 5 } =5.4\quad cm$

If area $(\Delta ABC)=36 cm^2, area (\Delta DEF)=64 cm^2$ and $DE=6.4 cm$. Find AB if $\Delta ABC\sim \Delta DEF$

1. $3.6$ cm

2. $7.2$ cm

3. $4.8$cm

4. None

Correct Option: C
Explanation:

In similar triangles, $\dfrac {area\Delta ABC}{area \Delta DEF}=\dfrac {AB^2}{DE^2}=\dfrac {36}{64}$

$\Rightarrow \dfrac {AB}{6.4}=\dfrac {3}{4}\Rightarrow AB=4.8$.

If the areas of two similar triangles are equal then the triangles :

1. are congruent

2. have equal length of corresponding sides

3. (A) and (B)

4. None of these

Correct Option: C
Explanation:

Triangles are similar if they have the same shape, but not necessarily the same size.   It is same as keeping its basic shape but either "zooming in" or out making the triangle bigger or smaller .

For triangles to be congruent if one triangle slides over the other , rotate them, and flip them over in various ways so they will exactly fit over each other.

So, if the triangles are similar, i.e. they are "zoomed-in" or "zoomed-out" versions of each other, and if they have equal area, then they are congruent, and hence have equal corresponding sides.

Sides of two similar triangles are in the ratio of $5 : 11$ then ratio of their areas is

1. $25 : 11$

2. $25 : 121$

3. $125 : 121$

4. $121 : 25$

Correct Option: B
Explanation:

Since, ratio of area of two similar traingles = ratio of square of corresponding sides

ratio of sides = 5 : 11
$\therefore$ ratio of their areas = $(5)^2 : (11)^2 = 25 : 121$

Sides of two similar triangles are in the ratio of $4 : 9$ then area of these triangles are in the ratio

1. $2 : 3$

2. $4 : 9$

3. $81 : 16$

4. $16 : 81$

Correct Option: D
Explanation:

$\displaystyle \because \Delta ABC\sim \Delta DEF$
$\displaystyle \therefore \dfrac{ar\Delta ABC}{ar\Delta DEF}=\dfrac{\left ( 4 \right )^{2}}{\left ( 9 \right )^{2}}=\dfrac{16}{81}=16:81.$

Similarity is represented by :

1. $\sim$

2. $=$

3. $\simeq$

4. none of these

Correct Option: A
Explanation:

$\sim$ is used to represent similarity.

So, answer is option $A.$

When the ratios of the lengths of their corresponding sides are equal, then the two figures are:

1. similar

2. congruent

3. equal

4. none of these

Correct Option: A
Explanation:

When the ratios of the lengths of their corresponding sides are equal, then the two figures are then those figure are similar by $SSS$ similarity.

Two triangles are $ABC$ and $PQR$ are similar, then symbolically it is represented as:

1. $ABC \sim PQR$

2. $ABC \simeq PQR$

3. $ABC = PQR$

4. none of these

Correct Option: A
Explanation:

Two similar triangles can be represented as $ABC∼PQR.$

All congruent figures are similar but the similar figures are not congruent.Is this statement true or false?

1. False

2. Both A and C

3. True

4. Not applicable

Correct Option: C
Explanation:

Congruent figures are of same shape and size. So they are similar but similar figures are not congruent as they might not be of the same size.
Therefore, C is the correct answer.

A tree of height 24m standing in the middle of the road casts a shadow  ofheight 16m. If at the same time a nearby pole of 48 m casts a shadow , what would the height of the shadow be?

1. 23 m

2. 32 m

3. 42 m

4. 24 m

Correct Option: B
Explanation:

If the ratio of the tree is 3 : 2 = 24 m : 16 m
The ratio of the pole is  same so 48m : 32m
Therefore, B is the correct answer.

There were three circular tracks made in a park having the same middle point but their radii was different. These tracks will be called

1. Not similar

2. Similar

3. Congruent

4. All of the above

Correct Option: B
Explanation:

They are said to be similar as the shape is same but size differs.
Therefore, B  is the correct answer.

All ......... triangles are similar.

1. Right angled

2. Isoscles

3. Equilateral

4. Reflex

Correct Option: C
Explanation:

All Equilateral  triangles , corresponding angles are equal .So, they are similar.
Therefore, C is the correct answer.

Anna went to the market to buy some boxes to store things. She was surprised to find boxes one inside the other. They were ....... boxes.

1. Not similar

2. Ambiguous

3. Same size

4. Similar

Correct Option: D
Explanation:

As the boxes were fitting one inside the other they were similar as they were of same shape ,but their size was different.
Therefore, D is the correct answer.

If two triangles are ____ they are similar.

1. Not equal

2. Equiangular

3. Different

4. Not proportionate

Correct Option: B
Explanation:

If the corresponding angles are equal then the triangles are similar .Hence if two triangles are Equiangular then the triangles are similar.

When one acute angle of a triangle is equal to one acute angle of other triangle, and the triangles are right angles, do you think the triangles are similar?

1. Not sure

2. Similar

3. Not similar

4. Cannot be possible

Correct Option: B
Explanation:

Given one acute angle is equal and both triangles are right angled . Hence one angle of both triangles is $90^{\circ}$ each.

Hence the triangles are similar by $AA$ similarity criteria.
Option $B$ is correct.

If corresponding angles of two triangles are equal, then they are known as

1. Equiangular triangles

3. Supplementary angles

4. Complementary angles

Correct Option: A
Explanation:

If corresponding angles of two triangles are equal, then they are known as Equiangular triangles.

Option $A$ is correct.

If the angles of one triangle $ABC$ are congruent with the corresponding angles of triangle $DEF$, which of the following is/are true?

1. The two triangles are congruent but not necessarily similar.

2. The two triangles are similar but not necessarily congruent.

3. The two triangles are both similar and congruent.

4. The two triangles are neither similar nor congruent.

Correct Option: B
Explanation:

Only the angles of two triangles being congruent meaning the same, implies that the triangles are similar, since there could be many triangles having those angles but of varied sizes, just enlarging the sides in proportion.

For congruency, atleast one side has to be taken into account while writing the congruency test.

Which of the following is true?

1. The ratio of sides of two similar triangles is same as the ratio of their corresponding altitudes.

2. The ratio of sides of two similar triangles is same as the ratio of their corresponding perimeters.

3. The ratio of sides of two similar triangles is same as the ratio of their corresponding area

4. The ratio of sides of two similar triangles is same as the ratio of their corresponding medians.

Correct Option: A
Explanation:

Option A is correct as it is the property of similar triangle

If the area of two similar triangles are equal, then they are

1. equilateral

2. isosceles

3. congruent

4. not congruent

Correct Option: C
Explanation:

They are congruent.

$Consider\triangle ABC\quad and\triangle PQR$
$\cfrac { ar(\triangle ABC) }{ ar(\triangle PQR) } =\cfrac { { AB }^{ 2 } }{ { PQ }^{ 2 } } =\cfrac { { AC }^{ 2 } }{ { PR }^{ 2 } } =\cfrac { { BC }^{ 2 } }{ { QR }^{ 2 } }$
$\implies\quad AB=PQ,\quad AC=PR,\quad BC=QR$
$\therefore The\triangle ABC\quad$ and $\triangle PQR$  are congruent.

Two polygons of the same number of sides are similar if all the corresponding interior angles are:

1. Equal

2. Proportional

3. Congruent

4. Cannot say

Correct Option: D
Explanation:

Two polygons of the same number of sides are similar, if:

(a) Their corresponding angles are equal.
(b) Their corresponding sides are in the same ratio (Proportional).
Hence, nothing can be said about two given polygons when only the angles are congruent, is known.

Triangle is equilateral with side$A$, perimeter $P$, area $K$ and circumradius $R$ (radius of the circumscribed circle). Triangle is equilateral with side $a$, perimeter $p$, area $k$, and circumradius $r$. If $A$ is different from $a$, then

1. $P : p = R : r$ only sometimes

2. $P : p = R : r$ always

3. $P : p = K : k$ only sometimes

4. $R : r = K : k$ always

5. $R : r = K : k$ only sometimes

Correct Option: B
Explanation:

Since the triangles are similar, we have $A:a = P:p = R:r = \sqrt {K}: \sqrt {k}$ always, so that (b) is the correct choice.

If in two triangles, corresponding angles are _______ and their corresponding sides are in the ______ratio and hence the two triangles are similar.

1. equal, same

2. unequal, same

3. equal, different

4. unequal, different

Correct Option: A
Explanation:

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

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

1. True

2. False

Correct Option: A
Explanation:

.The statement is true if in two triangles, corresponding sides are in the same ratio, then their corresponding angles are equal and hence the triangles are similar by $AA$ similarity criteria.

Which among the following is/are not correct ?

1. The ratios of the areas of two similar triangles is equal to the ratio of their corresponding sides.

2. The areas of two similar triangles are in the ratio of the corresponding altitudes.

3. The ratio of area of two similar triangles are in the ratio of the corresponding medians.

4. If the areas of two similar triangles are equal, then the triangles are congruent.

Correct Option: A,B,C
Explanation:

The theorem is that the ratio of the areas of two similar triangles is equal to the square of the ratio of the corresponding sides.
In options A, B, and C this condition does not hold.
So option A, B, and C are not true.
But option D is true because if the areas of the similar triangles are equal then the sides will also be equal.
So, the triangles will be congruent by SSS test .

State True or False.
If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are in the same ratio (proportional), then the triangles are similar.

1. True

2. False

Correct Option: A
Explanation:

If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are in the same ratio (proportional), then the triangles are similar by $SAS$ similarity criteria.

Therefore the statement is $True$.

The ratio of the areas of two  similar triangles is equal to the

1. ratio of corresponding medians

2. ratio of corresponding sides

3. ratio of the squares of corresponding sides

4. none of these

Correct Option: C
Explanation:

The area of triangle is proportional to the square of the side of the triangle.
ratio of areas of two similar triangles= ratio of the squares of corresponding sides

In two similar triangles ABC and PQR, if their corresponding altitudes AD and Ps are in the ratio 4:9, find the ratio of the areas of $\triangle ABC$ and $\triangle PQR$.

1. $16:81$

2. $9:16$

3. $81:16$

4. $16:9$

Correct Option: A
Explanation:
Since the areas of two similar triangles are in the ratio of the squares of the corresponding altitudes.

$\therefore$ $\dfrac { Area(\triangle ABC) }{ Area(\triangle PQR) } =\dfrac { { AD }^{ 2 } }{ { PS }^{ 2 } }$

$\Rightarrow$ $\dfrac { Area(\triangle ABC) }{ Area(\triangle PQR) } ={ \left( \dfrac { 4 }{ 9 } \right) }^{ 2 }=\dfrac { 16 }{ 81 }$              [$\because AD:PS=4:9$]

$\Rightarrow$ $\dfrac { Area(\triangle ABC) }{ Area(\triangle PQR) }$ = $\dfrac{16}{81}$

If $\triangle ABC$ is similar to $\triangle DEF$ such that BC=3 cm, EF=4 cm and area of $\triangle ABC=54 {cm}^{2}$. Determine the area of $\triangle DEF$.

1. $40\ cm^2$

2. $59\ cm^2$

3. $69\ cm^2$

4. $96\ cm^2$

Correct Option: D
Explanation:
Since the ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.

$\therefore$ $\dfrac { Area(\triangle ABC) }{ Area(\triangle DEF) } =\dfrac { { BC }^{ 2 } }{ { EF }^{ 2 } }$

$\Rightarrow$ $\dfrac { 54 }{ Area(\triangle DEF) } =\dfrac { { 3 }^{ 2 } }{ { 4 }^{ 2 } }$

$\Rightarrow$ $Area(\triangle DEF)=\dfrac { 54\times 16 }{ 9 } =96{ cm }^{ 2 }$

Two $\triangle sABC$ and DEF are similar. If $ar(DEF)= 243\ cm^2, ar(ABC)=108\ cm^2$ and $BC= 6\ cm$. Find $EF$.

1. $9$

2. $81$

3. $3$

4. $72$

Correct Option: A
Explanation:
Given:-
$\triangle{ABC} \simeq \triangle{DEF}$
$ar \left( DEF \right) = 243 {cm}^{2}$
$ar \left( ABC \right) = 108 {cm}^{2}$
$BC = 6 cm$

To Find:- $EF = ?$

As we know that,
$\because \; \triangle{ABC} \simeq \triangle{DEF}$

$\cfrac{ar \left( \triangle{ABC} \right)}{ar \left( \triangle{DEF} \right)} = {\left( \cfrac{AB}{DE} \right)}^{2} = {\left( \cfrac{BC}{EF} \right)}^{2} = {\left( \cfrac{AC}{DF} \right)}^{2}$

$\therefore \; \cfrac{ar \left( \triangle{ABC} \right)}{ar \left( \triangle{DEF} \right)} = {\left( \cfrac{BC}{EF} \right)}^{2}$

$\Rightarrow \; \cfrac{108}{243} = \cfrac{{6}^{2}}{{EF}^{2}}$

$\Rightarrow \; {EF}^{2} = \cfrac{243}{108} \times 36$

$\Rightarrow \; EF = \sqrt{81}$

$\Rightarrow \; EF = 9$

Hence, the correct answer is $9$.

$\Delta ABC$ and $\Delta DEF$ are similar and $\angle A=40^\mathring \ ,\angle E+\angle F=$

1. $140$

2. $40$

3. $80$

4. $180$

Correct Option: A
Explanation:

Since the triangles are similar.

$\angle A=\angle D$
$\angle D=40^{\circ}$
In triangle $\Delta DEF$
$\angle D+\angle E+\angle F=180^{\circ}$
$\angle E+\angle F=180^{\circ}-40^{\circ}=140^{\circ}$

STATEMENT - 1 : 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.
STATEMENT - 2 : 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. Statement - 1 is True, Statement - 2 is True, Statement - 2 is a correct explanation for Statement - 1

2. Statement - 1 is True, Statement - 2 is True : Statement 2 is NOT a correct explanation for Statement - 1

3. Statement - 1 is True, Statement - 2 is False

4. Statement - 1 is False, Statement - 2 is True

Correct Option: B
Explanation:

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.

If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar.
Both the statements are correct but $2$ is not the reason for $1$
If two corresponding angles are equal then the third corresponding become also equal , so the triangles are similar.
Option $B$ is correct

If $\triangle ABC$ and $BDE$ are similar triangles such that $2AB = DE$ and $BC= 8$ cm, then $EF$ is

1. $16$ cm

2. $17$ cm

3. $4$ cm

4. $8$ cm

Correct Option: A

Is the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians?

1. True

2. False

Correct Option: A

The areas of two similar triangles $\triangle{ABC}$ and $\triangle{DEF}$ are $144\ cm^{2}$ and $81\ cm^{2}$ respectively. If the longest side of larger $\triangle{ABC}$ be $36\ cm$, then, the largest side of the similar triangle $\triangle{DEF}$ is

1. $20\ cm$

2. $26\ cm$

3. $27\ cm$

4. $30\ cm$

Correct Option: A

The correspondence $ABC\rightarrow PQR$ is a similarity in $\Delta ABC$ and $\Delta PQR$. If the perimeter of $\Delta ABC$ is $24$ and the perimeter of $\Delta PQR$ is $40$, then $AB=PQ=$

1. $4:3$

2. $3:4$

3. $5:3$

4. $3:5$

Correct Option: A

$\triangle XYZ \sim \triangle DEF$ for the corresponding $XYZ-EFD$ if $mLX:mLY:mLz=2:3:5$ then in $\triangle DEF$_____ is a right angle.

1. $LD$

2. $LE$

3. $LF$

4. $LE$ or $LF$

Correct Option: A

The ratio of the angles in $\triangle ABC$ is $2 : 3 : 4$. Which one of the following triangles is similar to $\triangle ABC ?$

1. $\triangle DEF$ has angles in the ratio $4 : 3 : 2.$

2. $\triangle PQR$ has angles in the ratio $1 : 2 : 3.$

3. $\triangle LMN$ has angles in the ratio $1 : 1 : 1.$

4. $\triangle STW$ has sides in the ratio $1 : 1 : 1.$

5. $\triangle XYZ$ has sides in the ratio $4 : 3 : 2.$

Correct Option: A

The length of the sides of $\triangle DEF$ are $4,6,8$  $\triangle DEF \sim \triangle PQR$ for correspondence $DEF \leftrightarrow QPR$ if the perimeter of $\triangle PQR=36$, then the length of the smallest side of $\triangle PQR$ is_____

1. $2$

2. $4$

3. $6$

4. $8$

Correct Option: A

If $A={30}^{\circ},\,a=100,\,c=100\sqrt{2}$, find the number of triangles that can be formed.

1. $1$

2. $2$

3. $3$

4. $4$

Correct Option: B
Explanation:
Here $a, c$ and $A$ are given, $\therefore$ we will have to examine whether two triangle are possible or not. For two triangles
$(i)\,a>c\sin{A}$ and $(ii)a<c$
$\Rightarrow 100>100\sqrt{2}\sin{{30}^{\circ}}$
$\Rightarrow 100>100\sqrt{2}\times\dfrac{1}{2}$
$\Rightarrow 100>50\sqrt{2}$
and $a<c$
i.e., $100<100\sqrt{2}$
$\Rightarrow$ Two triangles can be formed.

In triangle ABC, AB = AC = 8 cm, BC = 4 cm and P is a point in side AC such that AP = 6 cm. Prove that $\Delta\,BPC$ is similar to $\Delta\,ABC$. Also, find the length of BP.

1. BP = 4 cm

2. BP = 8 cm

3. BP = 6 cm

4. BP = 12 cm

Correct Option: A
Explanation:

Given: $\triangle ABC$, $AB = AC = 8$, $BC = 4$ and $AP = 6$

In $\Delta\,ABC$,
$\displaystyle\,\frac{AB}{BC}\,=\,\frac{8}{4}\,=\,2$,
In $\Delta\,BPC$,
$\displaystyle\,\frac{BC}{CP}\,=\,\frac{4}{2}\,=\,2$

Now, in $\triangle ABC$ and $\triangle BPC$
$\displaystyle\,\dfrac{AB}{BC}\,= \displaystyle\,\dfrac{BC}{CP}$
$\angle\,ABC\,=\,\angle\,C.$
Therefore, by SAS, $\Delta\,ABC \sim \Delta\,BPC$

Thus, $\dfrac{AB}{BP} = \dfrac{AC}{BC}$

$\dfrac{8}{BP} = \dfrac{8}{4}$
$BP = 4$ cm

In the given figure, $DE$ is parallel to $BC$ and the ratio of the areas of $\triangle ADE$ and trapezium $BDEC$ is $4:5.$ What is $DE : BC: ?$

1. $1:2$

2. $2:3$

3. $4:5$

4. None of these

Correct Option: B

If in $\triangle$s $ABC$ and $DEF,$ $\angle A=\angle E=37^{\circ}, AB:ED=AC:EF$ and $\angle F=69^{\circ},$ then what is the value of $\angle B: ?$

1. $69^{\circ}$

2. $74^{\circ}$

3. $84^{\circ}$

4. $94^{\circ}$

Correct Option: B
Explanation:

In $\triangle ABC$ and $\triangle DEF$
$\angle A = \angle E = 37^{o}$
$\dfrac{AB}{ED} = \dfrac{AC}{EF}$
Thus, $\triangle ABC \sim \triangle EDF$ ....... (By SAS rule)
Thus, $\angle B = \angle D$

Now, $\triangle DEF$
$\angle D + \angle E + \angle F = 180$
$\angle D + 37 + 69 = 180$
$\angle D = 74^{\circ}$
Hence, $\angle B = \angle D = 74^{\circ}$

If two triangles are similar then, ratio of corresponding sides are:

1. unequal

2. equal

3. zero

4. none of these

Correct Option: B
Explanation:
Similar triangles have $:$
$i)$ All their angles equal
$ii)$ Corresponding sides have the same ratio

So, option $B$ is correct.

Two equilateral triangles with side $4 \ cm$ and $6 \ cm$ are _____ triangles.

1. similar

2. congruent

3. both

4. none of these

Correct Option: A
Explanation:

Any two equilateral triangles are similar by SSS criteria..
$SSS$ similarity states that if the lengths of the corresponding sides of two triangles are proportional, then the triangles must be similar.
Two equilateral triangles with side $4 \ cm$ and $6 \ cm$ are similar triangles by $SSS$ similarlty.

In $\triangle ABC \sim \triangle DEF$ such that $AB = 1.2\ cm$ and $DE = 1.4\ cm$. Find the ratio of areas of $\triangle ABC$ and $\triangle DEF$.

1. $36 : 50$

2. $49 : 50$

3. $36 : 49$

4. $1:2$

Correct Option: C
Explanation:

We know that area of two similar triangle is equal to the ratio of the squares of any two corresponding sides
$\dfrac {ar(\triangle ABC)}{ar (\triangle DEF)} = \dfrac {AB^{2}}{DE^{2}} = \dfrac {(1.2)^{2}}{(1.4)^{2}} = \dfrac {36}{49}$

The perimeter of two similar triangle are $30\ cm$ and $20\ cm$. If one side of first triangle is $12\ cm$ determine the corresponding side of second triangle.

1. $8\ cm$

2. $4\ cm$

3. $3\ cm$

4. $16\ cm$

Correct Option: A
Explanation:

Let the two similar triangles be $\triangle ABC$ and $\triangle DEF$

$\therefore \dfrac {AB}{DE} = \dfrac {BC}{EF} = \dfrac {AC}{DF} = \dfrac {P _{1}}{P _{2}}$

$\Rightarrow \dfrac {AB}{DE} = \dfrac {P _{1}}{P _{2}}$

$\Rightarrow \dfrac {12}{DE} = \dfrac {30}{20}$

$\Rightarrow DE = 8\ cm$

Which of the following is/are the property of similar figures?

1. Corresponding angles are congruent.

2. Corresponding sides are in the same ratio.

3. Both A and B

4. None

Correct Option: C
Explanation:

Shape can be different for similar figures be it circle, be it rectangles but if corresponding angles are equal and sides or radius in case of circle are in equal  ratio, then the corresponding two figures are similar.

$\displaystyle \Delta ABC$ and $\displaystyle \Delta DEF$ are two similar triangles such that $\displaystyle \angle A={ 45 }^{ \circ },\angle E={ 56 }^{ \circ }$, then $\displaystyle \angle C$ =___.

1. $\displaystyle { 56 }^{ \circ }$

2. $\displaystyle { 45 }^{ \circ }$

3. $\displaystyle { 101 }^{ \circ }$

4. $\displaystyle { 79 }^{ \circ }$

Correct Option: D
Explanation:

$\Delta ABC \sim \Delta DEF$        ...Given

$\Rightarrow \angle A = \angle D$                 ...C.A.S.T.
$\Rightarrow \angle B = \angle E$                 ...C.A.S.T.
$\Rightarrow \angle C = \angle F$                 ...C.A.S.T.
$\therefore \angle B = \angle E = 56^o$
In $\Delta ABC$,
$\angle A + \angle B + \angle C = 180^o$        ....Angle sum property of triangles
$\Rightarrow 45^o+56^o+\angle C = 180^o$
$\Rightarrow \angle C = 79^o$

If triangle $ABC$ has vertices as $(2, 1), (6, 1), (4, 7)$ and triangle $DEF$, with vertices as $(3, -1), (p,q), (5, -1),$ where $q<-1$, is similar to triangle $ABC$, then $(p,q)$ is equivalent to:

1. $(3, -4)$

2. $(3, -5)$

3. $(3, -1)$

4. $(4, -5)$

Correct Option: C

If a triangle with side lengths as $5, 12$, and $15$ cm is similar to a triangle which has longer side length as $24$ cm, then the perimeter of the other triangle is:

1. $38.4$

2. $44$

3. $51.2$

4. $58$

Correct Option: C
Explanation:

The longer side of the bigger triangle is $24$ cm.

The longer side of the smaller triangle is $15$ cm.
They are in ratio $24:15 = \cfrac{24}{15} = 1.6$
Thus, their perimeters also would be in the ratio $1.6$
The perimeter of the smaller triangle is $5 + 12 + 15 = 32$ cm
Implies the perimeter of the bigger triangle would be $32 \times 1.6 = 51.2$ cm

The perimeter of two similar triangles $\triangle ABC$ and $\triangle DEF$ are $36$ cm and $24$ cm respectively. If $DE=10$ cm, then $AB$ is :

1. $12$ cm

2. $20$ cm

3. $15$ cm

4. $18$ cm

Correct Option: C
Explanation:

Given that triangles $ABC$ and $DEF$ are similar.

Also given, $DE=10$ cm and perimeters of triangles $ABC$ and $DEF$ are $36$ cm and $24$ cm.
So, the corresponding sides of the two triangles is equal to the ratio of their perimeters.

Hence, $\dfrac {\text{perimeter of} \ ABC}{ \text{perimeter of } \ DEF}$ $=\dfrac {AB}{DE}$
Therefore, $\dfrac {36}{24}=\dfrac {AB}{10}$
$\Rightarrow AB=\dfrac {36\times 10}{24}$
$\Rightarrow AB=15$ cm

In $\Delta ABC$, DE is || to BC, meeting AB and AC at D and E. If AD = 3 cm, DB = 2 cm and AE = 2.7 cm, then AC is equal to:

1. $6.5$ cm

2. $4.5$ cm

3. $3.5$ cm

4. $5.5$ cm

Correct Option: B
Explanation:
In $\triangle$$ADE and \triangle$$ ABC$,

$\angle$$ADE=$$\angle$$ABC (corresponding angles) \angle$$AED=$$\angle$$ACB$    (corresponding angles)

so,$\triangle$$ADE \sim$$\triangle$$ABC so, \dfrac{AD}{AB}=\dfrac{AE}{AC} AC=2.7$$\times$$\dfrac{5}{3}$$=$$4.5$ cm

The sides of a triangle are $5$ cm, $6$ cm and $7$ cm. One more triangle is formed by joining the midpoints of the sides. The perimeter of the second triangle is:

1. $18$ cm

2. $12$ cm

3. $9$ cm

4. $6$ cm

Correct Option: C
Explanation:

Let the $\triangle ABC$ have sides $AB = 5$cm,

$BC = 6$ cm and $AC = 7$cm.
Let the midpoints of the sides AB and AC be points D and E respectively.
$\therefore \dfrac {AD}{DB} = \dfrac {AE}{EC}$         ...By B.P.T

$\therefore \dfrac {AD+DB}{DB} = \dfrac {AE+EC}{EC}$   ....By Componendo

$\therefore \dfrac {AB}{DB} = \dfrac {AC}{EC}$      ......(1)

Also, $\angle BAC \cong \angle DAE$    ....(2)

$\therefore \triangle ABC \sim \triangle ADE$      ....SAS test of similarity

$\therefore \dfrac {AB}{AD} = \dfrac {BC}{DE} = \dfrac {AC}{AE}$       ....C.S.S.T

But $\dfrac {AB}{AD} = \dfrac {AB}{\frac 12 AB} = \dfrac 12$

$\therefore \dfrac {BC}{DE} = \dfrac 12$

Perimeter $(\triangle ADE) = AD + DE + AE$
$= \dfrac 12 AB + \dfrac 12 BC + \dfrac 12 AC$

$= \dfrac 12 \left(AB + BC + AC \right)$

$= \dfrac 12 \times 18 = 9$ cm.

So, option C is correct.

Point L, M and N lie on the sides AB, BC and CA of the triangle ABC such that $\ell (AL) : \ell (LB) = \ell (BM) : \ell (MC) = \ell (CN) : \ell (NA) = m : n$, then the areas of the triangles LMN and ABC are in the ratio

1. $\dfrac{m^2}{n^2}$

2. $\dfrac{m^2 - mn + n^2}{(m + n)^2}$

3. $\dfrac{m^2 - n^2}{m^2 + n^2}$

4. $\dfrac{m^2 + n^2}{(m + n)^2}$

Correct Option: A

A man of height 1.8 metre is moving away from a lamp post at the  rate of 1.2 m/sec . If the height of the lamp post be 4.5 metre , then the rate at which the shadow of the  man is lengthening is

1. $0.4 m/sec$

2. $0.8m/sec$

3. $1.2m/sec$

4. None of these

Correct Option: B
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