Tag: maths

Questions Related to maths

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?

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

  1. 1:2

  2. 2:1

  3. 2:3

  4. 1:4

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Correct Option: A
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.

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.

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

  1. 3 cm

  2. 2 cm

  3. 5 cm

  4. 4 cm

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Correct Option: D
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.

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

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

  1. $\dfrac{11}{9}$

  2. $\dfrac{10}{9}$

  3. $\dfrac{9}{11}$

  4. $\dfrac{9}{10}$

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Correct Option: A
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 $=\sqrt{\left [ \dfrac{11}{9} \right ]^{2}}=\dfrac{11}{9}$

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 $9 : 16$?

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

  1. $4.5:8$

  2. $3:4$

  3. $4:3$

  4. $8:4.5$

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Correct Option: B

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.

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

  1. $8.75m$

  2. $6.75m$

  3. $7.75m$

  4. None of these

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Correct Option: A

If ratio of heights of two similar triangles is $4:9$, then ratio between their areas is?

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

  1. $2:3$

  2. $3:2$

  3. $81:16$

  4. $16:81$

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Correct Option: D
Explanation:

Altitudes of similar triangles are in ratio $4:9$
Hence, area of these triangles
$=$ square of the ratio of their heights or altitudes
$=(4:9)^2=16:81$
Option $(4)$.

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:

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

  1. 9.8 cm

  2. 7 cm

  3. 49 cm

  4. 25 cm

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Correct Option: B
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

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

 

$\Delta ABC\sim\Delta PQR.$ 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:

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

  1. 30 cm

  2. 0.5 m

  3. 50 cm

  4. 3 m

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Correct Option: A
Explanation:

$\triangle ABC\sim \triangle DEF$

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

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


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

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

  1. $100 cm^{2}$

  2. $125 cm^{2}$

  3. $150 cm^{2}$

  4. $200 cm^{2}$

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

In $XYZ$ and $\triangle PQR,XYZ\leftrightarrow PQR$ is similarity, $XY=8,ZX=16,PR=8$. So $PQ+QR$=______.

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

  1. $20$

  2. $10$

  3. $15$

  4. $9$

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Correct Option: A