Tag: maths

Questions Related to maths

If in trianges $ABC$ and $DEF$, $\cfrac{AB}{DE}=\cfrac{BC}{FD}$, then they will be similar, when:

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

  1. $\angle B=\angle E$

  2. $\angle A=\angle D$

  3. $\angle B=\angle D$

  4. $\angle A=\angle F$

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Correct Option: C
Explanation:

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..

In $\triangle PQR,$ $PQ=4$ cm, $QR=3$ cm, and $RP=3.5$ cm. $\triangle DEF$ is similar to $\triangle PQR.$ If $EF=9$ cm, then what is the perimeter of $\triangle DEF: ?$

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

  1. $10.5$ cm

  2. $21$ cm

  3. $31.5$ cm

  4. Cannot be determined as data is insufficient

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Correct Option: C
Explanation:

$PQ = 4, QR = 3$ 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

The perimeter of two similar triangles are $24$ cm and $16$ cm, respectively. If one side of the first triangle is $10$ cm, then the corresponding side of the second triangle is

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

  1. $9$ cm

  2. $\dfrac{20}3$ cm

  3. $\dfrac{16}3$ cm

  4. $5$ cm

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Correct Option: B
Explanation:
In similar triangles, ratio of the sides is equal to the ratio of the perimeters.
Thus, $\dfrac{p _1}{p _2} = \dfrac{s _1}{s _2}$
$\dfrac{24}{16} = \dfrac{10}{s _2}$
$s _2 = \dfrac{20}{3}$
Thus, side of the other triangle is $\dfrac{20}{3}$ cm.

In a $\triangle ABC$, $BC=AB$ and $\angle B={ 80 }^{ 0 }$. Then $\angle A$ is equal to?

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

  1. ${ 80 }^{ 0 }$

  2. ${ 40 }^{ 0 }$

  3. ${ 50 }^{ 0 }$

  4. ${ 100 }^{ 0 }$

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Correct Option: C
Explanation:

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

The area of two similar triangles $\displaystyle \Delta ABC$ and $\displaystyle \Delta DEF$ are 144 $\displaystyle cm^{2}$ and 81 $\displaystyle cm^{2}$ respectively If the longest side of larger $\displaystyle \Delta ABC$ be 36 cm then the longest side of the smaller triangle $\displaystyle \Delta DEF$ is

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

  1. 20 cm

  2. 26 cm

  3. 27 cm

  4. 30 cm

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Correct Option: C
Explanation:

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 perimeters of two similar triangles are $25\;cm$ and $15\;cm$ respectively. If one side of first triangle is $9\;cm$, then the corresponding side of the other triangle is

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

  1. $6.2\;cm$

  2. $3.4\;cm$

  3. $5.4\;cm$

  4. $8.4\;cm$

Show answer
Correct Option: C
Explanation:

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

If area $(\Delta ABC)=36 cm^2, area (\Delta DEF)=64 cm^2$ and $DE=6.4 cm$. Find AB if $\Delta ABC\sim \Delta DEF$

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

  1. $3.6$ cm

  2. $7.2$ cm

  3. $4.8 $cm

  4. None

Show answer
Correct Option: C
Explanation:

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


$\Rightarrow \dfrac {AB}{6.4}=\dfrac {3}{4}\Rightarrow AB=4.8$.

If the areas of two similar triangles are equal then the triangles :

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

  1. are congruent

  2. have equal length of corresponding sides

  3. (A) and (B)

  4. None of these

Show answer
Correct Option: C
Explanation:

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.

Sides of two similar triangles are in the ratio of $5 : 11$ then ratio of their areas is 

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

  1. $25 : 11$

  2. $25 : 121$

  3. $125 : 121$

  4. $121 : 25$

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Correct Option: B
Explanation:

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

 ratio of sides = 5 : 11
$\therefore$ ratio of their areas = $(5)^2 : (11)^2 = 25 : 121$

Sides of two similar triangles are in the ratio of $4 : 9$ then area of these triangles are in the ratio

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

  1. $2 : 3$

  2. $4 : 9$

  3. $81 : 16$

  4. $16 : 81$

Show answer
Correct Option: D
Explanation:

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