Tag: areas of similar triangles

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

If in $\displaystyle \Delta ABC$ and $\displaystyle \Delta DEF,\frac { AB }{ DE } =\frac { BC }{ FD } $, then they will be similar if :

  1. $\displaystyle \angle B=\angle E$
  2. $\displaystyle \angle A=\angle D$
  3. $\displaystyle \angle B=\angle D$
  4. $\displaystyle \angle A=\angle F$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If two sides of a triangle are proportional to the corresponding two sides in another triangle, and their included angles are equal, then the two triangles are similar by SAS rule.

If $\quad \dfrac { AB }{ DE } = \dfrac { BC }{ FD } $, then for two triangles ABC and DEF to be similar, the included angle must be equal. In this case, the included angles are $\quad \angle B\quad and\quad \angle D$.

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

Ratio of areas of two similar triangles is equal to :

  1. ratio of squares of the corresponding altitudes

  2. ratio of squares of corresponding medians.

  3. Either (A) or (B)

  4. (A) and (B) both
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Ratio of areas of two similar triangles is equal to ratio of squares of the corresponding altitudes and ratio of squares of corresponding medians. This means that if the ratio of either altitude or median is given and asked to find ratio of areas , then it will be the ratio of squares of the corresponding altitudes or ratio of squares of corresponding medians.

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

If $\Delta ABC\sim \Delta DEF$ such that area of $\Delta ABC$ is $9 cm^2$ and area of $\Delta DEF$ is $16 cm^2$ and $BC=1.8 cm$, then EF is

  1. 2.4 cm

  2. 1.35 cm

  3. 2.1 cm

  4. 3.2 cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$ar(\triangle ABC)=9cm^2,\,ar(\triangle DEF)=16cm^2$ and $BC=1.8cm$

$\triangle ABC\sim\triangle DEF$                [ Given ]

$\Rightarrow$  $\dfrac{ar(\triangle ABC)}{ar(\triangle DEF)}=\dfrac{(BC)^2}{(EF)^2}$                    [ Area of similar triangle theorem ]

$\Rightarrow$  $\dfrac{9}{16}=\dfrac{(1.8)^2}{(EF)^2}$
Taking square root on both sides,

$\Rightarrow$  $\dfrac{3}{4}=\dfrac{1.8}{EF}$

$\Rightarrow$  $EF=\dfrac{1.8\times 4}{3}$

$\Rightarrow$  $EF=\dfrac{7.2}{3}$

$\Rightarrow$  $EF=2.4\,cm$


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

The sides of two similar triangles are in the ratio $4:9$ Areas of these triangles are in the ratio

  1. $3 : 5$
  2. $4 : 9$
  3. $81 : 16$
  4. $16 : 81$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

If two triangles are similar to each other, then the ratio of the area of this triangle will be equal to the square of the ratio of the corresponding sides of this triangle.

$\therefore$ The ratio between area of these triangle$=\dfrac{(4)^2}{(9)^2}=\dfrac{16}{81}$

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 $\displaystyle 9\ { cm }^{ 2 }$ and $\displaystyle 16\ { cm }^{ 2 }$, respectively. The ratio of their corresponding heights is

  1. $3 : 4$
  2. $4 : 3$
  3. $2 : 3$
  4. $4 : 5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
In similar traingles: -

${(\dfrac{{h1}}{{h2}})^2} = \dfrac{{S1}}{{S2}}$

Where h1 and h2 are the heights 

and S1, S2 are the areas of similar traingles

${{\rm{(}}\dfrac{{h1}}{{h2}})^2} = \dfrac{{9c{m^2}}}{{16c{m^2}}}$

$\dfrac{{h1}}{{h2}} = \sqrt {\dfrac{9}{{16}}} $

$\dfrac{{h1}}{{h2}} = \dfrac{3}{4}\ or\  {3:4}$
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.