Tag: similarity

Quizzes (13) Questions (85)

Questions Related to similarity

Multiple choice maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

The area of two similar triangles ABC and PQR are $25\ cm^{2}\ & \  49\ cm^{2}$, respectively. If QR $=9.8$ cm, then BC is:

  1. 9.8 cm

  2. 7 cm

  3. 49 cm

  4. 25 cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\dfrac { ar(ABC) }{ ar(PQR) } =\dfrac { 25 }{ 49 } $

In two similar triangles, the ratio of their areas is the square of the ratio of their sides

$\Rightarrow { \left( \dfrac { BC }{ QR }  \right)  }^{ 2 }=\dfrac { 25 }{ 49 } \\ \Rightarrow \dfrac { BC }{ QR } =\dfrac { 5 }{ 7 } \\ \Rightarrow \dfrac { BC }{ 9.8 } =\dfrac { 5 }{ 7 } \\ \Rightarrow BC=\dfrac { 5 }{ 7 } \times 9.8=7$

 

Multiple choice maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

$\Delta ABC\sim\Delta PQR.$ If area$\left (ABC \right)= 2.25 m^{2}$, area$ \left (PQR \right)= 6.25 m^{2}$, $ PQ = 0.5 m $, then length of AB is:

  1. 30 cm

  2. 0.5 m

  3. 50 cm

  4. 3 m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\triangle ABC\sim \triangle DEF$

In two similar triangles, the ratio of their areas is the square of the ratio of their sides

$\Rightarrow \dfrac { ar(ABC) }{ ar(PQR) } ={ \left( \dfrac { AB }{ PQ }  \right)  }^{ 2 }\ \Rightarrow \dfrac { 2.25 }{ 6.25 } ={ \left( \dfrac { AB }{ .5 }  \right)  }^{ 2 }\ \Rightarrow \dfrac { AB }{ .5 } =\dfrac { 15 }{ 25 } \ \Rightarrow AB=.3m\ \Rightarrow AB=.3\times 100=30cm$


Multiple choice maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

In $ \triangle ABC\sim \triangle DEF$,  BC $ = $ 4 cm, EF $ =$ 5 cm and area($\triangle $ABC)$ = $ 80 $cm^2$, the area($\triangle$ DEF) is:

  1. $100 cm^{2}$
  2. $125 cm^{2}$
  3. $150 cm^{2}$
  4. $200 cm^{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given $\triangle ABC\sim \triangle DEF$

In two similar triangles, the ratio of their areas is the square of the ratio of their sides
$\Rightarrow \dfrac { ar(ABC) }{ ar(DEF) } ={ \left( \dfrac { BC }{ EF }  \right)  }^{ 2 }\ \Rightarrow \dfrac { 80 }{ ar(DEF) } ={ \left( \dfrac { 4 }{ 5 }  \right)  }^{ 2 }\ \Rightarrow \dfrac { 80 }{ ar(DEF) } =\dfrac { 16 }{ 25 } \ \Rightarrow ar(DEF)=125{ cm }^{ 2 }$

Multiple choice maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

Given $\Delta ABC-\Delta PQR$. If $\dfrac{AB}{PQ}=\dfrac{1}{3}$, then find $\dfrac{ar\Delta ABC}{ar\Delta PQR'}$.

  1. $\dfrac{1}{9}$
  2. $\dfrac{1}{8}$
  3. $\dfrac{8}{9}$
  4. $\dfrac{9}{1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\dfrac{AB}{PQ}=\dfrac{1}{3}$
$\dfrac{ar\Delta ABC}{ar\Delta PQR}=\left(\dfrac{AB}{PQ}\right)^2=\left(\dfrac{1}{3}\right)^2=\dfrac{1}{9}$.
Multiple choice maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

A point taken on each median of a triangle divides the median in the ratio 1:3 reckoning from the vertex . then the ratio of the area of the triangle with vertices at these points  to that of the original triangle is :  

  1. 5 : 13

  2. 25 : 64

  3. 13 : 32

  4. none
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using the properties of medians and area ratios, a point dividing a median in ratio 1:3 creates a triangle with vertices at these points that has an area ratio of 13/32 relative to the original triangle.

Multiple choice maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

$\Delta DEF -\Delta ABC$; If DE $:$ AB $=2:3$ and ar($\Delta$DEF) is equal to $44$ square units, then find ar($\Delta$ABC) in square units.

  1. $99$
  2. $33$
  3. $11$
  4. $66$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The ratio of areas of similar triangles is the square of the ratio of their corresponding sides. (DE/AB)^2 = (2/3)^2 = 4/9. Area(DEF)/Area(ABC) = 4/9. 44/Area(ABC) = 4/9, so Area(ABC) = 44 * 9 / 4 = 99.

Multiple choice maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

Given, $\Delta$ABC$-\Delta$PQR. If $\dfrac{ar(\Delta ABC)}{ar(\Delta PQR)}=\dfrac{9}{4}$ and $AB=18$cm, then find the length of PQ.

  1. $19$
  2. $12$
  3. $32$
  4. $44$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The ratio of areas is the square of the ratio of corresponding sides. Area(ABC)/Area(PQR) = (AB/PQ)^2. 9/4 = (18/PQ)^2. Taking the square root, 3/2 = 18/PQ. PQ = 18 * 2 / 3 = 12.

Multiple choice maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

ABC is an isosceles triangle right angled at B. Similar triangles ACD and ABE are constructed in sides AC and AB. Find the ratio between the areas of $\triangle ABE$ and $\triangle ACD$.

  1. $2:1$
  2. $1:1$
  3. $1:2$

  4. none
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In $\triangle ABC$

$\implies { AB }^{ 2 }+{ BC }^{ 2 }={ AC }^{ 2 }$
$\implies\quad { AB }^{ 2 }+{ AB }^{ 2 }={ AC }^{ 2 }$
$\implies\quad { AC }^{ 2 }={ 2AB }^{ 2 }\quad -(1)$
Ratio of areas of similar triangle is equal to ratio of squares of their corresponding sides.
$\implies\quad \cfrac { Area\quad (\triangle ABE) }{ Area\quad (\triangle ACD) } =\cfrac { { AB }^{ 2 } }{ { AC }^{ 2 } } $
 using(1)
$\implies\quad \cfrac { Area\quad (\triangle ABE) }{ Area\quad (\triangle ACD) } =\cfrac { { AB }^{ 2 } }{ { 2AB }^{ 2 } } =\cfrac { 1 }{ 2 } $