Introduction to similarity - class-IX

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

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

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.

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

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

In similar triangles, the ratio of corresponding medians is equal to the ratio of corresponding sides. Thus, CM/RN = AB/PQ is a true statement.

In square ABCD, angle BAC = 45 degrees. The bisector of BAC makes angles of 22.5 degrees. Through geometric properties and angle chasing, it can be shown that triangles ACY and ABX are similar.

Given triangle RST similar to triangle XYZ, the ratios of corresponding sides are equal: RS/XY = ST/YZ = RT/XZ. The provided option B correctly completes the ratios.

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

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.

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..

$PQ = 4, QR = 3$ 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

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, $\angle A = 50^{\circ}$

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

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

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

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.

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

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

$\sim$ is used to represent similarity.

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

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.

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.

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

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

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

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

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

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.

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

They are congruent.

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

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

.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.

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 .

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.

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 

Since the triangles are similar.

 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 triangles ABC and BDE are similar, the ratio of sides is constant. If 2AB = DE, then DE/AB = 2. If BC = 8, then the corresponding side EF must be 8 * 2 = 16.

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides, which is also equal to the square of the ratio of their corresponding medians.

Similar triangles must have the same ratio of angles. The ratio 2:3:4 is equivalent to 4:6:8 or any scalar multiple, but the order of the ratio matters for similarity. Option A provides the same ratio 4:3:2, which represents the same set of interior angles as 2:3:4.

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

The ratio of area(ADE) to area(ABC) is 4/(4+5) = 4/9. Since the ratio of areas of similar triangles is the square of the ratio of their corresponding sides, DE/BC = sqrt(4/9) = 2/3.

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, $\angle B = \angle D = 74^{\circ}$

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

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

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

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

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

$\Rightarrow DE = 8\ cm$

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.

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

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

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

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