Tag: congruence

Questions Related to congruence

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:

maths congruence introduction to shapes similarity of triangles introduction to similar triangles

  1. $(3, -4)$

  2. $(3, -5)$

  3. $(3, -1)$

  4. $(4, -5)$

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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:

maths congruence introduction to shapes similarity of triangles introduction to similar triangles

  1. $38.4$

  2. $44$

  3. $51.2$

  4. $58$

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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 :

maths congruence introduction to shapes similarity of triangles introduction to similar triangles

  1. $12$ cm

  2. $20$ cm

  3. $15$ cm

  4. $18$ cm

Show answer
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:

maths congruence introduction to shapes similarity of triangles introduction to similar triangles

  1. $6.5$ cm

  2. $4.5$ cm

  3. $3.5$ cm

  4. $5.5$ cm

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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:

maths congruence introduction to shapes similarity of triangles introduction to similar triangles

  1. $18$ cm

  2. $12$ cm

  3. $9$ cm

  4. $6$ cm

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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

maths congruence introduction to shapes similarity of triangles introduction to similar triangles

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

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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 

maths congruence introduction to shapes similarity of triangles introduction to similar triangles

  1. $0.4 m/sec$

  2. $0.8m/sec$

  3. $1.2m/sec$

  4. None of these

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Correct Option: B