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Questions Related to maths

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

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. $16\ cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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

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

$\Rightarrow DE = 8\ cm$

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

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
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

$\displaystyle \Delta ABC$ 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. $\displaystyle { 79 }^{ \circ }$
Reveal answer Fill a bubble to check yourself
D Correct answer
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
$\Rightarrow 45^o+56^o+\angle C = 180^o$
$\Rightarrow \angle C = 79^o$

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

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. $58$
Reveal answer Fill a bubble to check yourself
C Correct answer
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

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

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
Reveal answer Fill a bubble to check yourself
C Correct answer
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

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

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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
Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

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
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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. $\dfrac{m^2 + n^2}{(m + n)^2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Dividing the sides of a triangle in the ratio m:n creates smaller corner triangles. The area of each corner triangle relative to the main triangle can be found using the fraction of the sides bounding the vertex, resulting in the area ratio formula (m^2 - mn + n^2) / (m + n)^2 for the inner triangle LMN.

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

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
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let x be the distance of the man from the post and y be the length of the shadow. By similar triangles, y/1.8 = (x+y)/4.5. Simplifying gives 4.5y = 1.8x + 1.8y, so 2.7y = 1.8x, or y = (2/3)x. The rate of change dy/dt = (2/3) * dx/dt = (2/3) * 1.2 = 0.8 m/sec.