Tag: maths

Questions Related to maths

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.

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

  1. $8\ cm$

  2. $4\ cm$

  3. $3\ cm$

  4. $16\ cm$

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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?

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

  1. Corresponding angles are congruent.

  2. Corresponding sides are in the same ratio.

  3. Both A and B

  4. None

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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 $\displaystyle \angle C$ =___.

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

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

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

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

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

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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:

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

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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