Tag: similarity of triangles

Questions Related to similarity of triangles

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