Tag: areas of similar triangles

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Questions Related to areas of similar triangles

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

A vertical pole of $5.6m$ height casts a shadow $3.2m$ long. At the same time find the height of a pole which casts a shadow $5m$ long.

  1. $8.75m$
  2. $6.75m$
  3. $7.75m$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The ratio of height to shadow length is constant for objects at the same time. Height/Shadow = 5.6/3.2 = 1.75. For a shadow of 5m, the height is 1.75 * 5 = 8.75m.

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.