Mathematics · Quantitative Aptitude

Triangle Properties

243 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

Two isosceles triangles have their corresponding angles equal and their areas are in the ratio $25 : 36$. Find the ratio of their corresponding heights

  1. $25 : 35$
  2. $36 : 25$
  3. $5 : 6$
  4. $6 : 5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
We know, Ratios of areas of similar triangles is equal to ratio of squares of their corresponding sides.
Hence,

$\dfrac{Area \ of \ \triangle _1}{Area \ of \ \triangle _2}=\dfrac{(height \ of \ \triangle _1)^2}{(height \ of \ \triangle _2)^2}$

Taking square root on both sides,

$\dfrac{(height \ of \ \triangle _1)}{(height \ of \ \triangle _2)}=\sqrt{\dfrac{Area \ of \ \triangle _1}{Area \ of \ \triangle _2}}$

$\dfrac{(height \ of \ \triangle _1)}{(height \ of \ \triangle _2)}=\sqrt{\dfrac{25}{36}}=\dfrac{5}{6}$

Option C

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

In similar triangles $\triangle ABC$ and $\triangle FDE, DE = 4 cm, BC = 8 cm$ and area of $\triangle FDE = 25 cm^2$. What is the area of $\Delta ABC$?

  1. 144 cm$^2$
  2. 121 cm$^2$
  3. 100 cm$^2$
  4. 81 cm$^2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

  $DE=4\,cm,\,BC=8\,cm$ and $ar(\triangle FDE)=25\,cm^2$                  [ Given ]


$\Rightarrow$  $\triangle ABC\sim\triangle FDE$             [ Given ]

$\Rightarrow$  $\dfrac{ar(\triangle ABC)}{ar(\triangle FDE)}=\dfrac{(BC)^2}{(DE)^2}$                       [ By area of similar triangle theorem ]

$\Rightarrow$  $\dfrac{ar(\triangle ABC)}{25}=\dfrac{(8)^2}{(4)^2}$

$\Rightarrow$  $ar(\triangle ABC)=\dfrac{64}{16}\times 25$

$\Rightarrow$  $ar(\triangle ABC)=4\times 25$
$\therefore$  $ar(\triangle ABC)=100\,cm^2$

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 $81\ cm^{2}$ and $49\ cm^{2}$. If the altitude of the bigger triangle is $4.5\ cm$, find the corresponding altitude of the smaller triangle.

  1. $3 cm$
  2. $2.5 cm$
  3. $4 cm$
  4. $3.5 cm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given: Area of $two$ similar triangle $81{cm}^{2}$ and $49{cm}^{2}$

Altitude of bigger triangle $=4.5cm$
For similar triangle,
${\text{Ratio on sides}}^{2}=\text {Ratio of their Area}$
$\therefore$ $\cfrac { { 4.5 }^{ 2 } }{ { x }^{ 2 } } =\cfrac { 81 }{ 49 } $
$\cfrac { 45\times 45 }{ { x }^{ 2 }\times 100 } =\cfrac { 81 }{ 49 } $
$100{x}^{2}=25\times 49$
${x}^{2}=\cfrac{25\times 49}{100}$
${x}^{2}=\cfrac{49}{4}$
$x=\cfrac{7}{2}$
$x=3.5cm$

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

What is the ratio of the areas of two similar triangles whose corresponding sides are in the ratio 15:19?

  1. $\sqrt{15} : \sqrt{19}$
  2. $15 : 19$
  3. $225 : 361$
  4. $125 : 144$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know ratio  of areas of two similar triangles is the ratio of the square of their corresponding sides.

Ratio of sides $\dfrac{15}{19}$
Ratio of areas $={ \left( \dfrac { 15 }{ 19 }  \right)  }^{ 2 }=\dfrac { 225 }{ 361 } $
So option $C$ is correct.

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

If the sides of two similar triangles are in the ratio $2 : 3$, then their areas are in the ratio:

  1. $9 : 4$
  2. $4 : 9$
  3. $2 : 3$
  4. $3 : 2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, sides of two similar triangles are in the ratio $2:3$

Thus, the ratio of their areas are $($ side $)^2$
$=\left (\dfrac {2}{3}\right)^2=\dfrac {4}{9}$ 
Therefore, the areas are in the raatio $4:9$.

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

D and E are the points on the sides AB and AC respectively of triangle ABC such that $ DE||BC$. If area of $ \triangle DBC =15 cm^2$, then area of $\triangle EBC $ is:

  1. $30cm^{2}$
  2. $7.5cm^{2}$
  3. $15cm^{2}$
  4. $20cm^{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given, $DE||BC$

Therefore, the altitudes of $\triangle EBC$ and $\triangle DBC$ are equal.
Also, they have a common base $BC$.
Thus, $ \text{Ar}(\triangle EBC)=\text{Ar}(\triangle DBC)$
$\Rightarrow \text{Ar}(\triangle EBC)=15 \ \ \text{cm}^2$

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

Triangles ABC and DEF are similar. If their areas are 64 $cm^2$ and 49 $cm^2$ and if AB is 7 cm, then find the value of DE.

  1. 8 cm
  2. $\dfrac{49}{8}$ cm
  3. $\dfrac{8}{49}$ cm
  4. $\dfrac{64}{7}$cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\Delta ABC \Delta DEF$
$\dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{AC}{DF}$.
We know that,
$\dfrac{Area of \Delta  ABC}{ Area of \Delta  DEF} = $ $\Rightarrow \dfrac{64}{49} = \left ( \dfrac{7}{DE} \right )^2$
$\Rightarrow \left ( \dfrac{8}{7} \right )^2 = \left ( \frac{7}{DE} \right )^2 \Rightarrow \left ( \dfrac{8}{7} \right )^2 = \left ( \dfrac{7}{DE} \right )^2 = \dfrac{49}{8}$cm 

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

The area of two similar triangles are $200$ and $128$, then the ratio of their corresponding altitude is __________

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

Since we know that ratio of areas of two similar triangles is equal to the square of the ratio of their altitude
therefore
Ratio of their altitude=$\sqrt {\dfrac{{200}}{{128}}} $
$ = \sqrt {\dfrac{{100}}{{64}}} $
$ = \dfrac{{10}}{8}$
$ = \dfrac{5}{4}$
$ = 5:4$

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

The area of triangle whose vertices are $(1, 2, 3), (2, 5, -1)$ and $(-1, 1, 2)$ is

  1. $150\ sq. units$
  2. $145\ sq. units$
  3. $\sqrt {155}/2\ sq. units$
  4. $155/2\ sq. units$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the vertices of triangle are 


$A(1,2,3),B(2,5,-1)$ and $C(-1,1,2)$ 

Then, 

$\begin{array}{l} \overrightarrow { AB } =\overrightarrow { OB } -\overrightarrow { OA } =\hat { i } +3\hat { j } -4\hat { k }  \  \ \overrightarrow { AC } =\overrightarrow { OC } -\overrightarrow { OA } =-2\hat { i } -\hat { j } -\hat { k }  \end{array}$

Then, 

$\begin{array}{l} \overrightarrow { AB } \times \overrightarrow { AC } =\left| { \begin{array} { *{ 20 }{ c } }{ \hat { i }  } & { \hat { j }  } & { \hat { k }  } \ 1 & 3 & { -4 } \ { -2 } & { -1 } & { -1 } \end{array} } \right|  \  \ =-7\hat { i } +9\hat { j } +5\hat { k }  \  \ \left| { \overrightarrow { AB } \times \overrightarrow { AC }  } \right| =\sqrt { { { \left( { -7 } \right)  }^{ 2 } }+{ { \left( 9 \right)  }^{ 2 } }+{ { \left( 5 \right)  }^{ 2 } } }  \  \ =\sqrt { 49+8+25 }  \  \ =\sqrt { 155 }  \end{array}$

Area of triangle $ABC=\frac{1}{2}|\vec {AB} \times \vec {AC}|$

$=\frac{1}{2} \times \sqrt {155}$

$=\frac{\sqrt {155}}{2}$ sq. units 

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

The perimeter of the triangle formed by the points $(1,0,0),(0,1,0),(0,0,1)$ is 

  1. $\sqrt 2 $
  2. $2\sqrt 2 $
  3. $3\sqrt 2 $
  4. $4\sqrt 2 $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given points $A(1, 0, 0), B(0, 1, 0), C(0, 0, 1)$ is
$AB=\sqrt{1+1}=\sqrt{2}$
$BC=\sqrt{1+1}=\sqrt{2}$
$CA=\sqrt{1+1}=\sqrt{2}$
Perimeter of the triangle is $AB+BC+CA=\sqrt{2}+\sqrt{2}+\sqrt{2}=3\sqrt{2}$.
Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

Perimeter of triangle whose vertices are $(0,4,0), (3,4,0)$ and $(0,4,4)$, is

  1. $10$
  2. $12$
  3. $25$
  4. $15$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

 Vertices of triangle are $(0,4,0),(3,4,0)$ and $(0,4,4)$

then perimeter=?
here AB=$\sqrt{(3-0)^2+(4-4)^2+(0-0)^2}$
$=3$
BC$=\sqrt{(3-0)^2+(4-4)^2+(0-4)^2}$
$=\sqrt{9+16}$
$=5$
CA$=\sqrt{(0-0)^2+(4-4)^2+(0-4)^2}=4$
used distance formula b/w two points
$(x _1,y _1,z _1) and (x _2,y _2,z _2)$
$=\sqrt{(x _2-x _2)^2+(y _2-y _1)^2+(z _2-z _1)^2}$
$ perimeter =ABC+BC+CA$
$=3+5+4$
$=12\ units$

Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

If a pair of perpendicular straight lines drawn through the origin forms an isosceles triangle with the line $2x+3y=6$, then area of the triangle so formed is?

  1. $36/13$
  2. $12/17$
  3. $13/5$
  4. $17/3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

As the lines are perpendicular and form an isosceles triangle the other two angles must be $45^\circ$

Let the slope of the line be m
$tan 45^\circ = \Bigg|\cfrac{-\cfrac{2}{3}-m}{1-\cfrac{2m}{3}}\Bigg| = 1$
$m = -5$ and the other line slope =$\cfrac{1}{5}$
Lines 
$y +5x= 0$ and $5y =x$
Intersection points $(0,0)$ , $(-\cfrac{6}{13} , \cfrac{30}{13})$ and $(\cfrac{30}{13} , \cfrac{6}{13})$
Perpendicular distance from origin to line $2x+3y=6$ is $\cfrac{|0+0-6|}{\sqrt{2^2+3^2}} = \cfrac{6}{\sqrt{13}}$
Distance between the points are $\sqrt{\Bigg(\cfrac{36}{13}\Bigg)^2+\Bigg(\cfrac{24}{13}\Bigg)^2} = \cfrac{\sqrt{1872}}{13} = \cfrac{12\sqrt{13}}{13}$
Area = $\cfrac{1}{2} \times \cfrac{12\sqrt{13}}{13} \times \cfrac{6}{\sqrt{13}} = \cfrac{36}{13}$ 

Multiple choice maths constructions mid-point formula midpoints division of a line segment

Find the area of the triangle formed by joining the mid points of the sides of the triangle whose vertices are $(0.-1), (2, 1) and (0, 3)$

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

$\\Area\>of\>triangle\>=4\times\>of\>triangle\>formed\>using\>mid-point\>\\=4\times(\frac{1}{2})[x-1(y _2-y _3)+x _2(y _3+y _1)+x _3(y _1-y _2)]\\=2[0+2(3-1)+0]=8sq\>unit$

Multiple choice maths ratio, proportion and unitary method converting to ratios finding ratios other quantities

Sides of two similar triangles are in the ratio $4:9$.Area of these triangles are in the ratio

  1. $2:3$
  2. $4:9$
  3. $81:16$
  4. $16:81$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given:Ratio of sides of similar triangles$=\dfrac{4}{9}$

We know that if two triangles are similar, 

ratio of areas is equal to the ratio of squares of corresponding sides.

So, $\dfrac{area \,\, of\,\, triangle\,\, 1}{area \,\, of\,\, triangle\,\, 2}=\dfrac{{\left(side\,\, of\,\, triangle\,\, 1\right)}^{2}}{{\left(side\,\, of\,\, triangle\,\, 2\right)}^{2}}$

$\dfrac{area \,\, of\,\, triangle\,\, 1}{area \,\, of\,\, triangle\,\, 2}={\left(\dfrac{4}{9}\right)}^{2}=\dfrac{16}{81}$

$\dfrac{area \,\, of\,\, triangle\,\, 1}{area \,\, of\,\, triangle\,\, 2}=\dfrac{16}{81}$