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 properties of parallel lines and their transversal introduction to shapes similarity of triangles introduction to similar triangles

Ratio of two corresponding sides of two similar triangles is $4:9$. Then ratio of their area is ___.

  1. $\dfrac{16} {81}$
  2. $\dfrac{34} {81}$
  3. $\dfrac{81} {16}$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Ratio of areas of two similar triangles is equal to the squares of the ratio of their sides.

Ratio of sides $=\dfrac{4}{9}$
Ratio of areas $=\left( \dfrac { 4 }{ 9 }  \right) ^{ 2 }=\dfrac { 16 }{ 81 } $

Multiple choice maths properties of parallel lines and their transversal introduction to shapes similarity of triangles introduction to similar triangles

$\triangle PQR \sim \triangle XYZ, \dfrac{XY}{PQ}=\dfrac{3}{2}$ then $\dfrac{Area\ of\ \triangle PQR}{Area\ of\ \triangle XYZ}=$____.

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

$\dfrac{{XY}}{{PQ}} = \dfrac{3}{2}$


$ \Rightarrow \dfrac{{PQ}}{{XY}} = \dfrac{2}{3}$


Now,  $\dfrac{{Area{\rm{ of  }}\Delta {\rm{PQR}}}}{{Area{\rm{ of  }}\Delta {\rm{XYZ}}}} = {\left( {\dfrac{2}{3}} \right)^2} = \dfrac{4}{9}$

Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

The perimeter of two similar triangles are $24$ cm and $16$ cm, respectively. If one side of the first triangle is $10$ cm, then the corresponding side of the second triangle is

  1. $9$ cm
  2. $\dfrac{20}3$ cm
  3. $\dfrac{16}3$ cm
  4. $5$ cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
In similar triangles, ratio of the sides is equal to the ratio of the perimeters.
Thus, $\dfrac{p _1}{p _2} = \dfrac{s _1}{s _2}$
$\dfrac{24}{16} = \dfrac{10}{s _2}$
$s _2 = \dfrac{20}{3}$
Thus, side of the other triangle is $\dfrac{20}{3}$ cm.
Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

The area of two similar triangles $\displaystyle \Delta ABC$ and $\displaystyle \Delta DEF$ are 144 $\displaystyle cm^{2}$ and 81 $\displaystyle cm^{2}$ respectively If the longest side of larger $\displaystyle \Delta ABC$ be 36 cm then the longest side of the smaller triangle $\displaystyle \Delta DEF$ is

  1. 20 cm

  2. 26 cm

  3. 27 cm

  4. 30 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In similar triangle ABC & DBF
$\frac{AB}{De}=\frac{BC}{EF}=\frac{AC}{DF}=\frac{ratio ofArea of triangleABC}{ratio ofArea of triangleDEF}$ 
THEN $\frac{9}{12}=\frac{x}{36}$  (where x is longest side of the smaller triangle )
So x=27 cm

Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

The perimeters of two similar triangles are $25\;cm$ and $15\;cm$ respectively. If one side of first triangle is $9\;cm$, then the corresponding side of the other triangle is

  1. $6.2\;cm$
  2. $3.4\;cm$
  3. $5.4\;cm$
  4. $8.4\;cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$The\quad perimeter\quad of\quad triangle\quad is\quad 25cm\quad and\quad 15cm.\ The\quad ratio\quad of\quad Perimeter\quad of\quad triangle\quad is\quad 25:15=5:3\ The\quad first\quad side\quad is\quad 9cm\quad ,let\quad the\quad other\quad side=x\ Hence,\quad \dfrac { 9 }{ x } =\dfrac { 5 }{ 3\  } \ \Rightarrow x=\dfrac { 3\times 9 }{ 5 } =\dfrac { 27 }{ 5 } =5.4\quad cm$

Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

Sides of two similar triangles are in the ratio of $5 : 11$ then ratio of their areas is 

  1. $25 : 11$
  2. $25 : 121$
  3. $125 : 121$
  4. $121 : 25$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since, ratio of area of two similar traingles = ratio of square of corresponding sides

 ratio of sides = 5 : 11
$\therefore$ ratio of their areas = $(5)^2 : (11)^2 = 25 : 121$

Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

Sides of two similar triangles are in the ratio of $4 : 9$ then 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

$\displaystyle \because \Delta ABC\sim \Delta DEF$
$\displaystyle \therefore \dfrac{ar\Delta ABC}{ar\Delta DEF}=\dfrac{\left ( 4 \right )^{2}}{\left ( 9 \right )^{2}}=\dfrac{16}{81}=16:81.$

Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

In two similar triangles ABC and PQR, if their corresponding altitudes AD and Ps are in the ratio 4:9, find the ratio of the areas of $\triangle ABC$ and $\triangle PQR$.

  1. $16:81$
  2. $9:16$
  3. $81:16$
  4. $16:9$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Since the areas of two similar triangles are in the ratio of the squares of the corresponding altitudes.

$\therefore $ $\dfrac { Area(\triangle ABC) }{ Area(\triangle PQR) } =\dfrac { { AD }^{ 2 } }{ { PS }^{ 2 } } $

$\Rightarrow $ $\dfrac { Area(\triangle ABC) }{ Area(\triangle PQR) } ={ \left( \dfrac { 4 }{ 9 }  \right)  }^{ 2 }=\dfrac { 16 }{ 81 } $              [$\because AD:PS=4:9$]

$\Rightarrow $ $\dfrac { Area(\triangle ABC) }{ Area(\triangle PQR) }$ = $\dfrac{16}{81}$
Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

If $\triangle ABC$ is similar to $\triangle DEF$ such that BC=3 cm, EF=4 cm and area of $\triangle ABC=54 {cm}^{2}$. Determine the area of $\triangle DEF$.

  1. $40\ cm^2$
  2. $59\ cm^2$
  3. $69\ cm^2$
  4. $96\ cm^2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Since the ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.

$\therefore $ $\dfrac { Area(\triangle ABC) }{ Area(\triangle DEF) } =\dfrac { { BC }^{ 2 } }{ { EF }^{ 2 } } $

$\Rightarrow $ $\dfrac { 54 }{ Area(\triangle DEF) } =\dfrac { { 3 }^{ 2 } }{ { 4 }^{ 2 } } $

$\Rightarrow $ $Area(\triangle DEF)=\dfrac { 54\times 16 }{ 9 } =96{ cm }^{ 2 }$
Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

The areas of two similar triangles $\triangle{ABC}$ and $\triangle{DEF}$ are $144\ cm^{2}$ and $81\ cm^{2}$ respectively. If the longest side of larger $\triangle{ABC}$ be $36\ cm$, then, the largest side of the similar triangle $\triangle{DEF}$ is

  1. $20\ cm$
  2. $26\ cm$
  3. $27\ cm$
  4. $30\ cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Therefore, the ratio of their sides is the square root of the ratio of their areas, which is sqrt(144/81) = 12/9 = 4/3. Setting up the proportion 36/x = 4/3 yields x = 27 cm for the smaller triangle's corresponding side.

Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

In $\triangle ABC \sim \triangle DEF$ such that $AB = 1.2\ cm$ and $DE = 1.4\ cm$. Find the ratio of areas of $\triangle ABC$ and $\triangle DEF$.

  1. $36 : 50$
  2. $49 : 50$
  3. $36 : 49$
  4. $1:2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know that area of two similar triangle is equal to the ratio of the squares of any two corresponding sides
$\dfrac {ar(\triangle ABC)}{ar (\triangle DEF)} = \dfrac {AB^{2}}{DE^{2}} = \dfrac {(1.2)^{2}}{(1.4)^{2}} = \dfrac {36}{49}$

Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

The perimeter of two similar triangle are $30\ cm$ and $20\ cm$. If one side of first triangle is $12\ cm$ determine the corresponding side of second triangle.

  1. $8\ cm$
  2. $4\ cm$
  3. $3\ cm$
  4. $16\ cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the two similar triangles be $\triangle ABC$ and $\triangle DEF$

$\therefore \dfrac {AB}{DE} = \dfrac {BC}{EF} = \dfrac {AC}{DF} = \dfrac {P _{1}}{P _{2}}$

$\Rightarrow \dfrac {AB}{DE} = \dfrac {P _{1}}{P _{2}}$

$\Rightarrow \dfrac {12}{DE} = \dfrac {30}{20}$

$\Rightarrow DE = 8\ cm$

Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

The perimeter of two similar triangles $\triangle ABC$ and $\triangle DEF$ are $36$ cm and $24$ cm respectively. If $DE=10 $ cm, then $AB$ is :

  1. $12$ cm
  2. $20$ cm
  3. $15$ cm
  4. $18$ cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given that triangles $ABC$ and $DEF$ are similar.

Also given, $DE=10$ cm and perimeters of triangles $ABC$ and $DEF$ are $36$ cm and $24$ cm.
So, the corresponding sides of the two triangles is equal to the ratio of their perimeters.

Hence, $\dfrac {\text{perimeter of} \ ABC}{ \text{perimeter of } \ DEF}$ $=\dfrac {AB}{DE}$
Therefore, $\dfrac {36}{24}=\dfrac {AB}{10}$ 
$\Rightarrow AB=\dfrac {36\times 10}{24}$
$\Rightarrow AB=15$ cm

Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

The sides of a triangle are $5$ cm, $6$ cm and $7$ cm. One more triangle is formed by joining the midpoints of the sides. The perimeter of the second triangle is:

  1. $18$ cm
  2. $12$ cm
  3. $9$ cm
  4. $6$ cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the $\triangle ABC $ have sides $AB = 5$cm,

$BC = 6$ cm and $AC = 7$cm.
Let the midpoints of the sides AB and AC be points D and E respectively.
$\therefore \dfrac {AD}{DB} = \dfrac {AE}{EC}$         ...By B.P.T

$\therefore \dfrac {AD+DB}{DB} = \dfrac {AE+EC}{EC}$   ....By Componendo

$\therefore \dfrac {AB}{DB} = \dfrac {AC}{EC}$      ......(1)

Also, $\angle BAC \cong \angle DAE$    ....(2)

$\therefore \triangle ABC \sim \triangle ADE$      ....SAS test of similarity

$\therefore \dfrac {AB}{AD} = \dfrac {BC}{DE} = \dfrac {AC}{AE}$       ....C.S.S.T

But $\dfrac {AB}{AD} = \dfrac {AB}{\frac 12 AB} = \dfrac 12$

$\therefore \dfrac {BC}{DE} = \dfrac 12$


Perimeter $(\triangle ADE) = AD + DE + AE$ 
$ = \dfrac 12 AB + \dfrac 12 BC + \dfrac 12 AC$

$= \dfrac 12 \left(AB + BC + AC \right)$

$ = \dfrac 12 \times 18 = 9$ cm.

So, option C is correct.

Multiple choice maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

Find the perimeter of an isosceles right triangle with each of its congruent as 7cm.

  1. $7\sqrt 2$ cm
  2. $14$ cm
  3. $(2+ \sqrt 2)$ cm
  4. $7(2+ \sqrt 2)$ cm
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the other side of triangle is x cm

Then in isosceles right angle triangle two congruent sides are 7 cm
$x^{2}=(7)^{2}+(7)^{2}$
$\Rightarrow x^{2}=49+49$
$\Rightarrow x^{2}=198$
$\Rightarrow x=7\sqrt{2}$
Then perimeter of right angle isosceles triangle =$7+7+7\sqrt{2}=14+7\sqrt{2}=7(2+\sqrt{2})$ 

So, option D is correct.