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

Triangle A has a base of x and a height of 2x. Triangle B is similar to triangle A, and has a base of 2x. What is the ratio of the area of triangle A to triangle B?

  1. 1:2

  2. 2:1

  3. 2:3

  4. 1:4
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
1 to 4: Since you know that triangle B is similar to triangle A, you can set up a proportion to represent the relationship between the sides of both triangles:
$\dfrac{base}{height}=\dfrac{x}{2x}=\dfrac{2x}{?}$
By proportional reasoning, the height of triangle B must be 4x. Calculate the area of each triangle with the area formula:
Triangle A: $A=\dfrac{b\times h}{2}=\dfrac{(x)(2x)}{2}=x^2$
Triangle B: $A=\dfrac{b\times h}{2}=\dfrac{(2x)(4x)}{2}=4x^2$
The ratio of the area of triangle A to triangle B is 1 to 4.
Multiple choice maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

In $\displaystyle \Delta ABC\sim \Delta DEF$ and their areas are $\displaystyle { 36cm }^{ 2 }$ and $\displaystyle { 64cm }^{ 2 }$ respectively.If side AB=3 cm. Find DE.

  1. 3 cm

  2. 2 cm

  3. 5 cm

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

In $\displaystyle \Delta ABC\sim \Delta DEF$
$\displaystyle \frac { ar.\left( \Delta ABC \right)  }{ ar.\left( \Delta DEF \right)  } =\frac { { AB }^{ 2 } }{ { DE }^{ 2 } } =\frac { { AC }^{ 2 } }{ { DF }^{ 2 } } =\frac { { BC }^{ 2 } }{ { EF }^{ 2 } } $
$\displaystyle \frac { 36 }{ 64 } =\frac { { AB }^{ 2 } }{ { DE }^{ 2 } } $
$\displaystyle \frac { 6 }{ 8 } =\frac { 3 }{ DE } $
$\displaystyle DE=\frac { 8\times 3 }{ 6 } =\frac { 24 }{ 6 } =4cm$
Therefore, D is the correct answer.

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

The areas of two similar triangles are $121 cm^2$ and $81 cm^2$ respectively. Find the ratio of their corresponding heights.

  1. $\dfrac{11}{9}$
  2. $\dfrac{10}{9}$
  3. $\dfrac{9}{11}$
  4. $\dfrac{9}{10}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given the areas of two similar triangles are $121$ sq cm and $81$ sq cm

We know that, the ratio of areas of two similar triangles is equal to the ratio of the squares of the corresponding heights.

The ratio of area of triangles $= \dfrac{121}{81}=\dfrac{(11)^{2}}{(9)^{2}}$
Then ratio of height of triangle $=\sqrt{\left [ \dfrac{11}{9} \right ]^{2}}=\dfrac{11}{9}$

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

What is the ratio of the heights of two isosceles triangles which have equal vertical angles, and of which the areas are in the ratio of $9 : 16$?

  1. $4.5:8$
  2. $3:4$
  3. $4:3$
  4. $8:4.5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For similar triangles (or triangles with equal vertical angles), the ratio of areas is the square of the ratio of their corresponding heights. Since the area ratio is 9:16, the height ratio is sqrt(9):sqrt(16) = 3:4.

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
$\Rightarrow \dfrac { ar(ABC) }{ ar(DEF) } ={ \left( \dfrac { BC }{ EF }  \right)  }^{ 2 }\ \Rightarrow \dfrac { 80 }{ ar(DEF) } ={ \left( \dfrac { 4 }{ 5 }  \right)  }^{ 2 }\ \Rightarrow \dfrac { 80 }{ ar(DEF) } =\dfrac { 16 }{ 25 } \ \Rightarrow ar(DEF)=125{ cm }^{ 2 }$