Mathematics · Quantitative Aptitude

Triangle Properties

234 Questions

Triangle properties encompass the rules governing the sides, angles, and area of different types of triangles. Key areas include the Pythagorean theorem, similar triangles, and centroid calculations. These concepts form a foundational part of geometry in various competitive examinations.

Similar trianglesArea and perimeterPythagorean theoremTriangle inequalityEquilateral properties

Triangle Properties Questions

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

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

Area of similar triangles are in the ratio $25:36$ then ratio of their similar sides is _________?

  1. $5:7$
  2. $5:6$
  3. $6:5$
  4. $6:7$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The areas and sides of similar triangles are related as 

$\dfrac{Ar(\Delta ABC)}{Ar(\Delta PQR)}=\left(\dfrac {AB}{PQ}\right)^2\\dfrac {25}{36}=\left(\dfrac {AB}{PQ}\right)^2\\dfrac{AB}{PQ}=\sqrt {\dfrac {25}{36}}=\dfrac 56$

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

The perimeter of two similar triangles is 30 cm and 20 cm. If one altitude of the former triangle is 12 cm, then length of the corresponding altitude of the latter triangle is 

  1. 8 cm

  2. 10 cm

  3. 12 cm

  4. 15 cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\Delta$ABC and $\Delta$DEF be two similar triangle. Perimeter of first and second triangles are $30$cm and $20$cm respectively.
Then $\dfrac{AB}{DE}=\dfrac{BC}{EF}=\dfrac{AC}{DF}=k$ (say)
$\therefore AB=kDE, BC=kEF, AC=kDF$
$AB+BC+AC=k(DE+EF+DF)$
$\Rightarrow 30=k\times 20$
$\Rightarrow k=\dfrac{3}{2}$
$\Rightarrow \dfrac{AB}{DE}=\dfrac{3}{2}$
$\Rightarrow \dfrac{12}{DE}=\dfrac{3}{2}$
$\Rightarrow DE=8$.
Multiple choice maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

The perimeter of two similar triangles is 40 cm and 50 cm. Then the ratio of the areas of the first and second triangles is 

  1. 4 : 5

  2. 5 : 4

  3. 25 : 16

  4. 16 : 25
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
We know the ratio of perimeters of $2$ similar triangles are equal to ratio of corresponding sides

i.e., $\dfrac{perimeter \,of \,1^{st}}{perimeter\, of \,2^{nd}}=\dfrac{side\, of\, 1^{st}}{side\, of\, 2^{nd}}$

$\Rightarrow \dfrac{40}{50}=\dfrac{side\, of\, 1^{st}}{side\, of \,2^{nd}}=\dfrac{4}{5}$

As both the triangles are similar 

$\Rightarrow \dfrac{Area\, of\, 1^{st}}{Area\, of\, 2^{nd}}=\left(\dfrac{(side \,of\, 1^{st})^2}{(side\, of\, 2^{nd})^2}\right)=\dfrac{16}{25}$.
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 $49 \ {cm}^{2}$ and $64 \ {cm}^{2}$ respectively. The ratio of their corresponding sides is:

  1. $49:64$
  2. $7:8$
  3. $64:49$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Areas of two similar triangles are $49 $ cm $^2$ and $64$ cm $^2.$
For similar triangles the ratio of areas is equal to the ratio of square of corresponding sides.
Hence, $\dfrac{A _1}{A _2} = \dfrac{(s _1)^2}{(s _2)^2}$
$\Longrightarrow \dfrac{49}{64} = \dfrac{(s _1)^2}{(s _2)^2}$
$\Longrightarrow\dfrac{s _1}{s _2} = \dfrac{7}{8}$

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

Two isosceles triangles have equal vertical angles and their areas are in the ratio $16:25$. Find the ratio of their corresponding heights.

  1. $4:5$
  2. $25:16$
  3. $5:4$
  4. $16:25$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\triangle ABC$ and $\triangle DEF$ be the given triangles in which $AB=AC, DE=DF$, $\angle A=\angle D$
and $\cfrac{Area\quad (\triangle ABC)}{Area\quad (\triangle DEF)}=\cfrac{16}{25}$
Draw $AL\bot  BC$ and $DM\bot  EF$
Now, $\cfrac{AB}{BC}=1$ and $\cfrac{DE}{DF}=1$  ($\because \quad AB=AC;\quad DE=DF$)
$\Rightarrow \cfrac{AB}{AC}=\cfrac{DE}{DF}$,
$\therefore$ $\ln \triangle ABC$ and $\triangle DEF$, we have
$\cfrac{AB}{DE}=\cfrac{AC}{DF}$ and $\angle A=\angle D$
$\Rightarrow$ $\triangle ABC\sim \triangle DEF$ [By SAS similarity axiom)
But, the ratio of the areas of two similar $\triangle s$ is the same as the ratio of the squares of their corresponding heights.
$\cfrac{Area\quad (\triangle ABC)}{Area\quad (\triangle DEF)}=\cfrac {{AL}^{2}}{{DM}^{2}}$
$\Rightarrow$ $\cfrac{16}{25}={ \left( \cfrac {AL}{DM}  \right)  }^{ 2 }$
$\Rightarrow$ $\cfrac{4}{5}$
$\therefore$ $AL:DM=4:5$, i.e., the ratio of their corresponding heights$=4:5$

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

Let $\triangle ABC\sim \triangle DEF$ and their areas be, respectively $64\ {cm}^{2}$ and $121\ {cm}^{2}$. If $EF=15.4\ cm$, find $BC$.

  1. $11.2\ cm$
  2. $11.6\ cm$
  3. $11.4\ cm$
  4. $10.8\ cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\triangle ABC\sim \triangle DEF\quad $ (Given)
$\Rightarrow \cfrac { ar(ABC) }{ ar(DEF) } =\cfrac { { BC }^{ 2 } }{ { EF }^{ 2 } } $ (ratio of Areas of Similar triangles are equal to ratio of squares of corresponding sides)
$\Rightarrow \quad \cfrac { 64 }{ 121 } =\cfrac { { BC }^{ 2 } }{ { EF }^{ 2 } } \quad { \left{ \cfrac { BC }{ EF }  \right}  }^{ 2 }={ \left{ \cfrac { 8 }{ 11 }  \right}  }^{ 2 }$
$\Rightarrow \quad \cfrac { BC }{ EF } =\cfrac { 8 }{ 11 } \quad \Rightarrow \quad BC=\cfrac { 8 }{ 11 } \times EF$
$\Rightarrow \quad BC=\cfrac { 8 }{ 11 } \times 15.4cm=11.2cm$